Question

The acceleration due to gravity on moon is \( \frac{1}{6} \)th times the acceleration due to gravity on earth. If the ratio of the density of earth \( \rho_e \) to the density of moon \( \rho_m \) is \( \frac{5}{3} \), then the radius of moon \( R_m \) in terms of the radius of earth \( R_e \) is

• A. \( \frac{7}{6} R_e \)
• B. \( \frac{3}{18} R_e \)
• C. \( \frac{5}{18} R_e \)
• D. \( \frac{1}{2 \sqrt{3}} R_e \)

Hint:

In gravity-related problems, remember to relate mass and density, and use the formula \( g = \frac{GM}{R^2} \) along with the relationship between mass, density, and volume.

Answer 1

The Correct Option is C

Step 1: Using the relation between acceleration and radius.
The acceleration due to gravity is related to the radius and density of the planet by the formula: \[ g = \frac{G M}{R^2} \] Where \( M \) is the mass of the planet, \( R \) is the radius, and \( G \) is the gravitational constant. The mass of a planet is related to its density \( \rho \) by \( M = \rho V \), where \( V = \frac{4}{3} \pi R^3 \). Step 2: Comparing gravity on Earth and Moon.
The gravity on the moon is given by: \[ \frac{g_m}{g_e} = \frac{1}{6} \] Thus, comparing the gravity on Earth and the moon, we get: \[ \frac{R_m}{R_e} = \frac{5}{18} \] Thus, the radius of the moon is \( \frac{5}{18} R_e \), corresponding to option (C).