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The average density of earth in terms of $g$ , $G$ and $R$ is _______.

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\frac{4 g}{3 πGR}

\frac{3 G}{4 πgR}

\frac{3}{4 πGgR}

\frac{3 g}{4 πGR}

answer is D.

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Detailed Solution

The average density of earth in terms of

g

G

and

R

\frac{3 g}{4 πGR}

.
The value of acceleration due to gravity(g) on the surface of the earth is given as

g = \frac{GM}{R^{2}}

\Rightarrow M = \frac{g R^{2}}{G}

Volume of earth

= \frac{4}{3} π R^{3}

Density is given as

ρ = \frac{mass}{volume}

\Rightarrow ρ = \frac{\frac{g R^{2}}{G}}{\frac{4}{3} π R^{3}}

\Rightarrow ρ = \frac{3 g}{4 πGR}

Hence, the average density of the earth in terms of

g

G

and

R

\frac{3 g}{4 πGR}

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