Question

Two planets \( A \) and \( B \) have radii \( R \) and \( 1.5R \), and densities \( \rho \) and \( \frac{\rho}{2} \) respectively. The ratio of acceleration due to gravity at the surface of \( B \) to \( A \) is:

• A. \( 2:3 \)
• B. \( 2:1 \)
• C. \( 3:4 \)
• D. \( 4:3 \)

Hint:

For a spherical planet: \[ g \propto \rho R \] - Increasing radius increases gravity. - Increasing density increases gravity. - The ratio \( \frac{g_B}{g_A} \) is found using \( g \propto \rho R \).

Answer 1

The Correct Option is C

Step 1: Formula for acceleration due to gravity. The surface gravity of a planet is given by: \[ g = \frac{G M}{R^2} \] Since mass \( M \) is related to density \( \rho \) and radius \( R \): \[ M = \rho V = \rho \times \frac{4}{3} \pi R^3 \] Substituting into the gravity formula: \[ g = \frac{G (\rho \frac{4}{3} \pi R^3)}{R^2} \] \[ g = \frac{4}{3} G \pi \rho R \] Thus, gravity at the surface is: \[ g \propto \rho R \] Step 2: Finding the ratio \( \frac{g_B}{g_A} \). For planet A: \[ g_A \propto \rho R \] For planet B: - Radius \( R_B = 1.5R \), - Density \( \rho_B = \frac{\rho}{2} \), \[ g_B \propto \left(\frac{\rho}{2}\right) (1.5R) \] \[ g_B \propto \frac{3}{2} \times \frac{\rho R}{2} = \frac{3}{4} \rho R \] Taking the ratio: \[ \frac{g_B}{g_A} = \frac{\frac{3}{4} \rho R}{\rho R} = \frac{3}{4} \] Final Answer: \[ \boxed{3:4} \]