Assignment 1 (Due Sept 29) SOLN

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Astronomy 101 Instructor: Jaymie Matthews ASSIGNMENT 1 SOLUTIONS It was due: Wednesday, 29 September 2010 by 5 pm in your lab section slot beside Hennings 312 Total = 100 marks Late Penalty: 15% per day Format: Pages stapled together. Upper right corner of the first page should contain: Last and first name, and Student # A sense of scale In my lecture on 10 Sept., I talked about making a map which included the Sun, the Earth, and the next closest star to us after the Sun, proxima Centauri (which is close to the brighter star, alpha Centauri). Here are questions adapted from your textbook which are intended to help you appreciate these interstellar distances on a Galactic scale, and in terms of highway driving. You will need only to look up values in your textbook (or another reliable source) and to do some simple arithmetic to solve these problems. 1. Imagine shrinking the Milky Way Galaxy by a factor of 10 19 (10 million trillion) times, so that it would fit onto a football field. (a) On this scale, what would be the distance between the Sun and proxima Centauri? (b) What would be the diameter of the Sun on this scale? Compare the Sun‟s size on this scale t o the actual size of a typical atom (about 10 -10 metre in diameter). [ 10 marks ] (a) The distance between the Sun and proxima Centauri is approximately 4.24 light years, where 1 light year = 9.461 × 10 12 km, so the distance to the star is about 4.0 × 10 13 km. If you shrink this distance by a factor of 10 19 , then it becomes about 4.0 × 10 13 km / 10 19 = 4.0 × 10 6 km = 4 millimetres . (b) The diameter of the Sun is approximately 1.4 million km = 1.4 × 10 6 km. If you shrink this distance by a factor of 10 19 , then it becomes about 1.4 × 10 6 km / 10 19 = 1.4 × 10 13 km = 1.4 × 10 10 metre . This is comparable to the size of an atom. The values need only be approximate because I only want you to obtain a sense of the scale of the Galaxy and the relative size of the Sun in this problem. 5 marks marks 5 marks

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Astronomy Answers

Transcript of Assignment 1 (Due Sept 29) SOLN

Astronomy 101 Instructor: Jaymie Matthews

ASSIGNMENT 1 SOLUTIONS

It was due: Wednesday, 29 September 2010 by 5 pm in your lab section slot beside Hennings 312 Total = 100 marks

Late Penalty: –15% per day Format: Pages stapled together. Upper right corner of the first page should contain: Last and first name, and Student #

A sense of scale

In my lecture on 10 Sept., I talked about making a map which included the Sun, the Earth, and the next closest star to us after the Sun, proxima Centauri (which is close to the brighter star, alpha Centauri). Here are questions adapted from your textbook which are intended to help you appreciate these interstellar distances on a Galactic scale, and in terms of highway driving. You will need only to look up values in your textbook (or another reliable source) and to do some simple arithmetic to solve these problems.

1. Imagine shrinking the Milky Way Galaxy by a factor of 1019 (10 million trillion) times, so that it would fit onto a football field. (a) On this scale, what would be the distance between the Sun and proxima Centauri? (b) What would be the diameter of the Sun on this scale? Compare the Sun‟s size on this scale to the actual size of a typical atom (about 10-10 metre in diameter). [ 10 marks ]

(a) The distance between the Sun and proxima Centauri is approximately 4.24 light years, where 1 light year = 9.461 × 1012 km, so the distance to the star is about 4.0 × 1013 km. If you shrink this distance by a factor of 1019, then it becomes about 4.0 × 1013 km / 1019 = 4.0 × 10

–6 km = 4 millimetres.

(b) The diameter of the Sun is approximately 1.4 million km = 1.4 × 10 6 km.

If you shrink this distance by a factor of 1019, then it becomes about 1.4 × 106 km / 1019 = 1.4 × 10

–13 km = 1.4 × 10 –10 metre. This is comparable to the size of an atom. The values need only be approximate because I only want you to obtain a sense of the scale of the Galaxy and the relative size of the Sun in this problem.

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2. Imagine that you could drive a car at a constant speed of 100 kph (km/hr) even across oceans and through space. How long would it take to drive (a) Once around the Earth‟s equator? (b) From the Earth to the Sun? (c) From the Sun to Neptune? (d) From the Earth to proxima Centauri? (Give your answers in the most „natural‟ units to describe the trip. For example, I wouldn‟t tell somebody the driving time from Vancouver to Kelowna in seconds, or years, or centuries, but in hours. In other words, don‟t give all your answers in hours or seconds.) [ 10 marks ]

(a) The circumference of the Earth on the equator is 40,075.0 km (although I will accept 40,000 km or values close to that). At 100 kph, it would take 40,000 km / 100 km/hr = 400 hr (about 17 days) to circumnavigate the globe.

