Question

The Earth may be regarded as a spherically shaped uniform core of density ${\rho }_1$ and radius $\frac{R}{2}$ surrounded by a uniform shell of thickness $\frac{R}{2}$ and density ${\rho }_2$. Find the ratio of $\frac{{\rho }_1}{{\rho }_2}$ if the acceleration due to gravity is the same at surface as at depth $\frac{R}{2}$ from the surface.

Answer 1

At surface:
$fiel{d}_1=\frac{G\left(\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3{P}_1\right)}{R^2}+\frac{G\frac{4}{3}\pi \left(R^3{\left(\frac{R}{2}\right)}^3\right)}{R^2}$
$=\frac{4\pi G}{3R^2}(\frac{R^3}{8}{P}_1+\frac{7R^3}{8}{P}_2)$
$=\frac{4\pi G}{3R^2}\times \frac{}{}(P_1+7P_2)$
$=\frac{\pi G}{6}(P_1+7P_2)$
At $r=\frac{P}{2}$
$Fiel{d}_2=G\frac{\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3{P}_1}{{\left(\frac{R}{2}\right)}^2}$
At $r=\frac{R}{2}$
$=\frac{4\pi GR{P}_1\times 4}{3\times 8}$
$=\frac{2}{3}\pi G{P}_{1}R$
$Fiel{d}_1=Fiel{d}_2$
$⟹\frac{\pi GR}{6}(P_1+7P_2)=\frac{2}{3}\pi G{P}_{1}R$
$⟹\frac{P_1+7P_2}{6}=\frac{2P_1}{2}$ $\frac{P_1}{P_2}=\frac{7}{3}$
$⟹P_1+7P_2=4P_1$
$⟹\frac{P_1}{P_2}+7=\frac{4P_1}{P_2}$
$\frac{3P_1}{P_2}=7$
$\frac{P_1}{P_2}=\frac{7}{3}$