Question

A satellite moves in elliptical orbit about a planet. Its maximum and minimum velocities of satellites are $3 \times {10^4}\,m{s^{ - 1}}$ and $1 \times {10^3}\,m{s^{ - 1}}$ respectively. What is the minimum distance of a satellite from a planet if maximum distance is $4 \times {10^4}\,km$?

Answer 1

Here we have to use the concept of angular momentum.
Angular momentum is a calculation of the body’s momentum in rotational motion. The angular momentum of rigid bodies is maintained, thus the rotating sphere will continue to rotate until the external force acts. The change in angular momentum is equivalent to a torque.

Complete step by step solution:
Given,
Maximum velocity of the satellite, ${v_{\max }} = 3 \times {10^4}\,m{s^{ - 1}}$
Minimum velocity of the satellite, ${v_{\min }} = 1 \times {10^3}\,m{s^{ - 1}}$
Maximum distance of the satellite from the planet, ${r_{\max }} = 4 \times {10^4}\,km$
Minimum distance of the satellite from the planet, ${r_{\min }} = ?$
According to law of conservation of angular momentum,
$mvr = {\text{constant}} \\\ \therefore {{\text{v}}_{\max }}{r_{\min }} = {v_{\min }}{r_{\max }} \\\ \Rightarrow \left( {3 \times {{10}^3}} \right) \times {r_{\min }} = \left( {1 \times {{10}^3}} \right) \times \left( {4 \times {{10}^4}} \right) \\\ \Rightarrow {r_{\min }} = 4 \times {10^3}\,km \\\$
Minimum distance of satellite from planet if maximum distance is $4 \times {10^4}$ is $4 \times {10^3}\,km$.

Additional information:
Angular momentum of the system is retained as long as there is no net external torque operating on the system; the planet rotates on its axis from the time the solar system was created due to the law of accumulation of angular momentum.
There are two ways to measure the angular momentum of some object, if it is a point object in a rotation, then our angular momentum is proportional to the radius times the linear momentum of the object.
If we have an extended object, the angular momentum is given by the moment of inertia, i.e. how much mass is in motion in the object, and how far it is from the centre, times the angular velocity.

Note: Here we have carefully observed the question and see which is the maximum and which is the minimum velocity. If we mix up the values of the velocities then we may get a different answer.