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 Earth's surface is calculated using the formula: \[g = \frac{GM}{R^2},\] where \( G \) is the gravitational constant, \( M \) is the Earth's mass, and \( R \) is the Earth's radius. If the diameter is reduced to one-third of its original size, the new radius \( R' \) becomes: \[R' = \frac{R}{3}.\] With the mass \( M \) remaining constant, the new acceleration due to gravity \( g' \) is: \[g' = \frac{GM}{(R/3)^2} = \frac{GM}{R^2} \times 9 = 9g.\] Therefore, the acceleration due to gravity increases by a factor of 9. Final Answer: \( 9g \).