Question

A planet has a core and on outer shell of radii $R$ and $2R$ respectively. The density of the core is $x$ and that of outer shell is $y$. The acceleration due to gravity at the surface of planet is same as that at depth $R$. The ratio of $x$ and $y$ is $\frac{n}{3}$. Find $n$.

Answer 1

Calculating the acceleration due to gravity at depth R

Consider a spherical shell of radius r and thickness $dr$

Mass of this shell, $m=4\pi r^{2}x\partial r$

acceleration due to this mass at distance R from center = $\partial a=\frac{G4\pi r^{2}x\partial r}{R^2}$

$a_R={\int }_0^R\frac{G4\pi r^{2}x\partial r}{R^2}=\frac{4\pi xGR}{3}$

Calculating acceleration due to gravity at surface

acceleration due to this mass at distance $2$R from center = $\partial a=\frac{G4\pi r^{2}x\partial r}{4R^2}$

$a_{2R}={\int }_0^R\frac{G4\pi r^{2}x\partial r}{4R^2}+{\int }_R^{2R}\frac{G4\pi r^{2}x\partial r}{4R^2}=\frac{4\pi xGR}{4\ast 3}+\frac{4\pi yG}{4R^2}[\frac{(2R{)}^2}{3}-\frac{R^3}{3}]$

Therefore:

$\frac{a_R}{a_{2R}}=\frac{1}{1}$

=> $\frac{x}{y}=\frac{7}{3}$

So, $n=7$