A spherical planet far out in space has a mass `M_(0)` and diameter `D_(0)`. A particle of mass m falling freely near the surface of this planet will experience an accelertion due to gravity which is equal to

Options

A. `(4mGM_(0))/(D_(0)^(2))`
B. `(4GM_(0))/(D_(0)^(2))`
C. `(GmM_(0))/(D_(0)^(2))`
D. `(GM_(0))/(D_(0)^(2))`

Answer

B

Explanation

By definition : `g=(GM)/(R^(2))` for the earth.
For the planet, `g=g', M = M_(0)`
and `R=(D_(0))/(2) " " therefore R^(2)=(D_(0)^(2))/(4)`
`therefore g'=((GM_(0))/(D_(0)^(2)))/(4)=(4GM_(0))/(D_(0)^(2))`