Testing for an inverse-square law - as-a2-physics ... 11.pdf...1 Advancing Physics Testing for an...

1 Advancing Physics

Testing for an inverse-square lawQuestion 10W: Warm-up Exercise

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There are several ways you can test for an inverse-square law. You can see this using the databelow, which give the intensity of radiation at different distances from a star.

Intensity of radiation /

W m–2

Distance /million km

2000 100

500 200

222 300

125 400

80 500

56 600

41 700

31 800

25 900

20 1000

Simple testIf these data obey an inverse-square law then doubling the distance will cause the intensity to drop by

a factor of 22 (i.e. to a quarter of its initial value).

1. Try this for initial values of 100 million km and 500 million km. What do you conclude?

A numerical testIf the data obey an inverse-square law then the intensity I is related to the distance r by an equation ofthe form

2r

kI

so that k = I r 2 should be constant.

2 Advancing Physics

2. Calculate k = I r 2 for each set of data and see if it is constant. Try it. What do you conclude?

A graphical testIf the data obey an inverse-square law then a graph of I against 1 / r 2 should be a straight linethrough the origin.

3. Plot a graph of I against 1 / r 2. What do you conclude?

Here are some other data sets to test. Show which of these obey an inverse-square law.

4. Data set A: The pressure and volume of a gas at constant temperature

Pressure / kPa Volume / cm3

100 40.0

110 36.4

120 33.3

130 30.8

140 28.6

150 26.7

160 25.0

170 23.5

180 22.2

190 21.1

200 20.0

210 19.0

220 18.2

230 17.4

240 16.7

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Pressure / kPa Volume / cm3

250 16.0

260 15.4

270 14.8

280 14.3

290 13.8

300 13.3

5. Data set B: The resistance and diameters of 10 cm lengths of resistance wire

Resistance / Diameter / mm

1.27 0.10

0.57 0.15

0.32 0.20

0.20 0.25

0.14 0.30

0.10 0.35

0.08 0.40

0.06 0.45

0.05 0.50

6. Data set C: The amplitude of oscillation of a mass on a spring and time since release

Time / s Amplitude / cm

0 4.0

2 3.8

4 3.6

6 3.4

8 3.3

4 Advancing Physics


1.27 0.10

0.57 0.15

0.32 0.20

0.20 0.25

0.14 0.30

0.10 0.35

0.08 0.40

0.06 0.45

0.05 0.50



0 4.0

2 3.8

4 3.6

6 3.4

8 3.3

10 3.1

12 2.9

14 2.8

16 2.7

18 2.5

20 2.4

22 2.3

24 2.2

26 2.1

5 Advancing Physics




0 4.0

2 3.8

4 3.6

6 3.4

8 3.3

10 3.1

12 2.9

14 2.8

16 2.7

18 2.5

20 2.4

22 2.3

24 2.2

26 2.1

28 2.0

30 1.9

32 1.8

34 1.7

36 1.6

38 1.5

40 1.4

6 Advancing Physics

Orbital velocities and accelerationQuestion 20W: Warm-up Exercise

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Determining accelerationThis question considers an object moving in a circle of radius r at constant velocity v, taking Tseconds to complete one orbit. The magnitude of its acceleration is calculated by considering the(vector) change v in various decreasing time intervals, leading to an expression for the acceleration ofthe object at any moment.

Steps in the argumentFirst we consider half a circle:

v1

v2

where the velocity vector has simply reversed direction in the time interval t.

v1

v2

v

7 Advancing Physics

1. What is t, the time interval in terms of the period T ?

2. What is the size of v, the change in velocity in terms of v ?

3. Since acceleration a = v / t, derive an expression for a in terms of velocity v and time, T.

4. Since speed is distance travelled divided by the time taken, write an expression for T in terms of rand v.

5. Substitute for T in your answer to question 3 and deduce that the acceleration is 4 v 2 / 2r or

0.64 v2 / r .

Now consider a quarter circle

v1

v2

and the vector diagram:

v1

v2

v

6. What ist now, in terms of T?

8 Advancing Physics

7. From the vector diagram, use Pythagoras’s theorem to deduce v in terms of v.

8. Hence deduce an expression for a and, by substituting for T as above, express it in terms of v2 /

r .

For questions 9, 10 and 11 repeat questions 6, 7 and 8 using an angle of 60 (or revolution

These may help

60°

60°

v2

v1

v

v2

v1

Note that the vector diagram forms an equilateral triangle.

9.

10.

11.

Finally, consider movement through a small angle between the two vectors:

9 Advancing Physics

v1

v2

0

v2v1

v QP

12. What ist now in terms of T and

13. For the vector diagram above, express the arc length P Q in terms of v and . For a small angle this will be nearly be equal to the straight length v.

14. Deduce an expression for the acceleration and simplify it as before.

15. In what direction is the acceleration, relative to the object’s velocity?

Look at the expression for a as the time intervals have decreased. They should approach a limitingvalue.

16. Write down an expression for the centripetal acceleration of the object at any moment in itsmotion.

17. If the object has a mass m, write down the centripetal force, F, acting on it.

10 Advancing Physics

Radians and angular speedQuestion 70W: Warm-up Exercise


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In many physical situations you are concerned with the motion of objects moving in a circle, such asplanets in orbit around the Sun or more mundane examples like wheels turning on a bicycle orwashing drying in a spin drier. The measurement of speed can be expressed in several differentways; the following questions are designed to help you become confident in their use.

RotationSince a radian is the angle between two radii with an arc length equal to the radius, there are 2 radians in one complete circle. In the first set of questions, consider a simple example of an objectmoving in a circle at constant speed.

1. Work to two decimal places. Write down the angle in radians if the object moves in one completecircle, then deduce the number of radians in a right angle.

2. The object rotates at 15 revolutions per minute. Calculate the angular speed in radian persecond.

A rotating restaurantA high tower has a rotating restaurant that moves slowly round in a circle while the diners are eating.

3. The restaurant is designed to give a full 360 view of the sky line in the two hours normally takenby diners.

Calculate the angular speed in radians per second.

4. The diners are sitting at 20 m from the central axis of the tower.

Calculate their speed in metres per second.

Do you think they will be aware of their movement relative to the outside?


Newton’s gravitational lawQuestion 80W: Warm-up Exercise


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These questions are intended to give you practice in using the gravitational law. They will give you afeeling for typical forces with a range of masses and also how sensitive force is to distance.

Useful dataG = 6.67 10–11

N m2 kg–2

Earth’s mass = 5.97 1024 kg

Moon’s mass = 7.34 1022 kg

Sun’s mass is 2.00 1030 kg

Radius of the Moon = 1.64 106 m

Radius of the Earth = 6.37 106 m

Earth–Moon distance = 3.8 105 km

Earth–Sun distance = 1.5 108 km

1. You may sometimes find it difficult to get up from the sofa after watching a TV programme.Assuming the force of gravity acts between the centre of your body and the centre of the sofa,estimate the attraction between you and your sofa.

2. Calculate the size of the gravitational pull of a sphere of mass 10 kg on a mass 2.0 kg when theircentres are 200 mm apart.

What is the force of the 2.0 kg mass on the 10 kg mass?

3. At what distance apart would two equal masses of 150 kg need to be placed for the force

between them to be 2.0 10–5 N?


4. Calculate the gravitational pull of the Earth on each of the following bodies:

the Moon;

satellite A with mass 100 kg at a distance from the Earth’s centre 4.2 107 m;

and satellite B mass 80 kg at a distance from the Earth’s centre 8.0 106 m.

5. Show that the unit for G, the universal gravitational constant, can be expressed as m3 s–2 kg–1.

6. Calculate the weight of an astronaut whose mass (including spacesuit) is 72 kg on the Moon?

What is the astronaut's weight on Earth?

Comment on the difference.

7. Show that pull of the Sun on the Moon is about 2.2 times larger than the pull of the Earth on the


Moon.

8. Why then does the Moon orbit the Earth?

Change in momentum as a vectorQuestion 140W: Warm-up Exercise

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1. A runner of mass 70 kg accelerates from 3 m s–1 to 5 m s–1 in the same direction.

Draw three vectors to show the runner’s initial momentum, final momentum and change inmomentum.

Label each vector with its magnitude and ensure that the directions of the vectors are consistent.

2. A hare of mass 4 kg decelerates from 8 m s–1 to 2 m s–1 in the same direction.

Draw three vectors to show its initial momentum, final momentum and change in momentum.

Label each vector with its magnitude and ensure that the directions of the vectors are consistent.


3. A tennis ball of mass 60 g strikes a racket at 20 m s–1 and rebounds along the same line at 15 m

s–1.

Draw a vector diagram to show how the change in momentum can be found from the initial andfinal values.

Label the magnitude of all vectors you draw.

4. A car of mass 400 kg is travelling due north at 15 m s–1. It turns a corner so that it is travelling duewest at the same speed.

Draw a vector diagram to show how the change in momentum can be found from the initial andfinal values.

Label the magnitude of all vectors you draw, and indicate the values of any relevant angles.

5. A yacht of mass 1000 kg is travelling at a speed of 5 m s–1 due west.

Calculate its momentum.

The yacht is then blown off course and suffers a change in momentum of 500 kg m s–1 due north.

Calculate the magnitude and direction of its new momentum.

Pole vaultingQuestion 200W: Warm-up Exercise


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This question is about the energy exchanges from the time a pole vaulter makes her initial runtowards the jump point holding the flexible pole, to her launch upwards and the point where sheclears the pole.

g = 9.8 m s–2.

The jump1. Discuss the energy transfers involved in this situation, from the initial run to finally clearing the

bar.

2. The pole vaulter's mass is 60 kg and her speed is 7.5 m s–1.

Calculate the kinetic energy of the pole vaulter just before she lifts off the ground.

3. If all of her kinetic energy is converted into gravitational potential energy, what height can shereach?

4. Is your answer to question 3 the maximum height above the ground for the bar which the polevaulter can clear?

Centripetal forceQuestion 30S: Short Answer


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This question is about the motion of passengers on a giant swing.

A theme park attraction‘Vertigo’ is a new theme park attraction. In groups of three, passengers are suspended byinextensible steel wires from a giant arch. They are winched to the top of a tower, and released to


swing until the end of the ride.

50 m

arch

A

B

45 m

tower

path

1. A typical group of three passengers has a combined mass of 200 kg.

Calculate the potential energy they gain as they are winched to their maximum height of 45 m atA.

g = 9.8 N kg–1.

2. Show that their speed at the bottom B of their first swing is about 30 m s–1. State anyassumptions you make.

3. The radius of the arc followed by the passengers is 50 m.

Calculate the centripetal acceleration of the passengers at the instant they are at B.

4. Show that the vertical force exerted on the passengers by the suspension wires as the


passengers pass through B is more than 2.5 times the passengers’ weight. Neglect the weight ofthe wires.

Circular motion – more challengingQuestion 40S: Short Answer

Teaching Notes | Key Terms | Hints | Answers | KeySkills

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This question is about the forces experienced on a theme park ride called the ‘Gravitron’.

The ‘Gravitron’ is a circular chamber rotating about a vertical axis. Passengers lie back on couches

which line the interior of the chamber, each inclined at 62 to the horizontal.

Each couch is mounted on a track allowing it to rise a limited distance of a few centimetres as theGravitron spins faster.


62°

trackcouchrollers

reaction of couch

passenger’s weight

This is how the speed, measured at a passenger’s centre of mass, 5.0 m from the centre of theGravitron, varies during the first 60 seconds of a typical ride.

0 10 20 30 40 50 60

0

2

4

6

8

10

12

time / s

10.8 m s–1

1. The ride’s maximum permitted rate of rotation is 22 revolutions per minute.

Show whether the ride is operating within this limit.

2. Think carefully about the forces acting on a passenger at the instant the couch starts to rise on itstrack.

Show that the reaction of the couch on a 76 kg passenger is almost 1600 N.

g = 9.8 N kg–1.

3. Show that the centripetal acceleration experienced by the passenger is more than 20 m s–2.


4. Calculate the speed of the passenger at the instant the couch starts to rise.

5. It is claimed that the passenger experiences a force more than twice his own weight throughoutthe time when the couch is in the risen position.

Show whether this claim is justified.

