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 $$\underline{\hspace{2cm}}\,g.$$

Correct Answer: 9

If diameter reduced to 1/3, radius R/3, mass same. g' = GM/(R/3)² = 9GM/R² = 9g.

The answer is 9.

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