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^(5) " m",pi=3.14,g=10" ms"^(-2)`
We know that the period of revolution of the earth satellite
`T=2pisqrt(((R_(e)+h)^(3))/(gR_(e)^(2)))" "["if h" lt ltR_(e), "then"(R_(e)+h=R_(e))]`
So, `T=2pisqrt((R_(e)^(3))/(gR_(e)^(2)))=2pisqrt((R_(e))/(g))=2xx3.14sqrt((6.14xx10^(6))/(10))`
`=2xx3.14xx0.8xx10^(3)`
`=5.024xx10^(3)=5024s=83.73 min`.