Finding the mass of a planet with a satelliteQuestion 110S: Short Answer


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How are the masses of planets known?Data for planets and the Sun, including their masses, are provided in reference books and also onthis CD-ROM (see File 30T 'Planetary orbit data'). How can their masses be known?

We can find the mass of Earth ME by placing a known mass m at a known distance r from its centreand measuring its weight F with a spring balance. Then

.2

E Gm

rFM

But we cannot make such measurements at the surface of the Sun or other planets. There we mustuse the motion of a satellite to calculate the mass. The following relationship is shown for ageostationary satellite in section 11.2 of the Advancing Physics A2 student’s book.

2

23

4

GMTR

where R is the mean orbit radius and T the orbit period.

This was derived by noting that the centripetal force on a satellite is provided solely by thegravitational force of attraction. The equation therefore applies to all satellites. Notice that M can befound provided we know R, T and G. Astronomers can measure R and T remotely, and we alreadyknow G.

A simpler way to see what is going on is to notice that given radius R of orbit and time T, you can atonce calculate the orbital speed v.


T

Rv

2

Then you can calculate the centripetal acceleration under gravity

R

vg

2

Knowing g, you can find the mass M of the planet from the law of gravitation:

2R

GMg

so

G

gRM

2

The equation

2

23

4

GMTR

does this in one step, but it is less obvious how the calculation works.

Questions1. Rearrange

2

23

4

GMTR

to make M its subject.

2. Oberon, a moon of Uranus, has a periodic time of 323 hours and its mean distance from the

planet is 5.82 105 km.

Calculate the mass of Uranus.

3. Earth’s Moon has a periodic time of 27.3 days, and its mean orbit radius is 3.8 105 km. What dothese figures suggest for Earth’s mass?

4. In 1978 a moon of Pluto was discovered, called Charon. That moon’s mean orbit radius is


1.9 107 m and its orbit period 6.4 days.

Use these data to calculate the mass of Pluto.

5. Earth orbits the Sun every 365 days, with a mean orbit radius of 1.5 108 km.

What do these data suggest as a value for the mass of the Sun?

6. Can we use a similar method to find the mass of a moon (which has no natural satellite)? Explainyour answer.

Impulse and momentum in collisionsQuestion 150S: Short Answer


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(Note: g = 9.8 m s–2)

1. A powerboat is going at 80 km h–1 when its motor is turned off. In 5.0 seconds its velocity

decreases to 50 km h–1. The boat and its passengers weigh 1 tonne.

Calculate the average force exerted on the boat by the combined drag of the air and the water.

2. In a test of an energy-absorbing bumper, a 1250 kg car is driven into a barrier at 8.0 km h–1. The

duration of the impact is 0.4 seconds and the car bounces back from the barrier at 2.0 km h–1.

Calculate the size of the average horizontal force exerted on the car during the impact


A fragile object dropped onto a hard surface breaks because it is subjected to a large impulsive force.Suppose you drop a 50 g watch from 1.0 m above the floor, and the duration of impact is 0.001seconds and the watch bounces 5 cm above the floor.

Use arguments based on changes in kinetic and gravitational potential energy to find:

3. The speed with which the watch hits the floor.

4. The initial speed of rebound.

5. The average value of the impulsive force.

A railway wagon travelling at 1.0 m s–1 catches up with and become coupled to another wagon

travelling at 0.5 m s–1 in the same direction. The faster moving wagon has 1.7 times the mass of theslower one. The slower moving wagon is half-full of liquid in a tank.

6. Immediately after impact, what is the speed of the coupled wagons?

7. As a result of the impact the liquid in the part-full wagon sloshes back and forth violently.

Describe the effect this will have on the movement of the wagons.

8. A 400 kg satellite travelling at 7.0 km s–1 is hit head on by a 1.0 kg meteorite travelling at 12 km

s–1. The meteorite is embedded in the satellite by the impact. By how much is the satellite sloweddown?

Source

Open the JPEG image


9. An American spacecraft mass 18 tonnes docks with a Russian module mass 6.6 tonnes. Assume

the relative speed before docking is 0.2 m s–1 and a successful ‘soft’ docking is achieved firsttime.

Calculate how much faster the combined object is travelling than the original speed of the

Russian space vehicle?

Collisions of spheresQuestion 160S: Short Answer


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Calculation using the principle of conservation of momentumThese questions are about the collision of two spheres. One sphere B is sitting on a shelf at the verytop of the path taken by a sphere A as it is projected upwards. Sphere B rests on the shelf. Anyeffects of air resistance and friction should be ignored. The two spheres have equal mass.

B

A

Y C

The first impact


1. Assume that on the first impact the two spheres stick together.

Draw, on this copy of the diagram, the path they would follow after impact.

Explain why you have selected the path you have drawn.

B

A

Y C

The second impact2. Now assume that the collision is perfectly elastic.

State what you think would happen to the two spheres after impact.

3. Is it possible for sphere A to land at Y after a collision with B?


Jets and rocketsQuestion 180S: Short Answer


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These short questions increase in difficulty and ask you to relate impulse to change of momentum.

1. Thrust SSC is a supersonic car powered by 2 jet engines giving a total thrust of 180 kN.

Calculate the impulse applied to the car when the engines run for 4 seconds.

Assume the thrust is the only force acting on the car, which has a mass of 10 000 kg.

Calculate the increase in speed of the car after the 4 s.

2. One suggestion for powering spacecraft engines of the future is an ion engine. A beam of ions(charged atoms) is fired backwards, propelling the spacecraft forwards. In one test, xenon ionswere used.

Consider how using xenon ions would compare with using krypton ions, which are lighter.

If equal numbers of each ion were propelled back per second, at the same speed, which type ofion would you expect to give more thrust? Explain your answer

The mass of a xenon ion is 2.2 10–25 kg, and it can be ejected at a speed of 3.1 104 m s–1.

Calculate the number of ions that would have to be emitted per second to generate a thrust of 0.1N (a typical value of the thrust from such an engine).

3. When holding a hose firefighters need to ensure that they are not pushed backwards, especially ifthe water is ejected at a high speed.

Explain why firefighters experience a backwards force.


20 kg of water is ejected horizontally in 10 s, the speed of the water leaving the nozzle is 30 m

s–1.

Calculate the force experienced by a firefighter holding the hose.

4. A spacecraft is approaching the planet Zog and needs to slow down. To do so it fires a jet of gasforwards.

Explain how firing gas forwards slows the rocket down.

The rocket has a mass of 50 000 kg. The gas can be fired forward at a speed of 5 000 m s–1

relative to the rocket.

Calculate the mass of gas must the rocket eject to reduce its speed by 5 m s–1. Ignore the changein the rocket’s mass due to the ejection of gas.

5. Air of density 1.3 kg m–3 strikes a sail of area 15 m2. The air is initially moving at 5 m s–1 andassume it is brought to rest when it hits the sail.

Calculate the mass of air brought to rest in each second.

Hence calculate the force the air exerts on the sail.


6. A kestrel is a bird of prey which searches for prey by hovering above grassy areas.

Using your ideas about momentum, suggest how a kestrel is able to hover.

The kestrel has a mass of 200 g and it pushes down a column of air of area 600 cm2.

Estimate the downward speed given to the air by the kestrel

g = 9.8 N kg–1, density of air = 1.3 kg m–3.

Estimate the minimum power the kestrel needs to hover.

Suggest why there are no large birds which can hover. (Some large birds, such as buzzards andcondors, may appear to hover, but are not really doing so. They use upwards currents of air –‘thermals’ – to stay up.)

Gravitational potential energy and gravitational potentialQuestion 210S: Short Answer


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In a uniform fieldChapter 9 shows that changes in gravitational potential energy can be found as Ep = m g h.Gravitational potential is a scalar quantity. It is independent of the mass of the object in the field. The


change in gravitational potential is Vg = gh.

Take g = 9.8 N kg–1.

QuestionsClimbing a vertical rope is difficult. You have to lift your full body weight with your arms.

1. If your mass is 60 kg and you climb 2.0 m, by how much do you increase your gravitationalpotential energy?

2. What is the gravitational potential at 2.0 m?

3. What would the speed of your wristwatch be on reaching the ground if it dropped the 2.0 m?

4. What would your speed be on reaching the ground if you let go and fell?

A hod of bricks is raised vertically to a bricklayer at the top of a wall using a pulley system.

5. If the hod of bricks has a mass of 24 kg, what is its weight?

6. It is raised 3.0 m. Calculate its increase in gravitational potential energy at the top of the wall.

7. An accident takes place and one brick falls from the top. What speed does it have when itreaches the ground?

Travelling in a mountainous area, a bus of mass 3 tonnes reaches the edge of a steep valley. Thereis a 1 km vertical drop to reach the valley below, but 20 km of road to get there.

8. What gravitational potential energy will the bus lose in making its descent to the valley bottom?

9. Assuming its cruising speed does not change, where does the gravitational potential energy go?


10. Why is there a risk of brake failure in this situation?

A teacher releases a massive pendulum to swing freely, from the position shown in the diagram.

11. Describe the energy changes which take place through one full cycle.

12. Is the teacher in danger of being hit on the nose? Explain.

This is part of the gravitational field near the Earth’s surface. On such a local scale, the field can betaken as uniform.


A C147 J kg–1

98 J kg–1

49 J kg–1B

0 J kg–1

13. What is the distance apart of the equipotential surfaces shown?

14. How much work must be done to move a mass of 0.5 kg from B to A?

15. How much work must be done to move a mass of 1.0 kg from A to C?

16. On this diagram draw two possible paths of a ball thrown in the air, passing through points A andB, and describe what the paths have in common.

A C147 J kg–1

98 J kg–1

49 J kg–1B

0 J kg–1

17. Draw and label with values a set of equipotential surfaces separated by 2.0 m on the surface of

Mars, where the gravitational field strength is 3.8 N kg–1.


Summary questions for chapter 11Question 250S: Short Answer


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A satellite orbits Earth in a circular path with an orbital period of 90 minutes.

mass of Earth is 6.0 1024 kg, G = 6.7 10–11 m3 kg–1 s–2.

1. Calculate the distance of the satellite from the centre of Earth.

2. Is it possible to have the satellite orbiting at a larger distance but with the same orbital period?Explain your answer.

A person on the Equator is in circular motion about Earth’s centre. Assume the radius of Earth is

6.4 106 m.

3. Calculate the centripetal force necessary to keep a person of mass 60 kg in this path. Whatprovides this centripetal force?

4. Suppose Earth gradually started spinning more and more rapidly. Describe what you wouldexperience and what you would observe if we were on the Equator.


5. What would be different about your description in question 4 if you were at the North Pole?

Many airports have moving walkways to take passengers from one area to another. The speed of

such a walkway might be 2 m s–1 relative to the ground.

6. Estimate the total mass of a family of two children, two adults and their luggage for a flight.

7. Calculate the gain in momentum of the whole family if they walk onto the runway and keepwalking at the same speed relative to the runway that they were walking at before.

8. Estimate the average force exerted by the runway on the family in order to accelerate them.

9. If, on average, eight such families step onto the runway every minute, calculate the powerrequired to give extra kinetic energy to these people.


CentrifugesQuestion 50C: Comprehension


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Centrifuges are designed to separate particles in suspension from the liquid which contains them; forexample, separating the components of blood, or separating cream from milk. The mixture is pouredinto a tube in the centrifuge, which is then whirled around at a high speed in a horizontal circle.

p1 p2

p1 < p2

The closed end of the tube must exert the necessary force to make the tube’s contents move in acircle. Each part of the liquid suspension in turn exerts a force on the material next to it; this sets up apressure gradient with maximum pressure at the bottom of the tube and zero pressure at the surface.It is the pressure difference across each particular volume of mixture which provides a centripetalforce.

The centripetal force required is always F = mr .

1. Does the angular velocity depend on radius?

2. Let us say the centrifuge operates at 1000 rpm. What is the angular velocity?

3. At a particular radius r , consider two types of material. The less dense material has mass m1,and the more dense has mass m2, so that m1 < m2. Since force is determined only by radius andthe pressure gradient, they both have the same force acting on them. Explain why they cannotboth be in equilibrium.


4. At r = 10 cm, calculate the centripetal force is needed to keep a 1 g mass moving in a circularpath.

What about a 2 g mass?

