Q. A satellite moves in elliptical orbit about a planet. The maximum and minimum velocities of satellites are $3 \times 10^{4}{ms}^{- 1}$ and $\ 1 \times 10^{3}{ms}^{- 1}\ $ respectively. What is the minimum distance of satellite from planet, if maximum distance is $4 \times 10^{4}\ km?$
A) $4 \times 10^{3}\ km$
B) $3 \times 10^{3}\ km$
C) $4/3 \times 10^{3}\ km$
D) $1 \times 10^{3}\ km$
Explanation
(c) If no external torque acts on a system, then angular momentum of the system does not change.
$ie,\ If\ \ \ \ \ \ \ \ \ \ \ \tau = 0$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \frac{dL}{dt} = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = constant$
$Hence,\ \ {mv}_{\max}r_{\min} = {mv}_{\min}r_{\max}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ r_{\min} = \frac{v_{\min} \times r_{\max}}{v_{\max}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1 \times 10^{3} \times 4 \times 10^{4}}{3 \times 10^{4}} = \frac{4}{3} \times 10^{3}km$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.