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) $10m/s^2$ 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. $2\pi \times 7.2\pi \times {10}^{14}$J
• C. $1.8\pi \times {10}^{14}$J
• D. $0$J

Answer 1

Answer: (D)

$0$J

As the displacement of the satellite is always along the tangent to the circular orbit and the gravitational force acts towards the centre of the circular path. Hence the displacement is perpendicular to the gravitational force and thus the work done by gravitational force in moving the satellite is zero.