5. Explain how the materials are separated, using the answers to your questions.

How Cavendish didn’t measure G and Boys didQuestion 60C: Comprehension


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Newton’s lawNewton stated his law of gravity in terms of proportionalities; i.e. the force is proportional to the massand inversely proportional to the square of the distance. The ‘gravitational constant’ implicit in this lawdid not make its appearance in research papers and text-books until about 200 years after thePrincipia. We do not even know who first denoted this constant by the symbol G, or when. It was notNewton, and it was certainly not Henry Cavendish whose ‘experiment to measure G ’ falsely appearsin many physics text-books. What Cavendish set out to measure in 1798 was the mass of Earth inorder to estimate its mean density. His paper mentions neither G nor even g. His experiment wasimportant as being the first to depend on an accurate measurement of the small deflection producedby the tiny gravitational force between objects of everyday size.

The gravitational force between two 5 kg spheres whose centres are separated by 10 cm is a fractionof a millionth of a newton. This force, equivalent to less than 100 g weight would have been barelydetectable, let alone accurately measurable, by the best chemical balances of the day. However, theFrench physicist Coulomb had recently developed a highly sensitive ‘torsion balance’ for hisexperiments to establish the laws of attraction and repulsion of electric charges. An Englishinvestigator, Rev. John Michell (who was the first scientist to speculate about the existence of whatwe now call ‘black holes’) built a large-scale version of the torsion balance for experiments on thegravitational force. He died before these could be carried out, and his apparatus was passed on toCavendish, who installed a modified version of it in a specially built sealed room at his London home


Cavendish paid great attention to environmental conditions – freedom from draughts and temperaturestability (to avoid convection currents) was essential. He also made careful estimates of thecorrections to be made for the gravitational forces on the test spheres due to all the rest of theapparatus (e.g. the mahogany cabinet which enclosed the spheres).

The ‘Cavendish experiment’ described by most text-books was carried out by Charles Boys in Oxfordin or around 1893. Boys’ equipment was similar but far more compact – it could be fitted into a smallcupboard rather than a medium-sized room – and he really did want to measure G rather than themass of Earth.


m ml

small masses

torsion bar

torsion fibre

Side elevation

Torsion balance measurement of gravitational force

M

M

fibre attached here

Plan view

movement whenlarge masses areadded

fibre attached here

Plan view

Torsional oscillations in absence of large masses


The force F between known masses m and M at a known distance d apart is deduced from the anglethrough which the torsion bar twists in the presence of the large masses M. The only unknown inthe gravitational equation is then G:

2d

GMmF

so

.2

Mm

FdG

The twisting force in the suspension can be found by measuring the period of twisting of theoscillations of the bar.

Had Boys wished to find the mass of the Earth ME , he could have deduced it from a knowledge of G

and g (the local acceleration due to gravity) and the radius of the Earth R using the equation (seeAdvancing Physics A2 student’s book, chapter 11)

2R

GMg E

in the form

G

gRME

2

Cavendish, however, found ME without reference to either G or g or needing to know the value of thesmall mass m. He did not even work out forces explicitly – there was no agreed unit of force at thattime – but argued entirely in terms of ratios. His complete reasoning is ingenious and quite elaborate,but a key step was to relate the ratio ME / M to the ratio of the gravitational force FE exerted on m bythe Earth to the gravitational force F exerted on m by M. Newton’s laws of gravity told Cavendish that

FE is proportional to ME / R2 and F is proportional to M / d 2 so

2

d

R

F

F

M

M EE

F was the force needed to produce the torsional deflection which Cavendish observed (just overhalf a degree corresponding to an arc length of about 1 cm). FE was the weight of the small mass.Cavendish was able to deduce the ratio FE / F from the ratio of the period of the torsional pendulum tothat of a simple pendulum of the same length oscillating under gravity.

There were several attempts to measure G towards the end of the nineteenth century, but Boys’experiment was soon accepted as the most accurate, and more recent measurements have

essentially served to confirm a value close to 6.67 10–11 N m2 kg–2.

It is interesting to speculate about the sudden interest in the value of G 200 years after Newton and acentury after Cavendish. It would seem that geophysical measurements (e.g. the size, shape andmass of the Earth) were no longer of such great concern. Interest was growing in more fundamentalaspects of the forces of nature. Boys’ experiment belonged to an era of classic determinations of‘fundamental physical constants’ e.g. Michelson’s determination of the speed of light (1881–7) andMillikan’s measurement of the charge on the electron (1907–13). One of the motives for repeatedmeasurements of G was to test whether it was truly a universal constant, hence differentexperimenters used different materials (e.g. steel, platinum, gold, quartz) and different shapes (e.g.


cubes, cylinders).

Questions1. The text quotes the ‘gravitational force between two 5 kg spheres whose centres are separated

by 10 cm’.

Suggest suitable materials that could be used to make these spheres.

2. Cavendish quotes the length of a ‘seconds-pendulum’ in London in 1798 as 39.14 inches.

Use this value to calculate the local value of g at that time.

3. Cavendish quotes the mean diameter of the Earth as ‘41 800 000 feet’ and derives a meandensity of 5.48 relative to the density of water.

Calculate the mass in kg which this implies for the Earth.

Supposing Cavendish had wished to calculate G from his data, what value would he haveobtained?

Are there planets around other stars?Question 90C: Comprehension


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Do planets like our own orbit around other stars? The answer links intimately with the question ofwhether life exists beyond Earth. The most promising place for life to arise is on the surface of aplanet. A search for other planetary systems should therefore be a major component of any effort tounderstand the prospects for extraterrestrial life. This passage and the questions for you to answerexplore these ideas.

Looking for planetsEfforts to search for other planetary systems employ either direct or indirect methods. Direct methodsinvolve detecting reflected light or infrared radiation from the planets themselves. The primary


difficulty with this approach is that emission from a planet tends to be drowned out by the vastlybrighter emission of its nearby parent star.

Indirect methods involve scrutinising a star for signs that it is responding to the gravitational tug of anorbiting planet. As the planet moves from one side of the star to the other, it pulls the star back andforth. This pull manifests itself as an angular perturbation (a slight wobble) superposed on the star’soverall motion across the sky. It can also be detected as a slight periodic change in the star’s velocitywith respect to Earth.

Any motion towards or away from the Earth (radial velocity perturbation) causes the star’s light wavesto be slightly compressed or stretched, a phenomenon known as the Doppler effect. Carefulmeasurement of absorption lines in a star’s spectrum can, in principle, reveal any periodic changes inits motion.

Indirect searches can be made easier by some simplifying assumptions. Many researchers haveguessed that giant planets around other stars will have orbital periods similar to Jupiter’s, about 10years. Fixing the orbital period allows the size of the orbit to be estimated as a function of stellar (star)mass and thus calculate an expected angular or velocity perturbation (given the orbital distance andassuming a certain mass for the planet).

An alternative possibility is that giant planets form at distances from their stars where temperature isat or below the condensation point of water. In this case, the star’s luminosity determines the size ofthe orbit for giant planets. If temperature is the determining factor, then typical orbital periods of giantplanets may be much shorter than those commonly expected. The average star is cooler andtherefore less luminous than the Sun, so giant planets could orbit in relatively close, fast orbits. Onecan then proceed as before to determine how much the planet is likely to affect the star’s velocity andapparent path. Tables 1 and 2 show the interactions of giant planets and stars.

Table 1. If orbital periods are equal

Star mass (M = mass of Sun) 3.0M 1.0M 0.3M

Orbital distance R / AU(AU = Earth–Sun distance)

7.6 5.2 3.5

Temperature / K 293 135 57

Orbital period T / years 12 12 12

Angular perturbations / s of arc 76 157 351

Table 2. If distance depends on temperature

Star mass (M = mass of Sun) 3.0M 1.0M 0.3M

Orbital distance R / AU (AU = Earth–Sun distance)

22.6 3.3 0.4

Temperature / K 170 170 170

Orbital period T / years 61.6 6.0 0.5

Angular perturbations / s of arc 226 99 40


Radial velocity measurements are relatively impervious to the blurring caused by the Earth’s turbulentatmosphere. Astrometry (the ultraprecise measurement of stellar position), however, will gain greatlyby being conducted in space. The goal is to be able to measure stellar angular deflections as small as10 microseconds of arc. Such an ability would permit the detection of companions as small as 10Earth masses around any star within 30 light-years of the Sun. The Moon would be an ideal locationfor astronomical instruments. It is conceivable that the lunar far side will be a site of a suite ofscientific facilities that will search for other planetary systems.

Questions1. Calculate the distance in metres which is equivalent to 30 light-years.

c = 3.0 108 m s–1

2. Describe a direct method of searching for planets orbiting distant stars and explain why it isunsatisfactory.

3. Draw a diagram of a planet orbiting a star, showing the direction and relative magnitude of anyforces on each body. What is the origin of these forces?

The period of a planet around a star of mass Ms is connected to the radius of the orbit by the

relationship T2 is proportional to R3 / Ms.

4. Does the information provided in table 1 confirm this relationship for planets orbiting with thesame period around stars of different masses?

5. Does the information provided in both tables confirm this relationship for planets orbiting stars ofthe same mass?

6. Show that the ability to resolve 10 microseconds of arc in the measurement of angles wouldenable an astronomer on Earth to distinguish a 2 cm diameter coin on the Moon.

The Moon is approximately 400 000 km from Earth, 3600 seconds of arc = 1 degree, 1 radian is


approximately 60 degrees.

7. By choosing suitable pairs of values from tables 1 and 2 show that angular perturbation increaseswith increasing distance between a planet and its star.

Variations in gQuestion 130C: Comprehension


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The shape of the EarthNewton could explain the fact that the Earth is nearly a sphere. Since every bit of the mass of theEarth attracts every other bit, it takes up the smallest possible volume. However, because the Earthspins on its axis, he also predicted it would bulge slightly at the Equator. This is equally true of starsand other planets which spin.

Fg

Fg

Fg

Fg

Fg

Fg

Fc

Fc

Fc


r1

r2

N

S

r2 > r1

Much of the material which makes up the Earth is slightly elastic and so can stretch. All of the Earthspins at the same rate, so the pieces near the equator must move faster. Material between thesepieces and the axis of rotation stretches, providing an extra restoring force, keeping each piecemoving in a circle as the Earth spins:

.2

r

mvF

This extra radius caused by this stretching is about 21 km. In 1733 the French Academy of Sciencessponsored expeditions to Peru and to the Gulf of Bothnia to measure the distance for one degree oflatitude. Results confirmed Newton’s prediction. The Earth’s shape is properly called an ‘oblatespheroid’. The field strength g depends on latitude, being largest at the poles where r is smallest.


Fg

Fg

Fg

Stationary Earth

Fg

Fg

Fg

Spinning Earth

Fc

Fc

Fc

this spring stretches most to providelargest centripetal force

Weight and free fall at different latitudesIf the Earth were stationary the weight of a mass m would be m g, where g is the gravitational fieldstrength – g would also be the acceleration of free fall.

However, because of the Earth’s rotation, the measured gravitational pull is less than this and equalsm go where go is the observed acceleration of free fall.

Except at the poles, any mass spinning with the Earth’s surface needs a centripetal force acting on it.This force is supplied by part of the Earth’s gravitational pull on it. The centripetal force acting on the


body is m g – m go.

The centripetal force must be m 2 x – where x is the distance to the spin axis, zero at the Poles andr at the Equator. Again as a result, the value of go will depend on latitude, being largest at the Polesand smallest at the Equator.

QuestionsTake G = 6.67 10–11

N m2 kg–2 and the mass of the Earth = 5.98 1024 kg.

1. The centripetal force on a mass at the Equator is

o2 mgmgRm

Find g –go, which would be the difference between polar and equatorial values of the accelerationdue to gravity if the Earth were a perfect sphere.

2. Use the gravitational inverse square law to estimate the difference in values of g between the topof Mount Everest (8850 m) and sea level.

