Apples and Planets - The University of Arizonayelle/Classes/ptys206_2008/lectures/Feb28.pdf · Unit...
Transcript of Apples and Planets - The University of Arizonayelle/Classes/ptys206_2008/lectures/Feb28.pdf · Unit...
List of SymbolsList of Symbols
• F, force• a, acceleration (not semi-major axis in this
lecture)• v, velocity• M, mass of Sun• m, mass of planet• d, general distance• r,radius of circle, semi-major axis of orbit• R, radius of Earth
Newton’s LawsNewton devised a uniform and systematic method for
describing motion, which we today refer to as theScience of Mechanics. It remains the basic descriptionof motion, requiring correction only at very highvelocities and very small distances.
Newton summarized his theory in 3 laws:
1. An object remains at rest or continues in uniform motionunless acted upon by a force.
2. Force is equal to mass x acceleration (F=ma)
3. For every action there is an equal and opposite reaction.
Newton and Gravity
Link for animation
Cambridge was closed becauseof the Plague. As the story goes,Newton was sitting under theapple tree outside his farmhouse(shown right) and while watchingthe apples fall he realized that theforce that made the apples fallalso made the planets orbit thesun. Using his newly inventedCalculus, Newton was able toshow that Kepler’s 3 laws ofplanetary motion followed directlyfrom this hypothesis.
Falling Apples and Orbiting PlanetsFalling Apples and Orbiting Planets
Splat
What do these have in common?
Apples and PlanetsApples and PlanetsWe will know analyze the motion of terrestrial fallingbodies and orbiting planets in more detail. We willanalyze both phenomenon in the same way and showthat Newton’s theory explains both. The plan is tocombine Newton’s second law with Newton’s law ofgravitation to determine the acceleration.
The interesting thing here is that we are applying lawsdetermined for motion on Earth to the motion ofheavenly bodies. What an audacious idea!
Gravitational Force: UnitsGravitational Force: Units
According to Newton’s 2nd law, Force=mass x acceleration
The units must also match.
Units of mass = kilograms
Units of acceleration = meters/sec2
Unit of force must be kilograms-meters/sec2 = kg m s-2 (shorthand)
We define a new unit to make notation more simple. Let’s call it aNewton. From the definition we can see that
1 Newton = 1 kg m s-2
From now on we measure force in Newtons.
What are the units of G?What are the units of G?Newton’s law of gravitation
F = GMm/d2
Let’s solve for G (multiply by d2, divide by Mm)
G = Fd2/Mm
Examine the units
Fd2/Mm has units of N m2/kg2 or N m2 kg-2
Or, expressing Newtons in kg, m, and s (1 N = 1 kg m s-2)
Fd2/Mm has units of N m2 kg-2 = (kg ms-2)m2 kg-2= m3 s-2 kg-1
G has units of m3 s-2 kg-1
Numerically, G = 6.67×10-11 m3 s-2 kg-1
NewtonNewton’’s Law of Gravitys Law of Gravity• All bodies exert a gravitational force on each other.• The force is proportional to the product of their masses
and inversely proportional to the square of theirseparation.
F = GMm/d2
where m is mass of one object, M is the mass of theother, and d is their separation.
• G is known as the constant of universal gravitation.
NewtonNewton’’s Second Laws Second LawForce = mass x acceleration
F = ma
Falling Apples: Gravity on EarthFalling Apples: Gravity on Earth F = m a = G m M / R2
F = m a = G m M / R2 (cancel the m’s)
a = G M / R2
where: G = 6.67x10-11 m3kg-1s-2
M = 5.97x1024 kg On Earth’s surface:
R = 6371 kmThus:
a = G M / R2 = 9.82 m s-210 m s-2
a on Earth is sometimes called g.
•
The separation, d, isthe distance betweenthe centers of theobjects.
Newton Explains Galileo
The acceleration does not depend on m!Bodies fall at the same rate regardless of mass.