(b) The average distance between the Earth and the Sun is the value of the Astronomical Unit: about 150 million km = 1.5 × 10

8 km. At 100 kph, it would take 150,000,000 km / 100 km/hr = 1,500,000 hr or about 170 years to reach the Sun from the Earth. (c) The average distance between Neptune and the Sun is 30.06 Astronomical Units ~ 30.6 × 1.5 × 10

8 km ~ 4.5 × 10 9

km. At 100 kph, it would take ~ 45 million hours or a little over 5000 years.

(d) The distance between the Sun and proxima Centauri is approximately 4.24 light years, where 1 light year = 9.461 × 1012 km, so the distance to the star is about 4.0 × 1013 km. At 100 kph, it would take 4.0 × 1013 km / 100 km/hr = 4.0 × 1011 hr or about 46 million years to reach that star from the Solar System.

What’s your angle?

Angular size, physical size, and distance are all related in a simple way. Astronomers use this relationship to help determine the scales of objects in the Solar System and beyond. Measurements of angle can be made accurately with sophisticated instruments or approximately with your hand and your eye. On the next page are three questions on the theme of angular measurement in astronomy and in life: two adapted from your textbook and one Matthews original creation. In these questions, you‟ll need to look up information in the textbook, and in the third of these questions, you‟ll need to find a key piece of information from another sources. 3. Measure the width of your index fingernail and the length of your arm. Based on your measurements, estimate the angular width of your fingernail at arm‟s length. Express your answer in units of degrees.

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Does your result agree with the approximation shown in Figure 2.7c of the textbook (5th Edition), which is also shown in the notes from the lecture on 16 September 2010? Comment on the agreement or any difference you find. [ 10 marks ]

The width of my index finger near the tip is about 1.2 cm. At arm’s length, my index finger on my right arm is about 60 cm from my right eye.

θ = s / d

where s = 1.2 cm and d = 60 cm

θ ~ 1.2 / 60 ~ 0.020 radian 2π rad = 360º so 1 rad = 57.3º

So θ ~ 1.15º

Not bad agreement with what the text describes. My index fingernail would have an angular width of about a degree at arm’s length. You should discuss the possible sources of error in your measurements and variations from person to person.

You’ll get full marks for having made reasonable measurements and doing the calculation correctly. 4. The bright star Betelgeuse (pronounced “Beetlejuice”, as in the movie) can be seen in the constellation Orion the Hunter. Data collected with the Hubble Space Telescope led to an estimate of its angular diameter: 0.044 second of arc (arcsec). The distance to Betelgeuse is estimated to be about 430 light years. What is the diameter of Betelgeuse (in km)? Compare the size of this star to the size of the Sun and to the size of the Earth‟s orbit. [ 10 marks ]

θ = s / d

where θ = angular diameter of Betelgeuse, s = diameter of Betelgeuse and d = distance of Betelgeuse

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θ ~ 0.044 arcsec = 0.044 / 3600 degree ~ 1.2 × 10 -5 degree ~ 2.1 × 10

-7 rad d = 430 light years ~ 430 × ( 9.5 × 10

12 km ) ~ 4.1 × 10 15 km

So s = θ d ~ ( 2.1 × 10 -7

rad )( 4.1 × 10 15

km ) ~ 8.6 × 10

8 km

The radius of the Sun is about 700,000 km (7 × 10

5 km) so the diameter of the Sun is about 1.4 × 10

6 km. Therefore, Betelgeuse is about 610 times the diameter of the Sun. The average distance between the Earth and the Sun (1 Astronomical Unit = 1 AU) is about 150 million km (1.5 × 10

8 km). The diameter of Betelgeuse is almost 3 times the diameter of the Earth’s (nearly circular) orbit.

5. This photograph was taken when the (nearly) Full Moon was setting behind the Vancouver city skyline.

(a) What was the altitude of the Moon (in degrees) when this picture was taken? (Measure the altitude from the centre of the Moon‟s disk.) [ 5 marks ]

(b) Using the angular size of the Moon and the angular height of the Harbour Centre Tower (seen at the far right) in this image, how far away was the photographer from the Tower (in metres or km) when she took this picture? [ 10 marks ]

NOTE: Your assignment must include printed copies of the above image which show clearly what you measured on it to obtain your answers to both Questions 5(a) and 5(b).

