What is a period of revolution of earth satellite ? Ignore the height of satellite above the surface of earth.
Given :
The value of gravitational acceleration `g= 10 m s^(-2)`
Radius of earta `R_(E)= 6400 km`. Take `pi=3.14`.

Options

A. 83.73 minutes
B. 85 minutes
C. 90 minutes
D. 156 minutes

Answer

A

Explanation

Period of revolution of earth satellite is
`T= 2 pi sqrt(((R_(E)+h)^(3))/(gR^(2)E))`
Neglecting h, then
`T= 2 pi sqrt(R_(E))/(g)`
Substituting the given values, we get
`T=2xx3.14 sqrt((6400xx10^(3))/(10))s`
`=5024s=(5024)/(60)` minutes
`= 83.73` minutes