If $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth surface. Then the…
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
- $\frac{3 g}{4 \pi R G}$
- $\frac{4 \pi G}{3 g R}$
- $\frac{\pi R G}{12 g}$
- $\frac{3 \pi R}{4 g G}$
Solution
Acceleration due to gravity on the earth surface $g=\frac{G M}{R^2}$
$\begin{aligned} & \frac{g=G \frac{4}{3} \pi R^3 \rho}{R^2} \\ & g=G \frac{4}{3} \pi R \rho \\ & \Rightarrow \rho=\frac{3 g}{4 G \pi R}\end{aligned}$
Asked in: NEET 2023 (Manipur)
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