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.

Solution

Physics sir · 3 steps

Mass stays the same, so g depends only on 1/R²: shrink R to a third and g grows nine times.

1. Write g. On the surface, g = GM/R². Here M does not change, so g depends only on the radius.
2. New radius. Diameter and radius scale together. If the diameter becomes one third, the radius also becomes R' = R/3.

   g' = GM/(R/3)² = 9 GM/R² = 9g

3. So. The new acceleration due to gravity is 9g, so the blank is 9.

The common slip: 3. Forgetting to square the radius gives 3, not 9. Since g ∝ 1/R², a radius three times smaller makes g nine times larger.

The idea to keep. With mass fixed, surface gravity goes as 1/R². Square the factor by which the radius changes, and remember that diameter and radius change by the same factor.

Answer from the NTA final answer key. Solution by PrepWiser.