3. If a plumb bob is suspended near one side of a mountain, then it will be pulled very slightly fromthe vertical by the gravitational force which the mountain exerts on it. This is shown in a simplifiedway by the diagram.

sphere

string

Schiehallion

This experiment was done in 1774 by Nevil Maskelyne, the first Astronomer Royal, in order toestimate the Earth’s density. He used the mountain Schiehallion in Scotland, estimating its

volume was about 1.6 109 m3 and its density about 3000 kg m–3. At the height of the experimentthe bob was level with the centre of mass of the mountain and 2000 m from it.


Draw a vector diagram showing the gravitational forces exerted by the mountain Mm and by theEarth as a whole Me on a plumb bob mass m placed in the position shown in the diagram.

Write down expressions for these forces and use these to find the angle from the vertical throughwhich the plumb bob would be deflected if these estimates are true.

4. With the angle of deflection measured experimentally, calculations done in the reverse order canbe used to obtain a value for the mass of the Earth. Write down the algebraic expression used tofind the mass of the Earth.

These days geophysicists, using a very sensitive spring balance called a gravity meter, routinely

measure values of g to a precision of 10–7 (0.1 part per million). After making corrections for latitude,for altitude and for local topography the remaining variation in g is referred to as a gravitationalanomaly. This can give information about variations in density of rock beneath the surface, a methodused in prospecting for minerals and for oil.

Collision with spaceship EarthQuestion 170C: Comprehension


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In 1908, the Tunguska area in eastern Russia was devastated when a comet of mass 10 milliontonnes struck. In this remote and thinly populated area, trees and reindeer were destroyed for aradius of 30 miles around. The comet is thought to have had a diameter of a few hundred metres. Theexplosion released energy equivalent to several hydrogen bombs and sent shock waves around thewhole world. So much dust was thrown into the atmosphere that sunlight was scattered from thebright side of the globe right around into the Earth’s shadow. In London, 6000 miles away, themidnight sky was as bright as early evening. This happened well before the international travel andcommunications we take for granted were possible, so what caused the distant effects was notknown. Life in Tunguska took over 25 years to recover.

Earth may appear to be a lonely traveller through the solar system. But as well as small meteoroidsthere are two classes of large objects whose paths can cross ours: comets and asteroids. Cometsfollow eccentric (elliptical) orbits, some with short periods orbiting in the inner part of the solar system,others being long-period comets originating far out beyond Pluto. Asteroids normally orbit in theregion between Earth and Mars, but their paths can be disturbed by the gravitational pull of the largerplanets, Saturn and Jupiter; it is always possible that one could be diverted onto a collision course


with Earth.

The scarred and cratered surfaces of the Moon, Mercury and Mars show that collisions in the solarsystem are common enough. On Earth, the evidence of collisions is rapidly erased by erosion andcovered by growing vegetation. Traces of craters there certainly are; in a 1995 count 145 knowncraters were listed. Many scientists believe a collision on a much larger scale than the Tunguskaevent caused the extinction of dinosaurs 65 million years ago. A crater on the Yucatan peninsula,

Mexico is evidence of the ‘prime suspect’. Having an estimated mass of 1012 kg (8 miles in diameter),the colliding body would have thrown enormous amounts of dust and water vapour into theatmosphere. It would have darkened the day sky and disrupted the environment globally.

Small meteorites constantly arrive at our Earth. Falling through the atmosphere, they burn up as‘shooting stars’. Over a long enough time – millions of years – another major collision is probable. Inone lifetime, the probability is extremely small, but it’s not zero. Thinking about extremely unlikelyevents that have enormous consequences is difficult to do, even when you have facts and figures.

Questions1. With an orbit radius of 1.5 108 km, and an orbital period of a year, how fast does the Earth

travel along its path?

2. Imagine a pea-sized iron meteorite of mass about 10g collides with the Earth at the speed youjust calculated in question 1.

What kinetic energy would be involved in such a collision?

3. Calculate the speed in metres per second, of a 1000 kg car with the same kinetic energy.

Convert this speed to miles per hour.

‘Forewarned is fore-armed’Millions of people watched on TV as fragments of the Shoemaker–Levy comet fell into Jupiter in1994, scarring its surface. For many, that image will have demonstrated our vulnerability. If we are todo anything to prevent such a collision with Earth in our own time, we will need to have as muchadvance warning as possible.

We might hope to launch a rocket to intercept the object, before it gets near to Earth. In the popularmind, the purpose of this interception is simply ‘to blow the object up’. People have suggested usingnuclear weapons to do this. Never mind the hazards of launching a nuclear device, simple physicstells us an explosion will not work anyway.

Think of fireworks exploding in the air. The rocket is moving along a parabolic path when the bigexplosion takes place. The fragments scatter, but the path of their centre of mass continues to movealong the same path as before. This is consistent with the law of conservation of momentum. Sinceno external force has acted, there can be no change in momentum of the system.


Another question4. Let’s say you are out shooting clay pigeons, but the operator doesn’t know how to set up the

machine which launches ceramic targets. They are aimed in a graceful arc to fall where youstand! If you shoot and miss, you will be hit by the moving target. If you hit the target, whathappens?

A mass driverWhat we have shown so far is that the interception must deliver not energy (an explosion), but achange in momentum.

The correct answer to the impact problem is to use a machine called a mass-driver. A prototypeversion was built at Princeton University in the 1970s. It consisted of a magnetic accelerator pushingsmall buckets along a straight track. Each bucket carries a small cargo that you wish to accelerate. Atthe end of the track, the cargo flies off at high speed and the empty bucket returns to the start.

The small prototype machine was only a metre in length but could accelerate cargo to 100 m s–1!

To deflect a comet, you would need a larger model, perhaps a hundred metres long and acceleratingcargo to one kilometre per second. The cargo would be soil or ice from the comet. The cargo wouldfly off into space, carrying away its momentum, and an equal and opposite recoil momentum wouldbe delivered to the comet.

Final question5. Draw a diagram showing the path of the comet relative to the Earth if this strategy is successful.

On your diagram show the direction in which mass is being thrown off the comet.

Getting a satellite up to speedQuestion 190C: Comprehension


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The purpose of the launching phase is to take the satellite to the chosen height and to send it off

horizontally at a speed of roughly 8 km s–1. Hitting a golf ball off a cliff is not a practicable method,

because no cliff is high enough and no golfer drives faster than about 30 m s–1. There are bettermeans at our disposal, for man has, from time immemorial, spent too much of his energy devisingways of making projectiles fly faster, so that they will be more efficient as weapons of war. The


satellite has a grisly ancestry: the spear thrown by hand, the arrow shot from a bow, the cannon ball,the rifle bullet, the sixteen-inch naval gun, the V2 rocket (V stood for ‘Vergeltungswaffe’ -vengeance

weapon). The cannon ball is nowhere near satellite status, and the V2 with a speed of about 2 km s–1

is still a long way from the 8 km s–1 of the satellite. Man took thousands of years to go from 30 m s–1

(a spear) to 2 km s–1 and less than 15 years to make the step from 2 to 8 km s–1.

The agent which has brought about these spectacular advances is the rocket motor, which is muchbetter suited to the purpose than either a gun or an aeroplane engine. The disadvantage of a gun isthat although the missile fired may reach a very high speed on emerging from the muzzle, it ispromptly retarded by the severe air drag which it meets because it is going so very fast at low altitude,in dense air. A rocket missile on the other hand often starts quite slowly: if a racing car and anintercontinental rocket both started at the same moment, the car would probably go further than therocket in the first half-minute. The rocket only attains its high speed at a great height, where the air isvery thin and offers little resistance.

An air-breathing engine, like a turbo-jet or an ordinary piston engine, is not very useful for satellites,but a rocket thrives in airless space; air is merely a nuisance to it. The rocket is the mostself-possessed of engines: it carries along with it all that it needs for its proper working. Rocket fuelsmay be either solid, liquid or gaseous, but at present most big rockets use two liquid propellants,kerosene and liquid oxygen being a typical pair. These are kept in separate tanks and then pumpedinto a combustion chamber designed to withstand extreme heat. There they react together, producinglarge volumes of intensely hot gases, which are led through an exit nozzle. The difference in pressurebetween the hot compressed gas in the engine and the cold low density air outside drives a jet of gasout of the exit nozzle. The velocity of the jet is comparable to the average velocity of the molecules inthe hot gas. This exit jet drives the missile forward by the reaction of its momentum.

To illustrate this reaction principle, we can bring on some golf balls again. Suppose a lorry loadedwith golf balls is unable to move on a flat icy road. If the lorry driver is really desperate to get moving,he can do so by driving off not the lorry but the golf balls. If he has had the foresight to arm himselfwith a suitable golf club, he can drive the balls off backwards one by one, and there will be a smallforce pushing the lorry forward every time he strikes one. By the time the stock of balls (and thedriver) is exhausted, the lorry may be moving quite fast, providing air resistance can be neglected andthe ice is frictionless. The fusillade of golf balls corresponds to the stream of rocket exit gases, andthe lorry corresponds to the structure of the missile. This method of propulsion may not commenditself to every ice-bound lorry driver, but it is most useful outside the Earth’s atmosphere, where therereally is no air resistance and friction.

QuestionsLiquid fuels were initially preferred to solid fuels for rockets because they could yield faster jets of exit

gas (about 3 km s–1 rather than 1 km s–1).

1. Explain, in terms of momentum, the advantage of faster exit molecules.

2. Even so, in due course, solid rocket fuels became quite common; why do you think that this mighthave happened?


Suppose the ice-bound lorry weighed 2 tonnes, its cargo of golf balls 4 tonnes, and the balls could be

driven off at 30 m s–1. Estimate:

3. The number of golf balls.

4. The increase in speed given to the lorry when the first golf ball is driven off.

5. The increase in speed given to the lorry when the last golf ball is driven off.

6. The final speed of the lorry is about 33 m s–1; make a rough sketch of the speed–time graph forthe lorry during its acceleration assuming the balls are driven off at a rate of one every 10seconds.

7. What difference would it make to the calculations if the lorry had a cargo of peas and the driverwere equipped with a pea shooter.

Using Kepler's third lawQuestion 10D: Data Handling


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Take the radius of Earth to be 6370 km.

1. The radius of a geostationary orbit is 42 200 km. Use this fact together with the constancy of R3 /

T2 to estimate the height above the Earth’s surface of a satellite whose circular orbit is completedin 90 minutes. How many times a day would such a satellite orbit the Earth?


2. Low-orbiting Earth satellites usually have orbital periods in the range 90 to 105 minutes. Whatrange of heights does this correspond to?

3. 90 minutes is a typical orbital period for a military reconnaissance satellite, and 100 minutes for acivilian Earth observation satellite. Can you suggest a reason for this difference?

4. Kepler’s laws were formulated for elliptical orbits (of which the circular orbit is a simple specialcase). The ‘R’ of the third law is the semi-major axis (found as the average of the maximum andminimum distances between a satellite and the body it orbits). You can see how this works bylooking at data for Sputnik 1, the first artificial satellite, which was launched on 4 October 1957and, slowly losing energy due to the effects of atmospheric friction, spiralled back to Earth 3months later. Complete the following table of data:

4 October1957

25 October1957

25 December1957

Orbital period / minutes 96.2 95.4 91.0

Minimum height / km 219 216 196

Maximum height / km 941 866 463

Mean height / km

Mean radius / km

R3 / T2 two significant figures

Did the orbit become less elliptical as time passed?


The gravitational field between the Earth and the MoonQuestion 120D: Data Handling

Teaching Notes | Key Terms | Hints | Answers

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The dynamics of space travelThese questions pose a number of short problems about the dynamics of the motion of a spacecraft,returning from the Moon to the Earth. The mass of the spacecraft is 2000 kg.

E

radius at E= 20 000 km

380 000 km (Earth to Moon)

C

D

AB

v1 6000 m s–1

30º

MoonEarth

The diagram shows its path and is drawn to scale (except for the diameters of the Earth and theMoon which are shown larger than they should be). Distances given are to the centre of the Earth.

mean radius of Earth as 6.4 106 m, G M for the Earth = 4.0 1014 N m2 kg–1

Gravitational force and potential energy1. At point A the spacecraft has zero acceleration when its thrust motor is switched off. Suggest a

reason why.

2. At point B, again without any motor thrust, the speed of the spacecraft is constant, though it doeshave an acceleration. (At B, the craft is travelling at right angles to the line joining it to the Moon’scentre.) Explain this observation.

