Mean Density of Earth – Rankers Physics
Topic: Gravitation
Subtopic: Acceleration Due to Gravity and its variation

Mean Density of Earth

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
\(\frac{3g}{4\pi RG}\)
\(\frac{4\pi G}{3gR}\)
\(\frac{\pi RG}{12g}\)
\(\frac{3\pi R}{4gG}\)

Solution:

Acceleration due to gravity is \(g = \frac{GM}{R^2}\). Since mass \(M = \rho \times \frac{4}{3}\pi R^3\), we get \(g = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R^2} = \frac{4}{3}\pi \rho GR\). Rearranging gives \(\rho = \frac{3g}{4\pi RG}\).

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