The radius of a planet is `R`. A satellite revolves around it in a circle of radius `r` with angular velocity `omega_(0)`. The acceleration due to the gravity on planet's surface is

Options

A. `(r^(3)omega_(0))/R`
B. `(r^(3)omega_(0)^(2))/(R^(2))`
C. `(r^(3)omega_(0)^(2))/R`
D. `(r^(3)omega_(0)^(2))/(R^(2))`

Answer

D

Explanation

`(GMm)/(r^(2))=momega_(0)^(2)r`
`implies GM=omega_(0)^(2)r^(3)`
`implies gR^(2)=omega_(0)^(2)r^(3)`
`implies g=(omega_(0)^(2)r^(3))/(R^(2))`