Question

The ratio of the radii of a planet and the earth is \( 1:2 \), the ratio of their mean densities is \( 4:1 \). If the acceleration due to gravity on the surface of the earth is \( 9.8 \, ms^{-2} \), then the acceleration due to gravity on the surface of the planet is?

• A. \( 4.9 \, ms^{-2} \)
• B. \( 8.9 \, ms^{-2} \)
• C. \( 29.4 \, ms^{-2} \)
• D. \( 19.6 \, ms^{-2} \)

Hint:

The acceleration due to gravity depends on both the radius and density of a celestial body. When comparing different planets, use the ratio formula \( g \propto \rho R \).

Answer 1

The Correct Option is D

The acceleration due to gravity on the surface of a planet is given by: \[ g = \frac{4}{3} \pi G \rho R \] where: - \( G \) is the gravitational constant, - \( \rho \) is the mean density of the planet, - \( R \) is the radius of the planet. Taking the ratio of gravitational accelerations of the planet (\( g_p \)) and earth (\( g_e \)): \[ \frac{g_p}{g_e} = \frac{\rho_p}{\rho_e} \times \frac{R_p}{R_e} \] Given: \[ \frac{R_p}{R_e} = \frac{1}{2}, \quad \frac{\rho_p}{\rho_e} = 4 \] \[ \frac{g_p}{g_e} = 4 \times \frac{1}{2} = 2 \] Since \( g_e = 9.8 \, ms^{-2} \), \[ g_p = 2 \times 9.8 = 19.6 \, ms^{-2} \] Thus, the acceleration due to gravity on the planet is \( 19.6 \, ms^{-2} \).