Download Groupwork on Planetary Motion

Groupwork on Planetary Motion
Math 245, Section 2
Due October 3, 2016
1. Kepler’s Laws
Through a series of observations, Johannes Kepler (1571-1630) derived 3 laws of planetary motion.
1. Kepler’s First Law: The orbit of a planet is an ellipse with the Sun at one of the two foci.
2. Kelper’s Second Law: The line segment joining a planet and the Sun sweeps out equal areas
during equal intervals of time.
3. Kelper’s Third Law: The square of the orbital period of a planet is proportional to the cube
of the semi-major axis of its orbit.
Kepler derived these laws simply by looking at hundreds of calculations he had done and noticing
patterns in the data.
Amazingly, we can derive these laws using a little vector Calculus and two of Newton’s laws.
2. Newton’s Laws of Motion
Isaac Newton (1643-1727) had a different approach to deriving physical laws: he tried to determine
mathematical relationships between objects. We will use two of his best known laws.
1. Newton’s Second Law of Motion: F~ = m~a, where m is the mass of the object and ~a is its
acceleration.
G
2. Newton’s Law of Universal Gravitation: F~g = − mM
~r, where the sun of mass M is at the
|~
r |3
origin, ~r is the position vector of the planet with mass m, and G is the gravitational constant.
Newton’s Law of Universal Gravitation is an example of a central force being exerted on a moving
body.
3. Conservation of Angular Momentum
The first thing we will prove is that the angular momentum of an object moving under a central
force is preserved. The angular momentum is defined by
~
L(t)
= ~r × p~ = m ~r × ~r 0
If ~r(t) represents the position of the object at time t, then ~v = ~r 0 (t) is its velocity and ~a = ~r 00 (t) is
its acceleration. (~
p = m~r 0 is the linear momentum.)
We assume a central force, i.e. that the object experiences a force F~ = f (t)~r(t), which is directed
along the position vector ~r.
1. Set the expression for force in Newton’s Second Law equal to the expression for the central force
to derive the equation:
f (t)
~r(t).
~r 00 (t) =
m
1
2
2. Compute
~
dL
dt .
Use Problem 1 and your knowledge of cross products to conclude that
~
dL
dt
= ~0.
~
~ is a constant vector - it does not change with time. This means
The fact that ddtL = ~0 means that L
~ is also constant. Since L
~ = m ~r ×~r 0 , this implies that the position
that the plane with normal vector L
and velocity of the object always lie in the same plane!
We have just derived one of Kepler’s key observations: The motion of a planet is restricted to a fixed
plane, i.e. its path is described by a planar curve. With this observation, the formulation of Kepler’s
Laws becomes possible. From here on out, we will draw all our pictures in the standard xy-plane with
the sun at the origin.
4. Deriving Kepler’s Second Law
For this calculation, let’s say that at time t = 0, ~r(0) is on the x-axis and proceeds counterclockwise
as t increases. Let A(t) denote the area swept out by the radial vector between time 0 and time t.
3. Now fix t and consider a small time increment ∆t. Draw a diagram to show the area swept out
by the radial vector ~r from time t to time t + ∆t. Show that this is equal to the area given by
A(t + ∆t) − A(t).
When ∆t is very small, the area given by A(t + ∆t) − A(t) is very close to being a triangle.
4. Express the area of this triangle in terms of ~r(t) and ~r(t + ∆t).
3
A(t + ∆t) − A(t)
.
∆t
(Hint: you might find it useful to subtract ~r(t) × ~r(t) from your answer to Problem 4.)
5. Find A0 (t) = lim
∆t→0
6. Use conservation of angular momentum to conclude that A0 (t) =
~
|L|
2m
is a constant!
~ Since A0 (t) is a constant, we conclude that between any two times t1 and t2 , the area
Let L = |L|.
L
. This is precisely Kepler’s Second Law!
swept out is simply A(t2 ) − A(t1 ) = (t2 − t1 ) 2m
5. Deriving Kepler’s Third Law
For our derivation of Kepler’s Third Law, we will need to use the formula for the ellipse describing
the path of the planet that we derived in class,
L2
.
(1)
GM m2
q
2
Recall that e is the eccentricity of the ellipse, defined by e = 1 − ab 2 , where a and b are the major
and minor axis, respectively. In this coordinate system, the origin is at one of the foci of the ellipse,
not at the center of the ellipse. r = |~r| represents the distance to the origin and θ is the angle ~r makes
with the positive x-axis.
r(1 + e cos θ) =
7. Use algebra to show that b2 = a2 (1 − e2 ) and that e =
√
a2 −b2
.
a
√
The quantity a2 − b2 represents the distance f from the center of the ellipse to one of the foci. So
using problem 7 we have the simple relation,
f = ea.
(2)
4
2
L
8. Now use Equations (1) and (2) to show that a(1 − e2 ) = GM
.
m2
(Hint: Plug θ = 0 into (1) and note that when θ = 0, the planet lies on the positive x-axis, so r can
be expressed in terms of a and f . Then use (2). Remember the sun (origin) is at one focus, not at
the center of the ellipse.)
If T represents the orbital period of the planet, then the total area it sweeps out in one rotation is
Z T
A(T ) =
A0 (t)dt.
0
9. Use the fact that the area of an ellipse is πab and your answer to Problem 6 to derive a formula
for T in terms of a, b, m and L. Square this to obtain a formula for T 2 .
10. Combine Problems 7, 8 and 9 to prove T 2 =
4π 2 a3
GM .
This is Kepler’s Third Law!