# The density of the core of a planet is ρ1 and that of the outer shell is ρ2, the radii of the core and that of the planet….

Q: The density of the core of a planet is ρ1 and that of the outer shell is ρ2, the radii of the core and that of the planet are R and 2R respectively. The acceleration due to gravity at the surface of the planet is same as at a depth R. Find the ratio of ρ12

(a) 3/7

(b) 7/4

(c) 4/7

(d) 7/3

Ans: (d)

Sol: Let M = mass of planet
$\displaystyle M = \frac{4}{3}\pi R^3 \rho_1 + \frac{4}{3}\pi[(2R)^3 – R^3]\rho_2$

$\displaystyle M = \frac{4}{3}\pi R^3 (\rho_1 + 7\rho_2 )$

At the surface ,

$\displaystyle g_1 = \frac{GM}{(2R)^2}$

$\displaystyle =\frac{\pi G R}{3}(\rho_1 + 7\rho_2 )$

Mass of the core is

$\displaystyle m = \frac{4}{3}\pi R^3 \rho_1$

At depth R ,

$\displaystyle g_2 = \frac{Gm}{(R)^2}$

$\displaystyle = \frac{4}{3}\pi G R \rho_1$

As Acceleration due to gravity is same

$\displaystyle g_1 = g_2$

$\displaystyle \frac{\pi G R}{3}(\rho_1 + 7\rho_2 ) =\frac{4}{3}\pi G R \rho_1$

on solving ,
$\displaystyle 3\rho_1 = 7\rho_2$

$\displaystyle \frac{\rho_1}{\rho_2}=\frac{7}{3}$