Question

The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are \( \rho \) and \( \frac{\rho}{3} \) respectively. The ratio of acceleration due to gravity at their surfaces (i.e. \( g_A : g_B \)) will be:

• A. 2:1
• B. 2:3
• C. 3:4
• D. 4:3

Answer 1

The Correct Option is C

The acceleration due to gravity at the surface of a planet is given by: \[ g = \frac{GM}{R^2}. \] Using the relation \( M = \frac{4}{3} \pi R^3 \rho \), the acceleration \( g \) can be expressed as: \[ g = \frac{G \cdot \frac{4}{3} \pi R^3 \rho}{R^2} = \frac{4}{3} \pi R \rho G. \] For planet A, the acceleration due to gravity is: \[ g_A = \frac{4}{3} \pi R \rho G. \] For planet B, the radius is \( R_B = 1.5R \) and the density is \( \rho_B = \frac{\rho}{2} \). Therefore: \[ g_B = \frac{4}{3} \pi R_B \rho_B G = \frac{4}{3} \pi (1.5R) \left(\frac{\rho}{2}\right) G. \] Simplifying \( g_B \): \[ g_B = \frac{4}{3} \pi (1.5R) \cdot \frac{\rho}{2} \cdot G = \frac{4}{3} \pi R \rho G \cdot 1.5 \cdot \frac{1}{2}. \] The ratio of \( g_B \) to \( g_A \) is: \[ \frac{g_B}{g_A} = \frac{(1.5)^2}{1} \cdot \frac{\frac{\rho}{2}}{\rho} = \frac{(1.5)^2}{2} = \frac{2.25}{2} = \frac{3}{4}. \] Thus, the ratio of acceleration due to gravity at the surface of B to A is \( \boxed{3:4} \).

Answer 2

We are given that the radii of two planets 'A' and 'B' are \( R \) and \( 4R \), and their densities are \( \rho \) and \( \frac{\rho}{3} \), respectively. The formula for the acceleration due to gravity at the surface of a planet is: \[ g = \frac{4\pi G R \rho}{3} \] Since gravity is proportional to both the radius and density, the ratio of acceleration due to gravity at their surfaces can be written as: \[ g_A : g_B = \frac{\frac{4\pi G R \rho}{3}}{\frac{4\pi G (4R) (\frac{\rho}{3})}{3}} = \frac{R \cdot \rho}{(4R) \cdot \frac{\rho}{3}} = \frac{1}{4} \cdot 3 = \frac{3}{4} \] Thus, the correct ratio is \( g_A : g_B = 3:4 \).