3. At C, the velocity is 6000 m s–1, as shown. It is required to alter course by an angle of 1/100radian, without changing speed.


In what direction must the thrust be?

Calculate the length of time for which the motor thrust of 12 000 N be maintained?

4. In which direction is the spacecraft accelerating at D if the thrust motor is not in use?

Calculate the magnitude of the change its in momentum of the space craft over a time of 100seconds. Ignore any effect of the Moon in this question.

5. Make a rough calculation to show it is fair to ignore the effect of the Moon in question 4.

Ignore any effects of the Moon in questions 6–8.

6. The thrust motors are not used between C and E.

Calculate the change in gravitational potential energy of the spacecraft change between C and E.

Explain whether it is an increase or a decrease, and whether the answer would have beendifferent if the motors had been used.

7. At C the spacecraft’s velocity is 6000 m s–1.

Use your answer to calculate the kinetic energy of the spacecraft at E.

8. What kinetic energy would the spacecraft have at E if, instead of having come in from Moon orbit,the spacecraft were in permanent orbit around the Earth through E?


Gravitational potential difference, field strength and potentialQuestion 220D: Data Handling


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Voyager 1The Voyager space probe, launched on 5 September 1977, approached the planet Jupiter about 18months later. Its distance from the centre of Jupiter and its speed directly towards it, v, were recordedat weekly intervals:

Date / 1979 r / 109 m v / km s–1

30 January 35.10 10.90

6 February 28.08 11.00

13 February 21.06 11.14

20 February 14.04 11.40

27 February 7.02 12.17

G = 6.67 × 10–11 N m2 kg–2

1. Plot a graph of v2 against 1 / r for these data. Use the graph to find the speed of Voyager 1 whenit was a very long way from Jupiter.

2. Explain how the graph confirms that the space probe’s engines were not used during this period.

3. The gradient of the graph is equal to 2 G MJ. Deduce a value for the mass of Jupiter MJ.

Find the difference in gravitational potential Vg between a point on the Earth’s surface and a point 10m above the surface.


g = 9.8 N kg–1, mean radius of Earth as 6.4 106 m, G M for the Earth = 4.0 1014 N m2 kg–1

Do this calculation by the two different ways stated below:

4. By use of Vg = gh:

5. By use of the relationship Vg = – G M / r :

6. With a constant thrust, a rocket climbing away from Earth accelerates faster and faster. Onereason is that its mass is decreasing.

Use the idea of potential gradient to explain the other reason.

Rocket scientists need to know the amount of energy required to raise 1 kg from the Earth’s surfaceto various heights. They might calculate values of Vg between R (at the Earth’s surface) and variousdistances r (from the centre of the Earth). The results are shown in the following table:

r / 106 m Vg / MJ kg–1

6.371 0.00

7 5.64

10 22.7

14 34.1

18 40.5

22 44.5

30 49.3

40 52.7

50 54.7

7. Plot a graph of Vg against r (use graph paper).

Use the graph to estimate the amount of energy required to lift:

8. A 1000 kg probe from the Earth’s surface to a distance 20 106 m from the Earth’s centre.


9. A 200 kg satellite from the Earth’s surface to a height of 36 106 m above the surface.

10. A 1 kg mass from 40 106 m to 50 106 m from the centre of the Earth.

11. The graph seems to be approaching a limit – it does not continue to rise indefinitely. The extra

energy required to transport 1 kg from 50 106 m to a very great distance can be worked out: itcomes to about 7.8 MJ. What then is the total energy needed to lift 1 kg from the Earth’s surfaceto as great a distance as one would wish?

The following questions use values of the gravitational potential at different distances from the centreof the Earth to calculate energies required to escape from the Earth. A few values of Vg and r are

already given.

r / 106 m Vg / 106 J kg–1

6.37 – 62.5

6.38 – 62.4

10 – 40

400 (distance of the Moon) – 1

Infinity 0

12. How much energy is needed to raise 1 kg to a height of 10 km?

13. What force acting on 1 kg would transfer this amount of energy if it were uniform over the 10 km?Comment.

14. How much energy is needed to transport 1 kg from the Earth to a distance equal to that of theMoon?


15. Why is the energy not equal to the product of 10 N and this distance?

16. How much energy is needed for a mass of 1 kg to escape entirely from the Earth’s influence?

17. What velocity would the 1 kg object need at its launch to achieve this?

18. Why would a mass of any size need this same velocity?

19. Use the 1 / r variation of potential to deduce values of Vg at distances of 20, 40 and 80 106 m.

Hence plot a graph of Vg against r from r = (10–80) 106

m.

20. Find the gradients of tangents at r = 20 106 m and at 40 106 m.

21. What is represented by these gradients?

22. Find the ratio of the gradients and comment on the result.


Changing orbitsQuestion 230D: Data Handling


Quick Help

Launching a satellite into a circular geostationary orbit is a two-stage operation, involving a lowercircular orbit – called a parking orbit – and an elliptical transfer orbit, which is sometimes called aHohmann orbit.

Thrust causing change ofvelocity v1 takes spacecraftfrom low circular orbit intoelliptical transfer orbit.

v1

v2

Thrust causing change ofvelocity v2 takes spacecraftfrom elliptical transfer orbitinto high circular orbit.

low circular orbit ellipticaltransferorbit

high circular orbit

A satellite is launched into geostationary orbit (radius 42 200 km) via a parking orbit at a height 300km above Earth’s surface. Increases in speed to take the satellite into and out of the elliptical transferorbit can be regarded as instantaneous.

radius of Earth to be 6370 km.

A few questions1. Calculate the speed of the satellite in the geostationary orbit.


2. Use Kepler's third law (period2 mean orbit radius3 ) to calculate the period of the satellite in theparking orbit. Use the period to calculate the speed of the satellite in the parking orbit.

3. Is the kinetic energy of the satellite in the geostationary orbit greater or less than when it is in theparking orbit?

4. Find the average distance for the transfer orbit, by averaging largest and smallest distances.Hence, calculate the time taken for the satellite to travel from parking orbit to geostationary orbit.

5. The graphs show the energy of the satellite before, during and after the transfer between parkingand geostationary orbit. Identify which line refers to kinetic energy, which to potential energy andwhich to total energy.


time / hours

5 10 150

6. Sketch the path of the satellite around Earth for the 12 h shown.

Why is a ‘black hole’ black?Question 240D: Data Handling


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This set of questions is about one of the effects of gravity, and leads to the idea of a ‘black hole’.

Escape velocityWhen a projectile is fired vertically from the surface of the Earth with a small velocity, v, it reaches a


height which depends on v and then falls back to the surface. Above a certain velocity, called theescape velocity, it does not return to the Earth but carries on indefinitely. The escape velocity isfound by equating the projectile’s initial kinetic energy to the gain in gravitational potential energy asthe projectile moves from its starting point to ‘infinity’.

1. Explain why the escape velocity can be found by equating its initial kinetic energy to its gain ingravitational potential energy as the projectile moves from its starting point to ‘infinity’.

Show that the escape velocity vesc from the surface of a spherical body of mass M and radius R isgiven by:

vesc = (2 G M / R)1/2 where G is the universal gravitational constant.

2. Calculate the escape velocity for a projectile fired from the Earth

mass of the Earth = 6.0 1024 kg, mean radius of the Earth = 6.4 106 m, G = 6.7 10–11 N m2

kg–2

3. Show that the mass of the Sun is 2.0 1030 kg and the escape velocity from its surface is

6.2 105 m s–1.

mean radius of the Sun = 7.0 108 m, density of the Sun = 1.4 103 kg m–3

The making of a ‘black hole’In the later stages of a star’s life, when the star has used most of its energy resources, it can shrink toa very much smaller diameter. For example, if the Sun were to shrink until its density werecomparable to the density of an atomic nucleus it would have a radius of 12 km. A star in this state iscalled a neutron star.

4. Show that the escape velocity from the surface of the Sun, if it were to shrink to this size, wouldbe close to one half the speed of light.

speed of light in a vacuum, c = 3 108 m s–1

5. A black hole is an object with an escape velocity equal to the speed of light so that nothing, noteven light, can ever escape from its surface.

What would the radius of the Sun have to be in order for it to become a black hole?


Brahe and HamletReading 10T: Text to Read

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This reading builds links between history, science and literature.

Great Danes in fact and fictionThere are more things in heaven and earth, Horatio, than are dreamt of in your philosophy.

William Shakespeare Hamlet Act 1 Scene 5.

That’s it then is it?...The sun’s going down. Or the earth’s coming up, as the fashionable theoryhas it...Not that it makes any difference…

Tom Stoppard Rosencrantz and Guildenstern are Dead Act 5.

In the late sixteenth century, English traders bound for Copenhagen and the Baltic ports would havebecome aware of two splendid new landmarks as they passed through the ‘Sound’ – the narrowchannel between Denmark and present-day Sweden. The first, controlling the entrance to the Sound,and helping to enforce the levying of tolls on passing ships, was Frederick the Second’s beautifulcastle of Elsinore. The next, on the island of Hveen, was Tyge Brahe’s (Tycho was the fashionablyLatinised form of his first name) ornate domed astronomical institute which he called Uraniborg or‘castle of the heavens’. The island of Hveen and the ‘ton of gold’ needed to build and maintainUraniborg were King Frederick’s gifts to Brahe, an inducement to the already celebrated astronomerto stay in his native Denmark.

The subsequent fame of both Elsinore and Uraniborg was ensured at the opening of the seventeenthcentury by works which represent two pinnacles of human intellectual achievement, Shakespeare'sHamlet and Kepler’s New Astronomy. Kepler never visited Uraniborg, but Brahe’s data, gatheredthere over a period of 18 years, provided the basis for Kepler’s epoch-making discoveries. It isunlikely that Shakespeare ever visited Elsinore, but certain that several of his fellow actors in Londonspent time there in the 1580s as visiting players.

The story of Hamlet is from old Danish legend, but Shakespeare gave it a contemporary setting.Hamlet is a thoroughly modern prince, his wit as sharp as his rapier, a cultivated scholar in contrast tohis father, a grim and fearsome warrior. Hamlet, and the young noblemen of the court – Horatio,Rosenkrantz and Guildenstern – have studied at the great Protestant centre of the new learning, theUniversity of Wittenberg in Germany. This was the normal practice for the young Danish nobility inFrederick II’s time.

The young Tyge Brahe spent some time at Wittenberg. So did several of his kinsmen, the historical,as opposed to the fictional, Rosenkrantzes and Guildernsterns. Wittenberg, as well as being thecradle of Lutheranism, was also a centre of Copernican doctrine. Two professors of astronomy there


publicised and championed the theories of the Polish monk, in spite of the strong objections ofProtestant theologians that these were contrary to scripture. Another famous Protestant university, atTübingen in south Germany, was also a centre of Copernicanism, and it was here that JohannesKepler, as a young theology student, became convinced of the truth of the Sun-centred Universe.

The action of Hamlet opens on a clear winter’s night on the battlements of Elsinore. The sentinelstime the nightly appearances of the Ghost by the position of the stars. Horatio is reminded of the roleof the stars as portents of catastrophe (the word ‘disaster’ is derived from astrum the Latin word forstar); he speaks of ‘stars with trains of fire’ (comets) and ‘disasters in the sun’ (solar eclipses). Thelate sixteenth century was a time of well publicised astronomical turbulence; the nova of 1572(‘Tycho’s Star’) caused a sensation, as did the comet of 1577. The audiences for the early Londonperformances of Hamlet would have been only too familiar with solar and lunar eclipses, several ofwhich were visible in England in the years 1598 to 1601.

These years, coincidentally, were disastrous for Tyge Brahe, who fell out with the new king ofDenmark Christian IV and as a result went into exile in Bohemia. There he became ChiefMathematician (and Court Astrologer) to the Emperor Rudolph, a post which passed to his assistantKepler on Brahe’s death in 1601. Brahe travelled much during his lifetime but never visited England orScotland; he did, however, have important admirers at the English Court. Christian IV’s brother-in law,the scholarly James I, who succeeded Queen Elizabeth (as both monarch and Shakespeare’s patron)in 1603, had visited Uraniborg at the time when he was James VI of Scotland.

