Question

Acceleration due to gravity on the surface of earth is \( g \). If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is \_\_\_\_ g.

Hint:

Remember that the acceleration due to gravity depends inversely on the square of the radius of the Earth. When the radius decreases by a factor, the acceleration due to gravity increases by the square of that factor.

Answer 1

The acceleration due to gravity \( g \) on the surface of the Earth is given by the formula: \[ g = \frac{GM}{R^2}, \] where: - \( G \) is the gravitational constant, - \( M \) is the mass of the Earth, - \( R \) is the radius of the Earth. If the diameter is reduced to one third of its original value, the new radius \( R' \) becomes: \[ R' = \frac{R}{3}. \] Since mass \( M \) remains unchanged, the new acceleration due to gravity \( g' \) is: \[ g' = \frac{GM}{(R/3)^2} = \frac{GM}{R^2} \times 9 = 9g. \] Thus, the acceleration due to gravity increases by a factor of 9. Final Answer: \( \frac{g}{9} \).