An Elementary Proof of Kepler’s Law of Ellipses

ByArpan Saha,

Freshman Undergraduate,Engineering Physics with Nanoscience,

IIT Bombay

Saturday, 13th March, 2010F. C. Kohli Audi, KReSIT

Seminar Weekend Day 1

A Few ClarificationsWhat do I mean by ‘elementary’?Explicit use of sophisticated machinery from vector analysis is avoided.

Why bother with an elementary proof?It’s insightful to have some exposure to the style in which Newton et al worked.

Is this proof due to Newton?No, this was discovered independently by William Rowan Hamilton and Richard P. Feynman.

And now for the proof!

Make all the velocity vectors co-initial. What do we get?

A closed curve, right?Technically called a hodograph.

Looks like a circle?It is, in fact, a circle, but we must prove it.

Presumed orbit

Velocity vectors

We don’t know whether it is

elliptical.

Choose points on the hodograph such that |∆v| (difference between successive velocity vectors) is equal i.e. |UV| = |VW| = |WX|

By Newton’s ULoGr, we get∆v = a∆t = (GM/R2)∆t

Kepler’s Law of Ellipses aka Law of Conservation of Angular Momentum says areal velocity A is constant.

Now, A = R2∆θ/∆ti.e. ∆t = R2∆θ/A

Plugging the second equation into the first gives∆v = (GM/A)∆θ

What does this mean pictorially?

U

V

W

X

If in the orbit we choose points A, B, C, D, such that angles ASB, BSC, CSD (S being the Sun), then the corresponding points A’, B’, C’, D’ on the hodograph will satisfy |A’B’| = |B’C’| = |C’D’|.

But we’re looking for something more informative. For this we remember, ULoGr is a vectorial law.

Force is directed towards S, and so is Δv. Hence as, ‘radial’ vector rotates around by equal angular increments, so does Δv in the hodograph.

But, if equally long segments are placed end-to-end in such a manner that adjacent segments are differ by a constant angle, then the segments form a regular polygon, which in the limit of an infinite number of sides is a circle.

The hodograph is hence indeed a circle!

Orbit

HodographΔθ

Δθ

Δθ Δθ

Δθ

S

A’B’

C’

D’

Δv

Δv

A

B

C

D

Let’s take a look at the picture so far.

θ

θ

v

v

S

A

A’

Orbit

Hodograph

Since tangent to hodograph is parallel to the line joining A and S, the corresponding radius OA’ (whose length we take as v0) is perpendicular to the same.

OS’

This is all very pretty. But what does it have to do with an ellipse?A lot, as we shall soon see.

Construct the locus of a point the sum of whose distances to O and S’ is v0. The locus is, of course, an ellipse, let us denote which by Γ.

OP + S’P = OQ = OP + PQi.e. PQ = PS’

The ‘mirror property’ of ellipses tells us that if LMN is tangent to Γ at P, then

Angle LPS’ = Angle NPOi.e. Angle LPS’ = Angle LPQ

Hence we may regard Q as the reflection of S’ about LMN. LMN is thus the perpendicular bisector of S’Q.

Q

L MP N

S’ O

Hodograph

Let us compare corresponding points P and Q’ on Γ and the orbit respectively.

Q

L MP N

S’ O

Hodograph

We note that OP is perpendicular to SQ’, and the tangent LMN is perpendicular to velocity vector S’Q. We begin to suspect that if we were to rotate Γ by a right angle clockwise, and scale it with a dimensionful factor, it might serve as a valid trajectory as the directions of displacement and velocity vectors are consistent with ULoGr. To complete the proof all we need to show is that magnitudes are also consistent with ULoGr.

S

Q’

Orbit

For this we turn to the Law of Conservation of Energy.

LoCoEn says that the magnitudes of velocity v and displacement R are related by

(m/2)v2 – GMm/R = -Eeverything being defined in the usual way .

Does a similar relation hold in case of S’Q = v and OP = r? Let’s check.

Q

S’ O

M

P

r

v0 - r

v/2

v/2

f

First, we label the sides and angles as shown. On applying the Cosine Rule we get

2v0vCos(φ) = v2 + v02 + f2

i.e. 2v0v(v/2(v0 – r) = v2 + v02 + f2

i.e. (v0/(v0 – r))v2 – v2 = v02 + f2 = k

(say)i.e. (v0/(v0 – r) – 1)v2 = ki.e. v2 = k(v0 – r)/ri.e. (1/k)v2 – v0 /r = – 1

which has the same form as is required by LoCoEn.

φ

π/2

Quod Erat Demonstrandum!

Tying up the Loose EndsQ1: We have only shown that the ellipse is a solution. Is it the only solution?In general, no (parabolas and hyperbolas admitted), but for

closed trajectories, yes.This is because, Newtonian mechanics is deterministic. The

planet cannot be in a quandary as to which path it should follow. If ellipse is a solution, it must be the only solution.

Q2: This was Feynman and Hamilton’s proof. How did Newton go about it in his Principia (1687)?Instead of cutting up the orbit into arcs subtending equal

angles at the sun, he cut it up into arcs subtending equal areas at the sun.

Thank You!