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(a) What was the altitude of the Moon (in degrees) when this picture was taken? (Measure the altitude from the centre of the Moon‟s disk.) [ 5 marks ]

The Moon provides the angular scale of the photo. The angular diameter of the Moon as seen from Earth is about 0.5°. Altitude is angle above the horizon, shown as a yellow line for the centre of the disk of the Moon. It’s about 7⅓ Moon diameters above the waterline (the horizon) so the Moon’s altitude is ~3.5°.

(b) Using the angular size of the Moon and the angular height of the Harbour Centre Tower (seen at the far right) in this image, how far away was the photographer from the Tower (in metres or km) when she took this picture? [ 10 marks ]

Again the Moon provides the angular scale. The height of the Harbour Centre Tower in the photo is about 5½ times the diameter of the Moon, so about 2¾°.

We use the equation relating angular size, physical size and distance:

θ = s / d where θ = angular height of the tower (in radians), s = height of

the tower (in metres) and d = distance to the tower (in metres)

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You must convert the angle from degrees into radians:

θ = 2.75° ≈ 0.05 rad

θ = s / d so d = s / θ

You need to know s, the height of the tower. You can find that on-line; for example: http://en.wikipedia.org/wiki/Harbour_Centre The height to the pylon at the tower’s top is s = 177 metres.

Therefore, d = s / θ = (177 m) / 0.05 ≈ 3500 m or about 3½ km

Proving Lisa Simpson wrong

As I pointed out in the lecture on Wednesday, 15 Sept., even someone as brilliant as Lisa Simpson was sucked in by the myth that the water in sinks and toilets will drain in opposite senses in the Northern and Southern Hemisphere, and that this is due to the Coriolis Effect.

I‟m going to have you test for yourself whether this is really a myth. Can the Coriolis Effect govern the way water flows in your kitchen or bathroom sink? To answer this question (and in particular, part (b) of Question 6 below), you will have to either look for information from sources completely unrelated to astronomy or physics, or make your own simple measurements using water in a sink in your home or residence room.

6. The Coriolis Effect is real, due to the rotation of the Earth, but the force it seems to exert on freely moving objects is actually fictitious. The deflections are a result of our observing the motions in our rotating reference frame. The Coriolis force FCoriolis (in units of Newtons (N)) on an object of mass m (kg) moving with a velocity v (m/s) in a reference frame rotating with angular velocity ω (the lower-case Greek letter “omega”) is given by:

FCoriolis = – 2 m ( ω × v )

The angular velocity ω = dθ/dt of the rotating reference frame is given in radians per second (where the radian is a dimensionless unit so that in the equation above, F = ma, as expected).

a) What is the angular velocity ω (in rad/s) of the spinning Earth? [ 5 marks ]

The angular velocity of the rotating Earth (in radians/second) is

ωEarth = 2π / Protation

where Protation = rotation period of the Earth in seconds.

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The rotation period of the Earth is approximately 24 hours. The true rotation period of the Earth is 23 hr 56 min 4 sec = 23.9344 hr (a sidereal day). This differs by such a small amount (less than 0.3%) from the solar day of 24 hr that it will not affect the meaning of your result below in part (b) because of the approximations we make for the speed of the draining water. You’ll get full marks for calculating ωEarth with either a solar or a sidereal day.

P rotation = 24 hr = 24 hr × 60 min/hr × 60 sec/min = 86400 seconds = 8.64 × 10

4 sec

or

P rotation = 23.9344 hr = 23.9344 hr × 60 min/hr × 60 sec/min = 86164 seconds ≈ 8.62 × 10

4 sec

ωEarth = 2π / Protation ≈ 2π / (8.6 × 10 4 sec)

≈ 7.3 × 10 -5

rad/sec Now consider water flowing in a typical sink.

b) Estimate the velocity v of water moving horizontally in a sink as it drains. Provide the sources of and/or the reasoning behind the value you have chosen. What is the acceleration a the water will feel, deflecting it to the right (if the sink is in the Northern Hemisphere)? [ 15 marks ]

Depending on whether it is a kitchen sink or a bathroom sink, and on the sink design, the dimensions of sinks vary. But most will fall within a range of widths of, say, about 40 – 70 cm. I’ll take 60 cm for my solution but any reasonable answer (e.g., a measurement of the dimensions of your sink at home on in residence) will be accepted. So the distance from the edge of the sink to the centre (drain) is 30 cm in my example. That’s how far the water must travel in a straight line. The time it takes to flow across that distance may vary. I’ll accept answers from about 1 sec to 5 sec (for a slightly clogged sink) and will use 2 sec in my example here. You must describe how you estimated this time. For example, did you put a small piece of cork or paper in the water and measured how long it took to be carried from the edge of the draining sink to the centre?

