The acceleration due to gravity on the moon is `(1^(th))/(6)` that on the earth. If the earth and the moon are assumed to have the same density, then the radius of the moon is

Options

A. `6 R_(e )`
B. `(1)/(6)R_(e )`
C. `(1)/(3)R_(e )`
D. `(1)/(4)R_(e )`

Answer

B

Explanation

For the earth,
`g=(GM)/(R_(e )^(2))=G.(4)/(3)(pi R_(e )^(3)rho)/(R_(e )^(2))=((4)/(3)pi G d)R_(e )`
Similarly for the moon, `g'=((4)/(3)pi G d)R'`
where R' is the radius of the moon.
`therefore (g')/(g)=(1)/(6)=(R')/(R_(e ))" " therefore R'=(R_(e ))/(6)`