James I wrote a Latin poem in praise of the astronomer, and his learned Lord Chancellor, FrancisBacon, paid Brahe and Uraniborg another sort of compliment. Bacon is famous for probably not beingthe author of Shakespeare’s plays and, among other things, for being the prophet of modern scientificpractice. He advocated ‘putting Nature to the question’ in order to discover her innermost secrets,through systematic programmes of precise experimental measurement. This is exactly the approachthat Brahe had pioneered at Uraniborg, where he set up what was in effect the first modern scientificresearch institute. There is no record of Bacon visiting Uraniborg in person, but he was a keenstudent of astronomy and had dealings with Kepler. There can be little doubt that the House ofSolomon, the scientific research centre described in Bacon’s Utopia The New Atlantis was largelyinspired by Brahe’s Uraniborg.

HubbleReading 20T: Text to Read


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A poem by Diana Syder

It doesn’t look much, swathed in tin foil, its solar panels

all but juddering themselves to bits sixteen times a day

every time it passes from day to night, big as a Sheffield bus

shining in starlight and doing 500 mph over Madagascar

And it wasn’t easy getting it up there, all that hellfire steam


and thundering flame at Cape Canaveral to kick the earth away,

hardly subtle, hardly how you’d choose to get a piece

of precision engineering where you want it and hit

the exact place where the lures of space counteract

the attractions of earth, but here it is and it works, so I take

my hat off to the bright spark who dreamed it right at the start

and take it off again to what’s in us that wants to unzip

everything right up to the edge of time and only stop then

because we have to, but at least to go that far and see.

The photos are out of this world. Look at this one: a jet of gas

from a young star in Orion and it has it perfectly, just as it is,

or was, because we’re travelling in time by standing still,

seeing out and way back to whatever made us. If angels

do anything like this then sign me up because I’m a believer.

OK, an arms akimbo, mock-up angel, with bits sticking out

all over the place that would have no grace on earth

but I can extrapolate and this is all about imagination, in fact

I have to remind myself these are pictures of real things:

whole cities of stars, brown dwarfs, blue stragglers, plus exotics

I’ve never heard of. Eleven billion years for the Universe

to invent a way of looking at itself and I’m around to see it!

I could look for a lifetime and probably will. I’d love to tell them,

Galileo, Bruno and the others, right up to Hubble himself,

see what they all made of it – or would that not be fair?

Might it be too much to bear, your own beliefs coming true

because what price dreams then? Anyway, talking of interactions

between people and light makes it sound more grandiose

than it really is, when what goes on most days is stumbling around

heavy as lead, trying to find a good way of living and mostly

I’m tied up with that, but I sometimes get hit by something coming

out of the ether at me, like the phrase tomorrow is a new day,

just as I heard it drop ready formed into my mind as a consequence

of nothing related and took note because it sounded hopeful


and made me think of all the new days as they regularly fall,

the shadow backing off from the Pacific, America, the Atlantic.

All the people waking, yawning and sitting up, one group

after another as the light approaches, doing it in reverse

when night returns, oscillating between gravity, which could pin

us to the ground and imagination, which might spin us away –

and here’s a platform for imagination if ever there was one!

It’ll keep us busy for ages, working out how much is true

and how much more there is of what we need to know

to swim strong and well in our deep mysterious bell,

with bubbles rising slowly out of the past, present and future

and some of the dreams coming true for us, one secret at a time.

This poem by Diana Syder is the title poem in her volume Hubble (1997), Smith/Doorstop Books.

The problem of longitudeReading 30T: Text to Read


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Here a start is made in introducing some of the challenges faced in former times – just knowingwhere you were!

Latitude and longitudeTo locate any position on the surface of the Earth (or at sea!) two coordinates need to be specified.

Latitudes are imaginary lines that circle the Earth parallel to the Equator. The length of a day will tellyou your latitude approximately. By measuring how high the Sun or stars are in the sky at mid-day,compared to a vertical line obtained by hanging a plumb-line downwards, you can find your latitudeprecisely. The geometry of the Earth and sky defines latitude. The range of latitudes is 0–90 degreesNorth and 0–90 degrees South.

Longitudes are imaginary lines that start at the poles and circle the Earth. Since the lines passthrough common points (the poles), they cannot be parallel. The range is 0–180 degrees East and0–180 degrees West. East and West of what? A purely political decision is needed to define wherethe starting point is; at present, all countries agree the ‘prime meridian’, from which other longitudesare measured, is the line which passes through the old Greenwich Observatory in London. This iswhy Greenwich Mean Time is known around the world.


You need to know latitude and longitude accurately both to draw maps and to navigate.

Measuring longitudeYou can use the Earth’s rotation to work out longitude. Since the Earth takes 24 hours to completeone full revolution of 360 degrees, 1 hour marks 1/24 of a spin, or 15 degrees. So if you know thetime in London when the Sun is at mid-day your time, the time difference between you and Londoncan be converted into degrees of longitude. Wherever you are on Earth, every 4 minutes timedifference with London adds one degree of longitude.

Today we can take for granted the accuracy of even the cheapest of wristwatches and the GlobalPositioning System. But cast your thoughts back in time. The accuracy of the pendulum clock, notinvented by Galileo until the fifteenth century, is useless at sea where the rolling motion of a ship ruinsits timekeeping.

An alternative to using the Earth’s rotation is ‘dead reckoning’: the navigator records speed, time anddirection and plots a course by what you would call displacement vector addition. Speed is estimatedby throwing a log into the water and observing its motion; direction found by compass, and timemeasured very roughly in fractions of a day. Because of the (unknown) relative motions of wind andseawater, it is a terribly unreliable method. Many seafarers lost their lives as their ships ran agroundon rocks and shallow water, by simply not knowing where they were. Through good fortune some ofthose who lost their way survived; for example, famous explorers such as Magellan and Drake. Forsafety, ships often followed well-travelled routes along lines of latitude, though this made journeyslonger and exposed them to piracy.

In 1707 four British warships in a fleet of five ran aground off the Scilly Isles, and 2000 sailors losttheir lives. This so disturbed the British Government that an Act of Parliament was passed soonafterwards, offering a prize of £20 000 (the equivalent of millions of pounds today) to any person whocould solve the longitude problem. Essentially what was needed was a reliable and accurate clock.The prize-money made it a race among competitors.

Two rival campsThere were two basic approaches. One was that taken by the scientific elite of the time, astronomers,who aimed to harness the ‘celestial clockwork’. Large stakes of money, status and learning wereassociated with this approach. Major astronomical observatories were set up at Berlin, London andParis, constructing maps and tables so that the positions of stars, planets and moons could be usedto determine longitude. Great men such as Galileo, Cassini, Huygens, Newton and Halley contributedto this work.

The other approach was to invent better clockwork mechanisms, able to withstand variations oftemperature and of humidity as well as the violent rolling of ships at sea. This approach was pursuedalmost single-handedly by a remarkable man from Yorkshire called John Harrison. A man with noformal education, Harrison devoted his life to this challenge. He got rid of the pendulum and devisedclock movements which were almost friction-free, able to run at a constant rate in all conditions.Despite their prejudices, those adjudicating Parliament’s prize-money were eventually forced toaccept that Harrison the clockmaker had solved the problem.

A bestsellerA few years ago, a gripping book about the race to solve the longitude problem became aninternational bestseller. Its cover description reads ‘the dramatic story of an epic scientific quest, andof Harrison’s forty-year obsession with building his perfect timekeeper, known today as thechronometer. Full of heroism and chicanery, brilliance and the absurd, it is also a fascinating briefhistory of astronomy, navigation and clockmaking’.

You might like to read Longitude by Dava Sobel (1995, London, Fourth Estate).


Kepler’s second law and angular momentumReading 40T: Text to Read


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In the Advancing Physics A2 student’s book there is no explanation for why an orbiting planet mightsweep out equal areas in equal times. This behaviour is consistent with a deeper law of physics, forwhich you need to know about angular momentum. The deeper law applies to all rotations andcorresponds to the law of conservation of momentum for linear motions.

NOTE: Whereas the law of conservation of linear momentum is part of the course and is covered inthis chapter, the law of conservation of angular momentum is not required knowledge for this courseand will not be examined.

Conservation of angular momentumFor any rotating object, it proves useful to define angular momentum as the product of its rate ofrotation and its mass. For an isolated, rotating body angular momentum is conserved (remains thesame).

A figure skater makes use of this: by pulling in her arms and legs to reduce the distance, she makesher spinning action faster; spreading them to increase distance, her spinning slows down. The effectis also used in diving and gymnastics.

Why should this happen? To make an object spin, or to change its rate of rotation, a force must beapplied at some perpendicular distance from its centre (called a ‘torque’). A force through its centremay move the whole object, without changing its rotation state. People who play racquet sports suchas table-tennis understand this and use it to give top-spin, under-spin or no spin to the ball.


F

r = 0

no spin

Fr

Fr

For a point mass moving in a circle angular momentum = mass velocity distance to the turningpoint.

Then the rotational version of

force = rate of change of momentum

is

torque = rate of change of angular momentum.

No torque, no change in angular momentum.

Kepler’s second law

Sun

orbit

r

ms

area A = 1 sr2

v = st

The gravitational force keeping a planet in an elliptical orbit acts along a line between the centre ofthe planet and the centre of the Sun. There is therefore no torque on the planet, and its angularmomentum about the Sun stays constant.

As a planet of mass m moves nearer the Sun, r decreases so v must increase in the same proportion.In a short time t, the planet moves along a tangent s with velocity s / t.

The angular momentum is:

.2t

Am

t

srmr

t

smmvr

But r s is base times perpendicular height for the narrow triangle swept out by the planet in this time,or twice the area swept out. Because its angular momentum remains unchanged, the rate at whicharea is swept out will be constant.


Training for movement in spaceReading 50T: Text to Read


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Here you learn about how performers train for relative motions. Changing movement through spaceand time requires high accelerations and this in turn calls for specially trained performers.

Astronauts…Free fall, and the sensation of weightlessness it produces, are unusual experiences for humans. Towork effectively on a platform which is in orbit, and therefore in free fall, astronauts must becomeaccustomed to the sensation.

Astronaut free fall trainingspeed v

r

path radius r

acceleration

= v2

r

pilot flies a parabolic loopchoosing speed v andradius r to simulate free fall

To simulate free fall:downward acceleration = acceleration of free fall

= gv2

r

Example calculationg = 10 m s–2 approx

Suppose thatr = 1 km = 1000 mv2 = gr = 10 000 m2 s–2

Thus:v = 100 m s–1

v = 360 km hr–1

Michael Collins One of the Apollo 11 astronauts says:

The zero-g airplane was essentially a converted Boeing 707 which had all the seats removedfrom the cabin and the interior padded. It could imitate weightlessness for slightly over 20seconds at a crack by flying parabolic arcs. The pilots could push the airplane over the top atexactly the right rate to keep our bodies suspended between ceiling and floor in the aft cabin. At


least that was the theory. In practice, even the smoothest pilot wiggled a little bit, and we tendedto bang around in the back.

We were there to work. We flew so many parabolas! There would be 40 or 50 on a training flight.People not accustomed to it would get sick and throw up. I learned to hate the zero-g airplane.

During launch and descent stages, astronauts are subjected to large accelerations, or ‘g-forces’. Theyneeded to train for this too. Michael Collins continues:

If the zero-g airplane performs its task only briefly and with doubtful accuracy, then the high-gwheel is exactly the opposite – it does its specialised chore with deadly accuracy for as long as itspassengers can stand it. It swings an imitation cockpit around in a circle on the end of a fifty-footarm. As the arm spins faster and faster, his seat pushes into the occupant harder and harder toprovide the centripetal force. Routinely 7gs were reached during the Apollo return and it couldeasily soar to 10 or 15. So 15g was our limit on the wheel and it was no fun at all.

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At anything over 8g, I start feeling very uncomfortable, with difficulty in breathing and a paincentered below my breastbone. By 10g the pain has increased somewhat and breathing is nearlyimpossible; in fact an entirely different breathing technique is needed at high gs. The vision startsto deteriorate too, and darkness closes in toward the centre.

For astronauts, the seat is positioned so that accelerations are ‘eyeballs in’. This way they can enduremuch higher g-forces than fighter pilots without losing vision or consciousness.

It has been suggested that some people might want to pay for a short space flight as a new kind ofholiday experience. Would you?

