Question

If the acceleration due to gravity on the surface of Earth is \( g \), then the acceleration due to gravity on a planet whose diameter is \( \frac{1}{3} \) of that of Earth and same mass as that of Earth is \( g' = ng \), where \( n \) is

Hint:

The acceleration due to gravity is inversely proportional to the square of the radius of the planet. So, if the radius decreases, the gravity increases by the square of the ratio.

Answer 1

The acceleration due to gravity \( g' \) on a planet is given by the formula: \[ g' = \frac{GM}{R^2} \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. For Earth: \[ g = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \] Now, for the other planet with the same mass as Earth but with a diameter \( \frac{1}{3} \) of Earth's diameter, the radius of the planet \( R' \) is: \[ R' = \frac{1}{3} R_{\text{Earth}} \] Substituting into the equation for acceleration due to gravity: \[ g' = \frac{GM_{\text{Earth}}}{\left(\frac{1}{3} R_{\text{Earth}}\right)^2} = \frac{GM_{\text{Earth}}}{\frac{1}{9} R_{\text{Earth}}^2} = 9 \times \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} = 9g \] Thus, the acceleration due to gravity on this planet is: \[ g' = 9g \] Therefore, the value of \( n \) is \( \frac{1}{9} \).