Two planets A and B have the same average density. Their radii `R_(A) and R_(B)` are such that `R_(A): R_(B)=3:1`. If `g_(A) and g_(B)` are the acceleration due to gravity at the surfaces of the planets, the `g_(A): g_(B)` equals

Options

A. `3:1`
B. `1:3`
C. `9:1`
D. `1:9`

Answer

A

Explanation

Given, `(R_(A))/(R_(B))= (3)/(1)`…(i)
and `rho_(A) = rho_(B)` …(ii)
`because` Average density `rho= (3g)/(4pi RG)`
`therefore` From Eq. (ii), `rArr (3g_(A))/(4pi R_(A)G)= (3g_(B))/(4pi R_(B)G) rArr (g_(A))/(g_(B))= (R_(A))/(R_(B))` From Eq (i), `(g_(A))/(g_(B))= (3)/(1)`