Question

Two planets A and B have the same average density. Their radii $R_A$ and $R_B$ are such that $R_A : R_B = 3:1$. If $g_A$ and $g_B$ are the acceleration due to gravity at the surfaces of the planets, then $g_A : g_B$ equals

• A. $3:1$
• B. $1:3$
• C. $9:1$
• D. $1:9$
• E. $\sqrt{3}:1$

Hint:

Remember: If density is constant, $g \propto R$. If mass is constant, $g \propto 1/R^2$. Always check which parameter is being held constant in the problem.

Answer 1

The Correct Option is A

Concept:
The acceleration due to gravity ($g$) at the surface of a planet of mass $M$ and radius $R$ is: \[ g = \frac{GM}{R^2} \] Since $M = \text{density} (\rho) \times \text{volume} (V)$ and planets are assumed to be spheres ($V = \frac{4}{3}\pi R^3$): \[ g = \frac{G (\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}\pi G \rho R \]

Step 1: Identify the relationship between $g$ and $R$.
For planets with the same average density ($\rho_A = \rho_B = \rho$): \[ g \propto R \] This is because $G$, $\pi$, and $\rho$ are all constants in this scenario.

Step 2: Calculate the ratio $g_A : g_B$.
Given the ratio of radii $R_A : R_B = 3:1$. \[ \frac{g_A}{g_B} = \frac{R_A}{R_B} \] \[ \frac{g_A}{g_B} = \frac{3}{1} \] Thus, $g_A : g_B = 3:1$.