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If the radius of a planet is reduced by n% while its mass remains constant, the surface gravity increases by 2n%. Which of the following supports this relationship?

1. g = GM/R^2, where M = ρ × (4/3)πR^3
2. g = (4/3)πρR
3. At the planet's surface, g_p = (4/3)G(2ρ)πR
4. At Earth's surface, g_e = (4/3)GρπR

Correct answer: g = GM/R^2, where M = ρ × (4/3)πR^3

Solution

The acceleration due to gravity on the surface of a planet is given by g = GM/R^2. If the radius decreases by n%, the new radius becomes R' = R(1 - n/100). Substituting this into the formula for g, we find that g increases by approximately 2n% for small values of n. Thus, option A is correct.

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