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The escape velocity from a planet is vₑ. A particle starts from rest at a very large distance and falls toward the planet under gravity alone, then passes through a smooth tunnel bored through the planet's centre. What is its speed at the centre of the planet?
- sqrt(1.5)*vₑ
- vₑ/sqrt(2)
- vₑ
- zero
Correct answer: sqrt(1.5)*vₑ
Solution
Starting from rest at infinity (total energy 0), the speed at the centre is set by the gravitational potential there. For a uniform sphere V(centre) = -(3/2)(GM/R). Energy conservation: (1/2)v² = (3/2)(GM/R). Since vₑ² = 2GM/R, we get v² = 3GM/R = 1.5*vₑ², so v = sqrt(1.5)*vₑ.
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