Question

A geo-stationary satellite is orbiting around earth at 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 24 hours. Let the acceleration due to gravity (g) be $10m/s^2$ and mass of satellite 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

The correct option is D $0J$

Since $F_g$ is perpendicular to displacement at each instant due to circular motion of satellite so,
Work done $=F.ds=0$,