The ratio of cube of circumferences of the orbit of a satellite to the volume of the earth is `6xx10^(10)(g//R)`, where `g` is the acceleration of gravity. Find the time period of satellite (in secods). `R` is the radius of earth

Options

A. `2xx10^(5)`
B. `10^(5)`
C. `4xx10^(4)`
D. `10^(4)`

Answer

A

Explanation

`T=(2pir)/(R)sqrt(r/g)=K/Rsqrt(K/(2pig))` if `K=2pir`
`=sqrt(((2K^(3))/3R)/((4/3piR^(3))g))=sqrt(2/3(K^(3)R)/(Vg))`
Given , `(K^(3))/V=(6xx10^(10)g)/R`
So, `T=sqrt((2xx6xx10^(10))/3)=2xx10^(5)sec`