Three planets of same density have radii `R_(1),R_(2)` and `R_(3)` such that `R_(1) = 2R_(2) = 3R_(3)`. The gravitational field at their respective surfaces are `g_(1), g_(2)` and `g_(3)` and escape velocities from their surfaces are `upsilon_(1),upsilon_(2)` and `upsilon_(3)`, then

Options

A. `g_1//g_2=1//2`
B. `g_1//g_3=3`
C. `v_1//v_2=2`
D. `v_1//v_3 =1//3`

Answer

B, C, D

Explanation

`v=sqrt(2qR)=sqrt((2GM)/R)=sqrt((2G)/R(rho4/3piR^3))=sqrt((8piGP)/3)R`
Therefore , `rArr v_1/v_2=R_1/R_2 =2 , g_1/g_2=3 , g_1/g_3=R_1/R_3=3 , v_1/v_3=R_1/R_3=3`
`g=(GM)/R^2 =G/R^2 (rho 4/3 piR^3) =(5piGP)/3 R`