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 $rho/rho_2$ . ?

Question

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

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Answer 1 · 2018-03-15T13:25:48

$3$

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, $(rho_1)/(rho_2)=7/3$