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I was reading a Classical Mechanics book. The author was deriving Kepler's equation. He was changing variables for integrating the stuff later. Here's my reproduction of that figure.

$N$ is the normal drawn to the major axis of the ellipse (I forgot to label that, but it's just the dashed line passing through that big point). Since $CN=CF+FN$,

$$a\ \mathrm{cos}\ \psi=ae+r\ \mathrm{cos}\ \theta$$

Then, he uses the property of ellipse, $b^2=a^2(1-e^2)$ to get to

$$(1-e\ \mathrm{cos}\ \psi)(1+e\ \mathrm{cos}\ \theta)=\frac{b^2}{a^2}$$

I soon realized that he has also used the ellipse equation in polar coordinates to eliminate $r$.

$$\frac{1}r=\frac a{b^2}(1+e\ \mathrm{cos}\ \theta)$$

Finally, he does an implicit differentiation and arrives at a rather vague expression!

$$\frac{d\theta}{d\psi}=\frac b{a(1-e\ \mathrm{cos}\ \psi)}$$

He simply says, "we get to this expression after more manipulations."

But, I don't understand how he made that up. When I differentiated that, I can't do much progress beyond this expression,

$$\frac{d\theta}{d\psi}=\frac{b^2\ \mathrm{sin}\ \psi}{a^2(1-e\ \mathrm{cos}\ \psi)^2\ \mathrm{sin}\ \theta}$$

What am I missing here?

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In eccentric anomaly there is all you need.
There the eccentric anomaly $\psi$ of your textbook is called $E$.

You find $$\frac {\sin \psi}{\sin \theta}=\frac {\sqrt{1-e^2}}{1+e\cos \theta}$$ Equating the two expressions fo $r$, you get $$a(1-e \cos \psi)=\frac {a(1-e^2)}{1+e \cos \theta}$$ so that $$\frac {\sin \psi}{\sin \theta}=\sqrt{1-e^2} \cdot \frac {1-e \cos \psi}{1-e^2}=\frac ab \cdot (1-e \cos \psi)$$ Substitute in your last expression and you are done.

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  • $\begingroup$ Ah, nice indeed... Thanks for pointing that out! ;-) And, I wonder why I didn't get notified of your answer in my inbox o_O $\endgroup$ Sep 20, 2014 at 18:11

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