The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth be R, then radius of the planet would be

Options

A. 4 R
B. `(R )/(2)`
C. `(R )/(4)`
D. 2 R

Answer

B

Explanation

The acceleration due to gravity on the surface of the earth is
`g=(GM)/(R^(2))=G.(4pi)/(3)(R^(3)xx rho)/(R^(2))=((4)/(3)pi G)R rho`
`therefore g = KR rho " "` where `K = (4)/(3)pi G`
Similarly for the planet `g_(P)=KR_(P)rho_(P)`
But `g_(P)=g`
`therefore KR_(P)rho_(P)=KR_(rho)` but `rho_(P)=2rho`
`therefore R_(P)xx2rho=R rho " " therefore 2R_(P)` or `R_(P)=(R )/(2)`