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Two spherical planets of radii R and 2R, having masses M and 9M respectively, are fixed with their centres separated by 8R. A satellite of mass m is launched from the surface of the planet of mass M straight toward the centre of the heavier planet. The minimum launch speed v needed for the satellite to just reach the surface of the second planet is sqrt(aGM/(7R)). Find the value of a.
- a = 4
- a = 2
- a = 6
- a = 8
Correct answer: a = 4
Solution
Neutral point at distance x from planet M: GM/x² = 9GM/(8R - x)² => 8R - x = 3x => x = 2R (so 6R from the heavy planet). Launch surface is at R from M and 7R from 9M. Energy conservation: (1/2)v² - GM/R - 9GM/(7R) = -GM/(2R) - 9GM/(6R). RHS = -GM/(2R) - 3GM/(2R) = -2GM/R. So (1/2)v² = -2GM/R + GM/R + 9GM/(7R) = -GM/R + 9GM/(7R) = 2GM/(7R). Hence v² = 4GM/(7R), giving a = 4.
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