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

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 $g = \frac{GM}{R^2}$ and $M = \frac{4}{3}\pi R^3 \rho$. Substituting $M$, we get $g = \frac{G}{R^2} \times \frac{4}{3}\pi R^3 \rho = \frac{4}{3}\pi G \rho R$. Rearranging for density gives $\rho = \frac{3g}{4\pi GR}$.