Question

Considering earth to be a sphere of radius $R$ having uniform density $\rho$, then value of acceleration due to gravity $g$ in terms of $R,\rho$ and $G$ is

• A. $g=\sqrt{\dfrac{3\pi R}{\rho G}}$
• B. $g=\sqrt{\dfrac{4}{3}\pi \rho G R}$
• C. $g=\dfrac{4}{3}\pi \rho G R$
• D. $g=\dfrac{GM}{\rho R^2}$

Hint:

Physics Tip : For same density bodies, surface gravity is proportional to radius.

Answer 1

The Correct Option is C

Step 1: Use formula for gravity at surface. For a spherical body: $$ g=\frac{GM}{R^2} $$ where $M$ is mass of Earth.

Step 2: Write mass using density. Mass of sphere: $$ M=\rho \times \text{Volume} $$ Volume of sphere: $$ V=\frac{4}{3}\pi R^3 $$ Hence, $$ M=\rho\left(\frac{4}{3}\pi R^3\right) $$

Step 3: Substitute in formula of $g$. $$ g=\frac{G\left(\frac{4}{3}\pi \rho R^3\right)}{R^2} $$

Step 4: Simplify. $$ g=\frac{4}{3}\pi \rho G R $$

Step 5: Conclusion. $$ \therefore \text{Correct option is (C).} $$