…and a dancer

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Kitsou Dubois is a French dancer. Movement in space and time is how dancers think about dance,with music usually setting the time. Dubois was intrigued to explore the different sense of inner spaceand external space during weightlessness, exactly what disorients some people and makes them feelsick. Working with the French National Space Studies Centre, she joined in parabolic training flights


during September 1991. She enjoyed the challenge of adapting and learning to dance in the 20second periods of weightlessness. The experience inspired her to new choreographic forms on thetheme of weightlessness.

Forces on real objectsReading 60T: Text to Read


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By comparison with biology or chemistry, in physics we usually imagine the real and complex worldas made up of far simpler things which we can understand. When it comes to objects such as a rock,a sculpture, or even a galaxy we prefer to treat them all like tiny billiard balls or point particles, withtheir total mass all concentrated at one point.

Centre of massAny real object has size, so its mass is distributed over some volume: a mountain lake, an officebuilding, a set of keys. It is true of matter in all its states. But if you had to calculate what happens toany object when an external force acts by doing calculations for each of its separate bits of mass,problems would quickly become unsolvable. Fortunately, you can always find a fixed point at whichthe whole force seems to act, called the ‘centre of mass’. In the simplest cases, symmetry identifiesthis point.

sphere + midpoint rugby ball + midpoint

cylinder pair + midpoint doughnut ring + midpoint

ladder + midpoint

the centre of mass for some symmetric objects


If the shape is irregular or the density not uniform, it is harder to find the centre of mass bycalculation. Sometimes in these cases it can be found by experiment.

If a net external force acts in a direction passing through the centre of mass, the body will simplymove in a straight line. If this force does not pass through the centre of mass, the body will rotate aswell.

Centre of gravitySimilarly, the single point from which the weights of all the separate parts of an extended body appearto act is called the ‘centre of gravity’. In a uniform gravitational field, the centre of gravity is the sameas the centre of mass.

balance point

For small objects, you can use the balance point to identify the centre of gravity. But if the field isnon-uniform, the centre of gravity of a long or large object and its centre of mass will not be the same.

Gravity can pull things apartReading 70T: Text to Read


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In 1993 Comet Shoemaker-Levy came close to the massive planet Jupiter. Jupiter’s gravitational fieldtore it into 21 pieces. How? Because of the inverse square law, the side of the comet nearest toJupiter is pulled harder towards Jupiter than is the further side. Result: the comet is stretched alongthe line from it to Jupiter. It broke up under the stress and months later, one after the other, a line ofits fragments crashed into Jupiter at 216 000 km per hour. The image above shows the pieces asthey approach Jupiter.

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Looking at an image sent back from the Voyager 1 encounter with Jupiter, NASA scientist LindaMarabito discovered volcanoes on its moon Io (above). Io stretches and squashes because ofJupiter’s tidal forces, and this has heated up its interior to melting point. You can process this imagejust as she did (see chapter 1 of the Advancing Physics AS CD-ROM)

The difference in the pull of the Moon on the Earth across the diameter of the Earth is about 3%. Thatis enough to stretch the ocean water into a bulge at both ends. Result: twice-daily ocean tides, firstexplained in this way by Newton.

Starting from momentumReading 80T: Text to Read


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Most physics books tell you about Newton’s law F = m a, and about his third law that forces come inpairs of equal and opposite ‘action’ and ‘reaction’, and then show how these lead to the conservationof momentum. A more modern point of view is that momentum is the bedrock idea, and that theothers follow from it, rather than the usual way around.

This reading summarises how the modern argument goes.

A picture of the argumentHere is a picture showing the steps in the thinking. Read it from the top downwards. See how itarrives at F = m a at the end.


Thinking about momentum and forces

Principle 1 symmetry

+v –v

identical objects

result predictable from symmetry

Principle 2 invariance

+v –v

seen differently is the same as:

+2v

Conservation of momentum

–p +p

mass M

–p

change of momentum = –F t

force –F

acts fortime t

mass m

+p

change of momentum = F t

force +Facts fortime t

Split ‘crunch’ intoforces F on each

force F = pt

same time t.forces equaland opposite

if define force F = pt

then F = mv

= mat

thus F = ma

From symmetry and invariance (looking differently can’t change events):

1. momentum is conserved2. define mass from change of velocity in collision3. define force as rate of change of momentum, giving F = ma4. forces on interacting objects act in equal and opposite pairs

crunch

Starting from symmetryThe starting point is that the result of simple collisions can be predicted without knowing anythingmuch at all. If two objects come together at the same speed in front of you, and stick together, there is


no way the combined object can go off to the left or right.

Just moving makes no differenceThe second starting point is that observing a collision from a differently moving point of view can’tchange the actual collision. But it can change how it looks. An object which was previously seenmoving will be seen standing still, if you are travelling along with it. Result: predictions for somesimple collisions can be automatically extended to others.

Also, if just moving while observing a collision changes nothing essential, it must be that the rules forhow collisions work can depend only on changes of the velocity of objects, not on their velocities.Changes of velocity of an object don’t alter when you add or subtract an observer’s velocity from theobject’s velocity before and after the collision.

Changes of velocityThe argument so far is that the outcome of a collision must be described in terms of changes of thevelocity of the colliding objects. But this isn’t enough. Big hefty objects don’t change velocity muchwhen they collide with small light ones, which do then change velocity a lot.

Introducing massWe can define the idea of mass by the ratio of the changes of velocity of two objects in a collision.That is:

.1

2

2

1

v

v

m

m

The more massive of the two has the smaller velocity change. Knowing only ratios in this way, astandard unit is needed. It is the standard kilogram, a lump of metal kept in Paris.

Introducing momentumSymmetry says that two identical objects must have equal and opposite changes of velocity in acollision. When the masses are unequal, symmetry now says that the changes in the quantity m vmust be equal and opposite. In a collision there is no overall change in the quantity m v, calledmomentum. Give momentum the symbol p.

Inside a collisionInside a collision, the two objects deform one another. Forces act on each. To think about this, splitthe collision into two parts, one for each object, and for each part forget about the other object. Justrepresent the effect of the other object as a force acting on the object you are concerned with.

Force is rate of change of momentumThe effect of the collision is to change the momentum p of an object, over the time d t of the collision.Define the force F which represents the action of the other object as

.d

d

t

pF

This can be written

t

mvF

d

)(d

which if the mass is constant can be expressed as


.d

dma

t

vmF

Thus we reach Newton’s second law.

Equal and opposite forcesWe now see why ‘action’ and ‘reaction’ are equal and opposite. Two forces, one on each object,changing the momentum of both by equal but opposite amounts p, represent one single collision. Itmust last the same time for both objects so the pairs of forces introduced to split the collision must beequal and opposite.

Look for the other objectNewton’s law F = m a splits one interaction into two pieces. Talk of a force on an object should nowmake you ask where the other interacting object is. Sometimes it’s obvious – a fist for example.Sometimes it’s obvious when you think about it – the Earth pulling things down for example. Andsometimes it isn’t obvious at all. The discovery of outer planets by their effects on inner ones is oneexample. The inference that there may be a black hole at the centre of our galaxy is another.

Mechanics in modern physics‘Forces’ play little role in relativistic physics or in quantum mechanics. This is because interactionsare not regularly ‘taken apart’ or ‘split’. Instead, general principles such as the conservation ofmomentum are applied. It turns out that momentum, a bit modified by relativistic ideas, is a centralconcept in modern physics, while force is less so.

Watertight?The argument above seems difficult to fault. It relies in the end only on symmetry arguments and onGalilean invariance: that is, that uniform velocities of observers can’t change events. It is in the styleof modern arguments about symmetry. But interestingly, it is not in fact quite correct. The seeminglyinnocent assumption built in that velocities add in a simple way is not right. It fails to allow for therebeing a maximum possible relative velocity, the speed of light. But the concept of momentum,modified a bit, survives the changes.

‘Sling-shotting’ spacecraft or ‘gravity assist’Reading 90T: Text to Read


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To reach the distant outer planets, a spacecraft must climb steeply through the Sun’s gravitationalfield. Because there is a limit to how much fuel a rocket can carry, rockets powerful enough to launcha space probe to the outer planets are impossible.

Astronomers had long known that comets’ orbits were altered by encounters with planets. MichaelMinovitch, a UCLA graduate student working at the Jet Propulsion Laboratory in the 1960s, was thefirst to recognize that this same principle could be applied to spacecraft trajectories. A spacecraft cangain kinetic energy by interacting with one or more planets.


Something for nothing?The interactions between comets and planets, or spacecraft and planets, are elastic collisions. Thecomet or spacecraft falls into the gravitational field of the planet and then climbs out again. Twoconservation laws apply:

momentum is conserved in all collisions

kinetic energy is conserved in elastic collisions

How can the spacecraft gain kinetic energy by falling and then climbing back out? The first importantthing to recognise is that the planet is in motion.

First an experiment:

Take a strong bar magnet and two steel ball-bearings. One ball-bearing should be large and the otherrelatively small. Place the balls on a smooth, level surface so that they touch. Slowly bring the magnettowards the larger ball. Both balls will move towards the magnet and hit it. The larger ball will stick tothe magnet, but the smaller one will bounce off at high speed.

Here is a second experiment showing the same kind of interaction:

Take two superballs, one big and one small. Hold them in front of you with the smaller ball on top ofthe larger one. Release them so they fall to the ground. Be warned: the smaller ball may rebound to aheight many times where it started from – or hit the ceiling if you are doing this indoors! (This can alsobe done using a table tennis ball as the ‘small ball’ and a hard rubber ball as the large one.)

What both of these experiments have in common with a ‘gravity assist’ is that they too are elasticcollisions. The second important thing to recognise is that the two masses interacting are unequal.The mass of a planet is absolutely enormous in comparison with a spacecraft.

A qualitative explanationAssume the spacecraft and planet are approaching, each with a speed u. Imagine you are sitting onthe planet: the spacecraft appears to approach at speed 2 u. Physics tells you that, because it is anelastic collision, it will leave with the same speed 2 u. Yet to a distant stationary observer, the speedof the planet is always u, and the speed of the spacecraft after collision is u + 2 u = 3 u !

Something similar happens when you hit a ball with a bat – the ball leaves with much greater speedthan the bat has.

A visual explanation


path of spacecraft

planet v

u

v

u+u v

Imagine you see the collision from the planet. The spacecraft approaches and leaves in similar ways,falling and then climbing back out of the planet’s gravity field. Yet the planet too is moving. Anobserver in deep space would persuade you that the motion of the planet itself must be added to thespeeds you saw – a vector addition. In other words, the spacecraft leaves with greater speed than itarrived.

An explanation using algebraAssume the spacecraft and planet are approaching, each with a speed u. Call the large mass M andthe small mass m. After the collision M has speed V and m has speed v.

Start with equations stating the conservation laws.)1(mvMVmuMu

)2(2222 mvMVmuMu

It helps to re-arrange these equations by introducing = m / M:

)3()1( vVu

)4()1( 222 vVu

What we are aiming to do is to find out how V and v depend on the ratio . We eliminate V from thesetwo equations. Equation (3) gives:

)5(.)1( vuV

Equation (4) gives:

)6(.)1( 222 vuV


Substituting (5) into (6):

)7(.)1(])1([ 222 vuvu

Equation (7) can be expanded to:

.)1(2)21( 2222222 vuuvuu

If is small, we can ignore terms in 2.

.22 22222 vuuvuuu

So

222 22 vuvuu

and

.032 22 uvuv

The two solutions to this equation are

uv 3

and

.uv

The first one says the small ball can emerge with three times the speed (nine times the kineticenergy) of approach. Where does this energy come from? It is taken from the larger mass. Thediagram shows the movement of a spacecraft relative to planet. The extra kinetic energy carried awayby the spacecraft is too small to make a difference to the planet.

The second solution applies when the spacecraft approaches from the opposite direction, and giveskinetic energy to the planet.

Supernovae and black holesReading 100T: Text to Read


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Gravity is the weakest of the forces in Nature. Yet when masses get big, gravity takes over. Twodramatic astronomical phenomena show how important it can be.

