Acceleration due to gravity on moon is `1//6` of the acceleration due to gravity on earth. If the ratio of densities of earth `(rho_(e))` and moon `(rho_(m))` is `((rho_(e))/(rho_(m)))=5/3` then radius of moon `(R_(m))` in terms of `R_(e)` will be

Options

A. `5/18R_(e)`
B. `1/6R_(e)`
C. `7/18R_(e)`
D. `-1/(2sqrt(3))R_(e)`

Answer

A

Explanation

Acceleration due to gravity `g=4/3pi rho GR :. g prop rhoR`
or `(g_(m))/(g_(e))=(rho_(m))/(rho_(e)).(R_(m))/(R_(e))`
`[As (g_(m))/(g_(e))=1/6, (rho_(e))/(rho_(m))=5/3 (given)]`
`:. (R_(m))/(R_(e))=((g_(m))/(g_(e)))((rho_(e))/(rho_(m)))=1/6xx5/3 :. R_(m)=5/18R_(e)`