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A straight tunnel is bored through the Earth along a diameter, and a ball is released from one end. Using the fact that the gravitational potential at the Earth's center is -3GM/2R, what speed does the ball have on reaching the Earth's center?

1. √(GR)
2. √(2GR)
3. √(5/2 GR)
4. √(7.1GR)

Correct answer: √(GR)

Solution

Released from rest at the surface, by energy conservation (1/2)v^2 = V_surface - V_centre = (-GM/R) - (-3GM/2R) = GM/(2R). Hence v^2 = GM/R and v = sqrt(GM/R) = sqrt(gR) (with g = GM/R^2); the stored value with an extra factor of 2 is incorrect.

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