Supernovae‘Nova’, meaning a ‘new star’, is a term used by astronomers when they witness a star as it brightensand becomes visible for the first time. Supernovae are almost the biggest bursts of energyastronomers ever see. An energy of about 1046 J is released in just a few seconds, an amount whichis hard to comprehend. The rate at which energy is released matches that produced by a whole


galaxy. Supernovae occur suddenly and without warning; however, unless they are fairly near, theyare discovered only by chance.

In an average galaxy like our Milky Way, a supernova occurs at random every 30 years or so. Modernastronomers have seen only one local supernova, in 1987. The last supernova in our galaxy to beseen by the unaided eye was observed by Kepler, almost 400 years ago. Astronomers are stillexcited because Supernova 1987A provided an opportunity to test theories of how stars work againstreal data – and remarkably, theory and evidence agreed.

So how do astronomers explain supernovae? Stars are truly massive. For example, with a mass of2 1030 kg the Sun represents 99.8% of the mass of our solar system. Yet, on a universal scale, theSun is not a large star. Gravity tries to compress stars, but the chain of fusion reactions which drivethem produces sufficient radiation pressure to prevent their collapse – until the nuclear fuel runs out.When this happens a plasma of radius perhaps 6000 km will virtually free-fall until its radius is just10 km. This can happen in less than a second. You can calculate the enormous amount ofgravitational potential energy released.

When the star material hits ‘the bottom’, a core of hot plasma of iron nuclei, the fall is stoppedsuddenly. Inner parts bounce upwards, colliding with material still falling. Nuclei larger than iron areproduced by fusion, creating huge numbers of neutrinos, and the outer shell is blown away. Thesupernova remnant continues cooling years after the peak in visible light produced has passed.

Black holesThink of how Newton’s gravitational law works. The field strength at the surface of a body intensifiesas the mass is increased, or its size decreased. A useful way of characterizing the strength of gravityis to describe the speed needed for escape. To get away from the Earth, a rocket needs to leave thesurface travelling at 11 km s–1 (40 000 km h–1). For the Sun the escape speed is nearly 60 times

greater.

This tendency for the escape speed to become ever larger for more massive, or more compact,bodies raises a question. The largest speed possible is the speed of light. Escaping the Sun requiresa speed only 1/500th the speed of light. What would be the effect of gravity around a body for whichthe escape speed was as high as the speed of light?

The Rev. John Michell puzzled over this question in 1784. Arguing on the basis of Newton’s laws andassuming that light behave like particles of matter, Michell wrote a paper in which he noted

If the semi-diameter of a sphere of the same density as the Sun were to exceed that of the Sun inthe proportion of five hundred to one, and supposing light to be attracted by the same force inproportion to its inertial mass with other bodies, all light would be made to return towards it, by itsown proper gravity.

The French astronomer and mathematician Laplace arrived at a similar conclusion a decade later: themost massive objects in the Universe might give out no light, but still fiercely attract any nearbymaterial.

Almost 200 years later astronomers discovered two kinds of objects matching the speculations ofMichell and Laplace: neutron stars and black holes. A new theory of gravity, invented by AlbertEinstein early in the twentieth century, is used to explain them. Einstein’s general theory of relativitysays that energy gravitates (rest energy, or mass, included). But a gravitational field has energy andso itself gravitates. If the field is large enough, the field itself pulls itself inwards, in a run-awayprocess. A black hole results, around a point in space–time at which the field strength goes to infinity.


The Cassini–Huygens mission to SaturnReading 130T: Text to Read


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This article illustrates why the study of gravity and orbits is vital for the exploration of space. It gives asummary of the route and scientific objectives of the mission to Saturn.

The Cassini–Huygens missionLaunched in October 1997 at a cost of $3.2 billion, the Cassini–Huygens mission was one of thelargest and most complex space exploration missions then in progress. And also one of the mostcontroversial. As well as the huge cost of the project, the inclusion of 32 kg of plutonium as a powersource led NASA to be accused of recklessness and irresponsibility. An accident at launch couldhave showered a large area with radioactive material. Fortunately, the two critical phases of themission – the launch and the near-Earth flyby – passed without incident.

The mission, a collaboration between NASA and the European Space Agency (ESA), was designedto explore the Saturn system and all its elements – the planet Saturn and its atmosphere, its rings, itsmagnetic field, and many of the icy satellites. The mission paid special attention to Saturn’s largestmoon, Titan, the target of the Huygens probe.

The spacecraft was injected into a 6.7 year Venus–Venus–Earth–Jupiter gravity assist (VVEJGA)trajectory to Saturn. Included were gravity assists from Venus (April 1998 and June 1999), Earth(August 1999) and Jupiter (December 2000). Arrival at Saturn was in July 2004. (A gravity assist isthe use of a planet’s gravitational field to change the speed and direction of a satellite.)

VenusswingbyApril 21,

1998

Earth orbit

EarthswingbyAugust

17, 1999

Launch from EarthOctober 6, 1997

Saturn arrivalJuly 1, 2004

Jupiter swingbyDecember 30, 2000

Deep spacemanoeuvreJanuary 20,

1999

VenusswingbyJune 22,

1999

The Cassini orbiter made repeated close flybys of Titan, both for data about Titan and forgravity-assisted orbit changes. This ability to change orbits enabled close flybys of icy satellites,reconnaissance of the magnetosphere over a variety of locations and the observation of the rings and


Saturn from a variety of viewpoints.

The Huygens probeThe Huygens probe was released from the Cassini orbiter, to descend into Titan’s atmosphere.

Probe releaseNovember 6, 2004Titan probe entry

and orbiter flybyNovember 27, 2004

Saturn orbitinsertion

July 1, 2004

The Huygens probe had the task of entering Titan’s atmosphere, making in situ measurements of thesatellite’s properties during descent by parachute to the surface. The probe consisted of a descentmodule enclosed by a thermal-protection shell. The front shield of this shell is a very bluntly shapedconical capsule 2.7 m in diameter to provide a lot of drag as the probe descends towards Titan. Theshield is covered with a special thermal-ablation material to protect the probe from the enormous heatgenerated during atmospheric entry. On the rear side is a protective cover that is primarily designedto reflect away the heat radiated from the hot wake of the probe as it decelerates in Titan’s upperatmosphere. Atmospheric entry is a tricky affair – entry at too shallow an angle can cause the probeto skip out of the atmosphere and be lost. If the entry is too steep, the probe measurements wouldhave begun at a lower altitude than is desired.


1000

500

300

192

170

0

0 2.5Time / hours after entry

instruments’inletport opens

main chute deploys

drogue chute deploys

deceleratorjettisons

Once the probe had decelerated to about 1000 mph, the rear cover was pulled off by a pilotparachute. An 8.3 m diameter main parachute was then deployed to ensure a slow and stabledescent. The main parachute slowed the probe and allowed the decelerator and heat shield to fallaway, exposing the instruments to the – 180 C temperatures.

Science objectivesThe list of scientific objectives for Cassini–Huygens is extensive. There are specific objectives foreach of the types of bodies in the system – the planet itself, the rings, Titan, icy satellites and themagnetosphere.

The most critical phase of the mission following launch was the Saturn orbit insertion (SOI) phase,when the satellite was grabbed by Saturn’s gravity and pulled into orbit around the planet. Not onlywas it be a crucial manoeuvre, but it was also be a period of unique scientific activity, because at thattime the spacecraft was the closest it would ever be to the planet. The SOI phase of the trajectoryalso provided a unique opportunity for observing the rings.

When the Huygens probe reached Titan in 2004, the close-up work really started. Throughout thedescent of the Huygens probe towards Titan, the Huygens atmospheric structure instrument (HASI)measured more than half a dozen physical properties of the methane and nitrogen atmosphere, suchas temperature and pressure. The gas chromatograph and mass spectrometer (GCMS) determinedthe chemical composition of the atmosphere and the aerosol collector and pyrolyser (ACP) capturedaerosol particles, heated them and sent the effused gas to the GCMS for analysis.

The propagation of light in the atmosphere was measured in all directions by the descent imager andspectral radiometer (DISR). The DISR also imaged the cloud formations and the surface.

It was not even known whether the surface of Titan was solid or liquid. However, in the proximity of


the surface, the surface science package (SSP) activated a number of its devices to makemeasurements near and on the surface. It turned out that the touchdown area was solid. Theon-board camera took photographs of the local scenery, including a number of loose rocks.

Huygens was set to relay its information to Cassini for about half an hour after touchdown, to bere-sent to Earth for analysis. The Huygens transmitter had the same power as a mobile phone, socould not be read from Earth (though, amazingly, its carrier signal was detected). Unhappily, only oneof the two communication channels to Cassini worked, and half the data was lost.

Even so, the whole Cassini mission was one of the most exciting and informative scientific projects ofthe new millennium.

Gravitational potential due to a spherical massReading 110W: Well Actually/But Also


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We can find the change in gravitational potential by using calculus.

M

rP

A B

xx

The diagram shows a spherical mass M. To move a unit mass from A to B, a small distance x, aforce must be applied to the right.

Force applied to the unit mass = G M / x2 to the right.

The work done to move the unit mass from A to B will be:

.2

xx

GMW

The total work done in moving the unit mass from some point P to infinity will be:


.d2 r

GM

r

GMGM

x

GMx

x

GM

rr

Gravitational potential at infinity is therefore higher than at C. Since the definition implies potential atinfinity is zero, the potential at any point P will be given by:

.r

GMVg

If the sphere has a radius R, the gravitational potential at its surface will be:

.R

GM

Revision Checklist I can show my understanding of effects, ideas and relationships bydescribing and explaining cases involving:momentum as the product of mass × velocity

force defined as rate of change of momentum

conservation of momentum when objects interact; Newton's third law as aconsequence

A–Z references: momentum, Newton's Laws of motion

Summary diagrams: Conservation of momentum, Two craft collide, Morecollisions, Momentum, invariance, symmetry, Jets and rockets

work done (as force × distance moved in the direction of the force: includingcases where the force does not act in the direction of the resulting motion)

changes of gravitational potential energy to kinetic energy and vice versa whenobjects move in a gravitational field

motion in a uniform gravitational field

A–Z references: work, gravitational field, potential energy, kinetic energy,projectile

Summary diagrams: Graph showing g against h

the gravitational field and gravitational potential due to a point mass

A–Z references: gravitational field, gravitational potential


Summary diagrams: Field and potential

motion in a horizontal circle and in a circular gravitational orbit about a centralmass

A–Z references: circular motion, Kepler's laws

Summary diagrams: Centripetal acceleration, Geometry rules the Universe, Ageostationary satellite

I can use the following words and phrases accurately whendescribing effects and observations:force, momentum

A–Z references: momentum

kinetic energy, potential energy

A–Z references: kinetic energy, potential energy

gravitational field, gravitational potential, equipotential surface


Summary diagrams: Relationship between g and Vg

I can sketch, plot and interpret:graphs showing the variation of a gravitational field with distance, and know thatthe area under the graph shows the change in gravitational potential

A–Z references: gravitational field

Summary diagrams: Field and potential, Graph showing g against h

graphs showing the variation of gravitational potential with distance, and knowthat the tangent to the curve gives the gravitational field strength

A–Z references: gravitational potential

Summary diagrams: Field and potential

diagrams illustrating gravitational fields and the corresponding equipotentialsurfaces

Summary diagrams: Relationship between g and Vg, Field and potential

I can make calculations and estimates involving:kinetic energy ½ mv2, gravitational potential energy change mgh

energy transfers and exchanges using the idea: work done E = Fs cos(no


work is done when F and s are perpendicular)

A–Z references: kinetic energy, potential energy, gravitational potential,projectile

momentum p =mv and F = (mv)/t

A–Z references: momentum, Newton's laws of motion

circular and orbital motion: a = v2 / r ; F = mv2 / r

A–Z references: satellite motion, Kepler's laws

Summary diagrams: Centripetal acceleration, Speeds and accelerations in theSolar System, Acceleration of the Moon, A geostationary satellite

gravitational fields: for the radial components

2gravr

GmMF

, m

Fg grav

2r

GM

gravitational potential energy r

GmM

gravitational potential r

GMm

EV grav

grav


Summary diagrams: Field and potential, Apollo returns from the Moon

Testing for an inverse-square law - as-a2-physics ... 11.pdf...1 Advancing Physics Testing for an...

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