A particle describes an ellipse as a central orbit about a focus. Show that the velocity at the end of the minor axis is the geometric mean between the greatest and least velocities.
My attempt:
I am not sure that the elliptic path described under the inverse square law of distance or not. Is so, please give me idea how to identify?
If it is so, I know that $v^2=\mu \left(\frac{2}{r}-\frac{1}{a}\right)$ where $v$ is the velocity of the particle on the ellipse under attractive force $\frac{\mu}{r}$ at a distance $r$ from the center of force (the focus)and $a$ is the semi-major axis of the elliptic orbit.
Now the velocity at the end of minor axis = velocity at B, where B is end point of minor axis. But, if S be the focus and C be the center, then $SB^2=SC^2+BC^2=(ae)^2+b^2=a^2$ (since, $b^2=a^2(1-e^2)$). Therefore, $SB=a$.
Then, square of velocity at B = $V^2=\mu \left(\frac{2}{a}-\frac{1}{a}\right)=\frac{\mu}{a}$.
Now how to find greatest and least velocities.
Best Answer
The Kepler's first law states that: $$\rho(\theta) = \frac{b^2}{a}\frac{1}{1-e\cos\theta}\tag{1}$$ and since the angular momentum $\rho^2\dot{\theta}$ is preserved (we have a central force) it follows that the speed $v = \rho\, \dot{\theta}$ satisfies: $$ |v| = C\cdot\frac{a}{b^2}(1-e\cos\theta),\tag{2}$$ so the minimum and maximum speeds are given by $C\cdot\frac{a}{b^2}(1\mp e)$ (these speeds are attained in the endpoints of the major axis), while $\cos\theta$ in the endpoints of the minor axis equals $\frac{c}{a}=e$, so the claim follows from: $$ (1-e)(1+e)=(1-e^2).\tag{3} $$