Question

The density of core of a planet is `rho_1` and that of outer shell is `rho_2`. The radius of core is R and that of planet is 2R. Gravitational field at outer surface of planet is same as at the surface of core what is the ratio `rho/rho_2`. ?

(1)3/4
(2)5/3
(3)7/3
(4)3/5

Answer 1

`3`

Explanation:

Suppose, mass of the core of the planet is `m` and that of the outer shell is `m'`

So,field on the surface of core is `(Gm)/R^2`

And,on the surface of the shell it will be `(G(m+m'))/(2R)^2`

Given,both are equal,

so, `(Gm)/R^2=(G(m+m'))/(2R)^2`

or, `4m =m+m'`

or, `m'=3m`

Now,`m=4/3 pi R^3 rho_1` (mass=volume `*` density)

and, `m'= 4/3 pi ((2R)^3 -R^3) rho_2=4/3 pi 7R^3 rho_2`

Hence,`3m=3(4/3 pi R^3 rho_1)=m'=4/3 pi 7R^3 rho_2`

So,`rho_1 =7/3 rho_2`

or, `(rho_1)/(rho_2)=7/3`