The Earth may be regarded as a spherically shaped uniform core of density ρ1 and radius R2 surrounded by a uniform shell of thickness R2 and density ρ2 . Find the ratio of ρ1ρ2 if the value of acceleration due to gravity is the same at surface as at depth R2 from the surface

Total mass of earthM=43π(R2)3ρ1+43π(R3−R38)ρ2M=4πR324(ρ1+7ρ2)Acceleration due to gravity at earth's surface =GMR2=4πGR24(ρ1+7ρ2)Acceleration due to gravity at depth R2 from the surface is =G[43π(R2)2ρ1](R2)2=4πGρ1R6Now according to question4πGR(ρ1+7ρ2)24=4πGRρ16⇒ρ1ρ2=73

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The Earth may be regarded as a spherically shaped uniform core of density $ρ_{1}$ and radius $\frac{R}{2}$ surrounded by a uniform shell of thickness $\frac{R}{2}$ and density $ρ_{2}$ . Find the ratio of $\frac{ρ_{1}}{ρ_{2}}$ if the value of acceleration due to gravity is the same at surface as at depth $\frac{R}{2}$ from the surface

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answer is 2.33.

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Detailed Solution

Total mass of earth
$\begin{array}{l} M = \frac{4}{3} π {(\frac{R}{2})}^{3} ρ_{1} + \frac{4}{3} π (R^{3} - \frac{R^{3}}{8}) ρ_{2} \\ M = \frac{4 {πR}^{3}}{24} (ρ_{1} + 7 ρ_{2}) \end{array}$
Acceleration due to gravity at earth's $surface = \frac{GM}{R^{2}} = \frac{4 πGR}{24} (ρ_{1} + 7 ρ_{2})$

Acceleration due to gravity at depth $\frac{R}{2}$ from the surface is
$= \frac{G [\frac{4}{3} π {(\frac{R}{2})}^{2} ρ_{1}]}{{(\frac{R}{2})}^{2}} = \frac{4 {πGρ}_{1} R}{6}$
Now according to question
$\frac{4 πGR (ρ_{1} + 7 ρ_{2})}{24} = \frac{4 {πGRρ}_{1}}{6} \Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{7}{3}$