If $v$ is the true anomaly and $E$ the eccentric anomaly, how can I show that $$\frac{dv}{dE}=\frac{b}{r}=\frac{\sin v}{\sin E}~?$$
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$\begingroup$ It is enough to show that $\frac{dv}{dt}=\frac{b}{na}$ for a elliptical orbit $\endgroup$– SmurfCommented Feb 1, 2016 at 20:42
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$\begingroup$ Please define all the symbols. $\endgroup$– ProfRobCommented Feb 1, 2016 at 21:19
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$\begingroup$ a and b are the semiaxis of the elliptical orbit, $n=2\pi/P$ with P the period $\endgroup$– SmurfCommented Feb 1, 2016 at 21:21
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$\begingroup$ and what is $r$? $\endgroup$– ProfRobCommented Feb 1, 2016 at 22:13
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1$\begingroup$ I posted some time ago a some what related question about time (since periapsis passage) or mean anomaly. The first term within the brackets of the last expression for time is the same as the eccentric anomaly and the second term can be shown to be equal to the eccentricity times the sine of the eccentric anomaly, which is just Kepler's equation. $\endgroup$– fibonaticCommented Feb 2, 2016 at 1:22
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1 Answer
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Here is the proof: Please refer to the wikipedia page on eccentric anomaly for a diagram and a couple of intermediate formulae.
For an ellipse with the usual formula $x^2/a^2 + y^2/b^2=1$, it is the case that $\sin E = y/b$, and also by studying the figure on the wiki page you can see that $\sin (\pi-\nu) = \sin \nu = y/r$. Thus the two results you wish to derive follow from one another.
The relationship between the eccentric anomaly and true anomaly is $$ \tan \left(\frac{\nu}{2}\right) = \left(\frac{1+e}{1-e}\right)^{1/2} \tan \left(\frac{E}{2}\right) \tag{1}$$
Differentiating (1): $$\sec^2 \left(\frac{\nu}{2}\right) \frac{d\nu}{dE} = \left(\frac{1+e}{1-e}\right)^{1/2}\sec^2 \left(\frac{E}{2}\right)\tag{2}$$
But using (1) to replace the eccentricity term in (2) $$ \frac{d\nu}{dE} = \frac{\sec^2 (E/2) \tan (\nu/2)}{\sec^2 (\nu/2) \tan (E/2)}$$ $$ \frac{d\nu}{dE} = \frac{\sin (\nu/2) \cos (\nu/2)}{\sin (E/2) \cos (E/2)} = \frac{\sin \nu}{\sin E} = \frac{y/r}{y/b} = \frac{b}{r}$$