Therefore, the velocity of the water is vwater = distance / time = 30 cm / 2 sec = 15 cm/s = 0.15 m/s. (You must use units of m/s)

The Coriolis Force on the water in a sink is given by

FCoriolis = –2 m ( ωEarth × vwater )

where vwater = the horizontal velocity of the water flowing toward the drain.

ωEarth ≈ 7.3 × 10 -5 rad/sec from part (a) and vwater ≈ 0.15 m/s

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FCoriolis = ma = –2 m ( ωEarth × vwater )

The mass m cancels on both sides, so

a = –2 (7.3 × 10 -5 rad/sec × 0.15 m/s ) a ~ –2.2 × 10 -5 m/s2

The water should be deflected to the right by the Coriolis Effect with an acceleration of about 0.000022 metres per second squared.

c) Across the distance between the edge of the sink and the location of the drain, over what distance would the water be deflected to the right at this rate of acceleration? Do you think this is enough to force the water to drain in a counterclockwise sense? Comment on what factors dominate the water flow. [ 15 marks ]

We assume the water started flowing toward the centre of the sink (the drain) so it had no sideways motion to begin with. The distance d (metres) it moves sideways in time t (seconds), starting from rest and accelerating at a constant rate a (m/s2) is

d = ½ a t2

where a ~ –2.2 × 10 -5 m/s2

and t ~ 2 sec (Both values from part (b).)

Therefore, d = ½ ( –2.2 × 10 -5 m/s2 )( 2 s )2 = 4.4 × 10 -5 m = 0.044 mm (Don’t worry about the negative sign. The direction’s not important here.)

The water is deflected to the right by the Coriolis Effect by less than 1/20th of a millimetre! This is not enough to influence the direction of the water flow or make it drain one way or the other.

I’ll accept any reasonable brief discussion of the factors that really influence water flow in a sink (e.g., sink shape, friction, viscosity, the twist of your hand when you pull out the plug, etc.) Now consider air flowing from high to low pressure at a speed of 200 kph that will become part of a hurricane system about 1000 km across.

d) Estimate the acceleration of that air sideways due to the Coriolis force and the distance it will be deflected sideways after travelling forward 500 km. Does it make sense that the Coriolis Effect could create hurricanes and cyclones in our atmosphere? [ 5 marks ]

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This is done the same way as for parts (b) and (c) except you don’t have to estimate the velocity or distance traveled by the air.

a = –2 ( ωEarth × vair )

The unit of air speed must be m/s, not kph, for the Coriolis formula vair = 200 kph = 200 km/hr = (200,000 m) / (3600 sec) ~ 55.6 m/s ωEarth ≈ 7.3 × 10 -5 rad/sec from part (a)

a ~ –2 ( 7.3 × 10 -5 rad/sec )( 55.6 m/s ) ~ –8.1 × 10 -3 m/s2

The air travels forward 500 km at a speed of 200 km/hr so it takes 500/200 hr = 2.5 hr = 9000 sec to cover that distance.

In t = 9000 sec, it will be deflected sideways a distance d = ½ a t

2 = ½ ( –8.1 × 10 -3 m/s2 ) ( 9000 s) 2 ~ 328,000 m ~ 300 km (Don’t worry about the negative sign. The direction’s not important here.)

So the air is shoved sideways 300 km for every 500 km it moves forward. That’s certainly enough to cause an obvious cyclonic circulation, either clockwise or counterclockwise.

e) Find examples of how the Coriolis Effect has been observed on at least two other planets and describe them. (Cite your sources of information.) [ 5 marks ]

The most obvious example is the Great Red Spot on Jupiter – a cyclone (like a hurricane) which has been raging for at least 400 years, since it was first observed on the disk of Jupiter with a telescope. This cyclone is large enough to encompass several Earths.

Another more subtle example is the existence of cyclones on Mars (http://hubblesite.org/newscenter/archive/releases/1999/22/) discovered in images from the Hubble Space Telescope (and hinted at in earlier Viking orbiter images).

Saturn’s atmosphere also shows evidence of cyclonic activity, as does Venus’. In the case of Venus, because it rotates so slowly, the Coriolis effects are weak.

You’ll get 2 marks for each example (and a brief description or explanation) and 1 marks for providing the sources of your information,

TOTAL = 100 marks

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