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

1 Answer
Mar 15, 2018

#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#