Question

A geo-stationary satellite is orbiting around earth at a height of 30,000 km in circular orbit. The radius of the earth is taken as $6000 km$. The geo-stationary satellite comes back to its position after one revolution in exactly $24hours$.Let the acceleration due to gravity(g) be $10m/{s^2}$ and mass of satellite be be $1000 kg$; calculate the work done in 12 hours when moving under gravitational force.
(A) $3.6\pi \times {10^{14}}J$
(B) $2\pi \times 7.2 \times {10^{14}}J$
(C) $1.8\pi \times {10^{14}}J$
(D) $0J$

Answer 1

Hint: Here work done by earth on satellite is asked under the influence of gravity all you need to do is take the dot product of the force and displacement, Although in this case since force is always perpendicular to the displacement.

Complete step by step answer:
Here it is clearly mentioned that the satellite is moving in circular orbit, and gravitational force by earth acts as a centripetal force so work done by gravitational force on satellite due to earth is zero because centripetal force always acts towards centre.
Therefore force is always perpendicular to the movement of the satellite so work done will be zero at every instant it doesn’t matter at which time and height.

Hence Option-D is correct.

Note: Sometimes questions are completely theoretical like in this question all you need to do is apply the basic concepts and you will get the answer.(here data like height of satellite and radius of earth and time are given which has no use in solving this question. It is just given to confuse the student. You should be aware of these types of questions.