Physics · Acceleration due to gravity and its variation with altitude and depth
Suppose that the acceleration of a free fall at the surface of a distant planet
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:
- A. 4: 1
- B. 2: 1
- C. 1: 1
- D. 1: 2
Step-by-step solution
At the surface, g = GM/R². With M = (4/3)πR³ρ, we get g = (4/3)πGρR. Given g_planet = g_earth and R_planet = 2R_earth, we have ρ_planet * (2R_earth) = ρ_earth * R_earth, so ρ_planet/ρ_earth = 1/2.
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