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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It is enough to show that $\frac{dv}{dt}=\frac{b}{na}$ for a elliptical orbit – Smurf Feb 01 '16 at 20:42
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Please define all the symbols. – ProfRob Feb 01 '16 at 21:19
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a and b are the semiaxis of the elliptical orbit, $n=2\pi/P$ with P the period – Smurf Feb 01 '16 at 21:21
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and what is $r$? – ProfRob Feb 01 '16 at 22:13
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Norm of the position vector (with a reference system centered in the perihelion) – Smurf Feb 01 '16 at 22:21
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1I 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. – fibonatic Feb 02 '16 at 01:22
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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) $$ \frac{d\nu}{dE} = \frac{\sec^2 (E/2) \tan (\nu/2)}{\sec^2 (\nu/2) \tan (E/2)}$$ $$ \frac{d\nu}{dE} = \frac{\sin (\nu/2) \cos (\nu/2)}{\sin (E/2) \cos (E/2)} = \frac{\sin \nu}{\sin E} = \frac{y/r}{y/b} = \frac{b}{r}$$
ProfRob
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