The free-fall acceleration at the surface of planet 1 is 22 m/s2. The radius and the mass of planet 2 are twice those of planet 1.
What is the free-fall acceleration on planet 2?
1 Answer
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g1 = G M1/R1^2
g2 = G M2/R2^2
g2/g1 = [M2/M1] [R1/R2]^2
g2/g1 = [2 M1/M1] [R1/2 R1]^2
g2/g1 = [2] [1/4] = 1/2
g2 = g1/2 = 22/2
g2 = 11 m/s^2