Question

A satellite is revolving in a circular orbit distance of 2620 km from the surface of the earth. The time period of the revolution of the satellite is (Radius of the earth =6380 km , mass of the earth $=6\times {10}^{24}$ kg, G $=6.67\times {10}^{11}N-m^2/{kg}^2$

• A. 2.35 hours
• B. 23.5 hours
• C. 3.25 hours
• D. 32.5 hours

Answer 1

The correct option is A 2.35 hours

Orbital velocity,
$4v_0=\sqrt{\frac{{GM}_e}{R_e+h}}\sqrt{\frac{(6.67\times {10}^{-11})(6\times {10}^{24})}{6380+2620}}=6.67km/secTimeperiodofrevolution,T=2\pi \frac{{(R}_e+h)}{v_0}=\frac{2\times 3.14\times (6380+2620)}{6.67\times {10}^3}=8474sec=2.35hours$