F = GMm/d2Newton’s law of gravity:
a = GM/R2Cancel m on both sides ofthe equation
ma = GMm/R2Set forces equal
F = GMm/R2
The separation d is thedistance between thefalling body and the centerof the Earth d=R
F = maNewton’s 2nd Law:
Planetary motion is morecomplicated, but governed by
the same laws.First, we need to consider theacceleration of orbiting bodies
Circular AccelerationCircular Acceleration
Acceleration is any change in speed or directionof motion.Circular motion isaccelerated motionbecause direction ischanging. For circularmotion:
a = v2/r
Orbiting Planets ContinuedOrbiting Planets Continued
So, orbiting planets areaccelerating. Thismust be caused by aforce. Let’s assumethat the force is gravity.We should be able tocalculate the force andacceleration usingNewton’s second lawand Newton’s law ofgravity.
Orbits come in a varietyof shapes (eccentricities).In order to keep the mathsimple, we will considerin this lecture only circularorbits. All of our results also apply to ellipticalorbits, but we will not derive them that way.
Step 1: Calculate the VelocityStep 1: Calculate the VelocityWe take as given that acceleration and velocity in circular motion arerelated by
a = v2/r
According to Newton’s 2nd law
F = ma = mv2/r
According to Newton’s law of gravity
F = GMm/r2
Equating the expressions for force we have
mv2/r = GMm/r2
Solving for v2 gives
v2 = GM/r
Step 2: The Velocity is related to theStep 2: The Velocity is related to thesemi-major axis and periodsemi-major axis and period
The velocity is related to the semi-major axis and theperiod in a simple way: velocity = distance/time
distance = 2πr,
where r=semi-major axis, radius of circle
time = Period, P
v = 2πr/P = distance/time
Step 3: Relate the Period to theStep 3: Relate the Period to theOrbital RadiusOrbital Radius
We have
v2 =GM/r
And
v = 2πr/P
So it follows that
(2πr/P)2 = GM/r
Or
4π2r2/P2 = GM/r
How Does This Relate to How Does This Relate to KeplerKepler’’ssThird Law?Third Law?
We have
4π2r2/P2 = GM/r
Multiply both sides by r
4π2r3/P2 = GM
Multiply both sides by P2
4π2r3 = GM P2
Divide both sides by 4π2
r3 = (GM/4π2) P2
NewtonNewton’’s form of s form of KeplerKepler’’s s Third LawThird Law
We have
r3 = (GM/4π2) P2
Kepler’s third law was a3=P2, where a=semi-major axis(not acceleration). Since today we are using r=semi-major axis, this equation is the same as Kepler’s 3rd if
(GM/4π2) = 1 AU3/year2
Let’s check
Do Newton and Do Newton and Kepler Kepler Agree?Agree?
We want to know if
(GM/4π2) = 1 AU3/year2
Plug in G = 6.7×10-11 m3 s-2 kg-1, M=2.0×1030 kg
(GM/4π2) = 3.4×1018 m3 s-2
Recall 1 AU = 1.5×1011 m and 1 year = 3.1×107 s
So
1 AU3/year2 = (1.5×1011 m)3/(3.1×107 s)2
1 AU3/year2 = 3.4×1018 m3 s-2 Wow!!!
Using NewtonUsing Newton’’s Form of s Form of KeplerKepler’’ssThird Law: Example 1Third Law: Example 1
Planet Gabrielle orbitsstar Xena. The semimajor axis of Gabrielle'sorbit is 1 AU. Theperiod of its orbit is 6months. What is themass of Xena relative tothe Sun?
Using NewtonUsing Newton’’s Form of s Form of KeplerKepler’’ssThird Law: ExampleThird Law: Example 22
Planet Linus orbits starLucy. The mass of Lucy istwice the mass of the Sun.The semi-major axis ofLinus' orbit is 8 AU. Howlong is 1 year on Linus?
Using NewtonUsing Newton’’s Form of s Form of KeplerKepler’’ssThird Law: ExampleThird Law: Example 33
Jupiter's satellite (moon)Io has an orbital period of1.8 days and a semi-majoraxis of 421,700 km. Whatis the mass of Jupiter?