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If $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be
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The correct answer is:
$\frac{3 g}{4 \pi R G}$
Acceleration due to gravity on the earth surface $g=\frac{G M}{R^2}$
$\begin{aligned} & \frac{g=G \frac{4}{3} \pi R^3 \rho}{R^2} \\ & g=G \frac{4}{3} \pi R \rho \\ & \Rightarrow \rho=\frac{3 g}{4 G \pi R}\end{aligned}$
$\begin{aligned} & \frac{g=G \frac{4}{3} \pi R^3 \rho}{R^2} \\ & g=G \frac{4}{3} \pi R \rho \\ & \Rightarrow \rho=\frac{3 g}{4 G \pi R}\end{aligned}$
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