The orbit of geostationary satellite is circular, the time period of satellite depeds on (i) mass of the satellite, (ii) mass of earth, (iii) readius of the orbit and (iv) height of the satellite from the surface of the earth

Options

A. (i) only
B. (i) and (ii)
C. (i), (ii) and (iii)
D. (ii), (iii) and (iv)

Answer

D

Explanation

Orbital velocity, `v_(0)=sqrt((GM_(E))/(R_(E)+h))`
Time period, `T=(2pi(R_(E)+h))/(v_(0)) =(2pi(R_(E)+h)^(3//2))/((GM_(E))^(1//2))`
Thus, the time period of satellite in independent of mass of satellite but depends on mass of the earth, radius of the orbit `(R_(E)+h)`, height of the mass satellite from the surface of the earth.