Question

The acceleration due to gravity on moon is `(1/6)^"th"` times the acceleration due to gravity on earth. If the ratio of the density of earth 'ρ_e' to the density of moon 'ρ_m' is `5/3`, then the radius of moon 'R_m' in terms of the radius of earth 'R_e' is ______.

Options

• `(3/18)R_e`

• `(1/(2sqrt3))R_e`

• `(5/18)R_e`

• `(7/6)R_e`

Answer 1

The acceleration due to gravity on the moon is `(1/6)^"th"` times the acceleration due to gravity on earth. If the ratio of the density of earth 'ρe' to the density of moon 'ρm' is `5/3`, then the radius of moon 'Rm' in terms of the radius of the earth 'Re' is `underline((5/18)R_e)`.

Explanation:

`g_m/g_e=1/6`      `rho_e/rho_m=5/3`    `R_m/R_e=?`

`g_m/g_e=M_m/R_m^2xxR_e^2/M_e=(4/3piR_m^3rho_m)/R_m^2xxR_e^2/(4/3piR_e^3rho_e)=(R_mrho_m)/(R_erho_e)=1/6`

`therefore R_m/R_e=1/6xxrho_e/rho_m=1/6xx5/3=5/18`