Q. If satellite is revolving around a planet of mass $M$ in an elliptical orbit of semi-major axis $a$, find the orbital speed of the satellite when it is at a distance $r$ from the focus
A) $v^{2} = GM\left\lbrack \frac{2}{r} - \frac{1}{a} \right\rbrack$
B) $v^{2} = GM\left\lbrack \frac{2}{r^{2}} - \frac{1}{a} \right\rbrack$
C) $v^{2} = GM\left\lbrack \frac{2}{r^{2}} - \frac{1}{a^{2}} \right\rbrack$
D) $v^{2} = G\left\lbrack \frac{2}{r} - \frac{1}{a} \right\rbrack$
Explanation
(a) As in case of elliptic orbit of a satellite mechanical energy
$E = - (GMm/2a)$ remains constant, at any position of satellite in the orbit,
$KE\ + \ PE\ = - \frac{GMm}{2a}$ …(i)
Now, if at position $r,v$ is the orbital speed of satellite
$KE\ = \ \frac{1}{2}\ mv^{2}\ $ and $PE\ = \ - \frac{GMm}{r}$ …(ii)
So, from Eqs. (i) and (ii), we have
$\frac{1}{2}\ mv^{2} - \frac{GMm}{r} = - \frac{GMm}{2a},\ ie,\ v^{2} = GM\left\lbrack \frac{2}{r} - \frac{1}{a}\ \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.