What is a period of revolution of the earth satellite ? Ignore the height of satellite above the surface of the earth.
Given,
(i) the value of gravitational acceleration, `g = 10 ms^(-2)`
(ii) radius of the earth, `R_(g) =6400` km (take, `pi = 3.14`)

Options

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

Answer

C

Explanation

Given `R_(e )=6400 km =6.4xx10^(6)m`
`pi =3.14 g =10 m//s^(2)`
we know that the period of revolution of the earth satellite
`T=2pisqrt((R_(e)+h)^(3)/(gR_(e)^(c ))[if lt lt R_(e)` then `(R_(e)+h=R_(e))]`
so `T=2pi sqrt(R_(e)^(3))/(gR_(e)^(2))=2pisqrt(R_(e))/(g)=2xx3.14sqrt(6.4xx10^(6))/(10)`
`=2xx3.14xx0.8xx10^(3)=5.024xx10^(3)=5024s`
and `T=(5024)/(60)=83.73` min