Mean Density of the Earth – Rankers Physics

Acceleration Due to Gravity and its variation: Practice Problem & Solution

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 Explained:

To solve this problem, we apply the core principles of Acceleration Due to Gravity and its variation. Understanding the underlying formula is key to arriving at the correct answer below:

We know \(g = \frac{G M}{R^2} = \frac{G}{R^2} \left(\frac{4}{3}\pi R^3 \rho\right) = \frac{4}{3}\pi RG\rho\). Solving for density, \(\rho = \frac{3g}{4\pi RG}\).

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