This talk, given as a part of the Annual Seminar Weekend 2010, IIT Bombay, was centered around the hodograph proof of Kepler’s Law of Ellipses independently discovered by Maxwell, Hamilton and Feynman.

1. Seminar Weekend Day 1
By
Arpan Saha,
Freshman Undergraduate,
Engineering Physics with Nanoscience,
IIT Bombay
Saturday, 13th March, 2010
F. C. Kohli Audi, KReSIT

2. A Few Clarifications
What 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.

3. And now for the proof!

4. Velocity
vectors
Make all the velocity vectors co-initial.
What do we get?
A closed curve, right?
Technically called a hodograph.
Presumed orbit
We don’t know
whether it is
elliptical.
Looks like a circle?
It is, in fact, a circle, but we must prove it.

5. Choose points on the hodograph such that |∆v|
(difference between successive velocity vectors)
is equal i.e. |UV| = |VW| = |WX|
X
By Newton’s ULoGr, we get
∆v = a∆t = (GM/R2)∆t
W
Kepler’s Law of Ellipses aka Law of
Conservation of Angular Momentum
says areal velocity A is constant.
V
U
Now, A = R2∆θ/∆t
i.e. ∆t = R2∆θ/A
Plugging the second equation into the first gives
∆v = (GM/A)∆θ
What does this mean pictorially?

6. 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.
A’
D
B’
Hodograph
C
Δθ
Δv
C’
B
Δθ
Δθ
Δθ
Δv
A
Orbit
Δθ
D’
S
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!

7. Let’s take a look at the picture so far.
Hodograph
v
A
A’
v
S’
S
Orbit
θ
O
θ
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.

8. 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.
Q
L
P
M
N
OP + S’P = OQ = OP + PQ
i.e. PQ = PS’
The ‘mirror property’ of ellipses tells us that if
LMN is tangent to Γ at P, then
S’
O
Angle LPS’ = Angle NPO
i.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.
Hodograph

9. Let us compare corresponding points P and Q’ on Γ and the orbit respectively.
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.
Q
Q’
L
P
M
S’
N
O
S
Orbit
Hodograph
For this we turn to the Law of Conservation of
Energy.

10. LoCoEn says that the magnitudes of velocity v and displacement R are related by
(m/2)v2 – GMm/R = -E
everything being defined in the usual way .
Does a similar relation hold in case of S’Q = v and OP = r? Let’s check.
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 = k
i.e. v2 = k(v0 – r)/r
i.e. (1/k)v2 – v0 /r = – 1
Q
v0 - r
φ
v/2
π/2
M
r
v/2
S’
which has the same form as is required
by LoCoEn.
P
f
O

11. Quod Erat Demonstrandum!

12. Tying up the Loose Ends
Q1: 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.