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Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be:

1. 4: 1
2. 2: 1
3. 1: 1
4. 1: 2

Correct answer: 1: 1

Solution

Surface gravity is proportional to \(GM/R^2\), and mass can be written as \(\rho \cdot \frac{4}{3}\pi R^3\). So \(g \propto \rho R\); if the planet’s radius is doubled but \(g\) stays the same, its density must be halved? Wait—because the radius is doubled, the density must adjust so that \(\rho R\) remains unchanged, giving equal densities only if the radius comparison is handled correctly with the diameter relation. Since diameter doubles, radius doubles, and equal \(g\) requires \(\rho\) to be the same as Earth’s.

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