Elliptical Orbits – Deriving Kepler’s Equation

Here is that drawing again, showing the eccentric anomaly E and the true anomaly f. What we’ve done so far in this post and in that post is just use Kepler’s equation

M = E – e Sin E

to move between position and time on an elliptical orbit.

Let’s derive the equation using geometry and trigonometry. (Both Conway & Prussing, and Bate, Mueller & White – see my bibliography page – have this derivation.)



We have an elliptical orbit in black. Think of it as a planet orbiting the sun — on an unrealistically eccentric orbit. (What I’ve drawn has eccentricity = \frac{\sqrt{3}}{2} = .866\ ; the eccentricity of Pluto’s orbit is just under .25 .) We have a circumscribed circle in red.

The symbols on the drawing are:

Z – is the center of the ellipse and of the circle;
F – is the occupied focus (the sun);
V – is periapse, the point of closest approach;
P – is the position of the orbiting object (the planet) at some time;
f – is the true anomaly, the angle VFP
S – is the foot of a perpendicular dropped from P to the line ZV.
Q – is the line segment PS extended to the circumscribed circle.
E – is the eccentric anomaly, the angle VZQ.

Not shown are:

t – is the unknown time from V to P;
a – is the semi-major axis, lengths ZV and ZQ;
c – is the distance from center to focus, length ZF;
M – is the mean anomaly;
e – is the eccentricity,

and e = c/a = length ZF / length ZV.

We want to find the time t since periapse passage. Well, we’ve actually solved two problems using Kepler’s equation – now I want to derive it.

We can describe the time t in terms of the fraction of the area spanned. Let T be the period of the orbit. Kepler’s Second law says that the object traverses equal areas in equal times (and we could prove that from other equations we’ve gotten). That means that the fraction of time t/T is equal to the fraction of area FVP / ellipse:

t/T= (Area Sector FVP) / (Area ellipse).

It turns out that there is a simple relationship between the area of the elliptical sector SVP and the circular sector SVQ. What? Why? Because for any pair of points P and Q, the ratio of heights QS/PS = a/b. That is,

PS = b/a QS.

(We saw that in this post.)

That in turn implies that

area sector SVP = b/a area sector SVQ.

That is, we have related the area bounded in red

to the area bounded (mostly) in black:

Now, how do we get the elliptical sector FVP from the elliptical sector SVP? Easy enough: they differ by the triangle SFP:

Area Sector FVP = Area Sector SVP – Area Triangle SFP.

What’s the area of the triangle SFP? Well, what the base? |SF|.

|ZF| = c = ae.

|ZS| = |ZQ| cos E = a cosE

so |SF| = ae – a cosE.

What’s the height? |PS| = b/a |QS|, but

|QS| = |ZQ| sinE = a sinE

so

|PS| = b/a |QS| = (b/a) a sinE

and then the area of the triangle is half the base times the height:

Area SFP = 1/2(ae – a Cos E) (a Sin E) b/a

We can also do some trigonometry to relate the area of the circular sector SVQ to the area of the wedge ZVQ. First, the area of the wedge is a fraction of the area of the circle — and that fraction is E / 2\pi\ . That is,

Area ZVQ = (E/2\pi)\ \pi r^2\ .

The radius is |ZV| = a, so

Area ZVQ = E/2\ a^2\ .

Now, the area of the sector SVQ is the area of the wedge ZVQ minus the area of the triangle ZSQ:

area sector SVQ = area wedge ZVQ – area triangle ZSQ.

What’s the area of the triangle ZSQ? base ZS has length

|ZS| = |ZQ| cos E = a cosE

and height

|QS| = |ZQ| sinE = a sin E

so the area ZSQ = 1/2 a cosE a sinE = a^2/2\ \sin(E)\ \cos(E)\ .

Putting it all together:

Area Sector FVP = Area Sector SVP – Area Triangle SFP.
Area Triangle SFP = 1/2(ae – a Cos E) (a Sin E) b/a
area
sector SVP = b/a area sector SVQ.
area sector SVQ = area wedge ZVQ – area triangle ZSQ.
Area wedge ZVQ = E/2 a^2.
area triangle ZSQ = a^2/2 sinE cosE.

Let Mathematica do all that….

That’s the area associated with the time t. And when we divide that area by the total area of the ellipse, namely \pi a b\ , we get

We have

t / T = (E-e \sin[E])/(2 \pi)

2\pi t /T = E - e \sin[E]\ .

We recognize the RHS as M, so what we have been calling M is

M = 2 \pi t / T

but 2\pi / T\ is the average angular speed (cycles per time); call it n. We have gotten

M = n t

M = E – e Sin E,

so n t = E – e Sin E;

and

n = 2 \pi / T = \sqrt{\mu/a^3}\ .

I remember how long it took me to sort out what I had to work with; that’s why I chose to work a couple of examples first.

Ah, but we still need to relate true anomaly f to eccentric anomaly E. We see that |ZS| can be written in terms of each angle:

ZS = a cos E = a e + r cos f ( = ZF – SF in this case, for which cos f is negative).

We use the equation of the orbit to eliminate r, then we simplify, divide both sides by a, and solve for E:

That’s the most straight-forward relationship between E and f. We can also show that

\tan (E/2) = \sqrt{\frac{1-e}{1+e}} \tan (f/2)

First, Mathematica verifies that for any angle x, we have the half-angle formula (no, I didn’t remember this!):

\tan^2 (x/2) = \frac{1-\cos(x)}{1+\cos(x)}\ :

Take that half-angle formula and plug in our relationship between E and f, and let Mathematica go to town:

The signs aren’t where I want them, but the equation is right – so take the square root of both sides (the LHS is \tan^2 (E/2)\ :

\tan (E/2) =\sqrt{\frac{1-\cos(E)}{1+\cos(E}} = \sqrt{\frac{1 - e}{1 + e}} \tan (f/2)\ .

Okay? I hope so… and that’s enough of orbits for a while. Remember that I’ve only shown you a time equation for elliptical orbits.

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4 Responses to “Elliptical Orbits – Deriving Kepler’s Equation”

  1. hasan Says:

    thanksss a lot!!

  2. Bob Kuriger Says:

    If I can ever verify this I will be so happy, maybe I will take you skydiving sometime.

  3. Amanda Salmoiraghi Says:

    Thanks :) Very helpful to have the trig/geometry laid out so nicely.


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