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A body of mass m is launched from a height R with a velocity ve, sufficient for it to escape the gravitational influence at that distance. If the potential energy at height R is given as -GMm/(2R), determine the escape velocity ve.

1. ve = √(GM/R)
2. ve = √(2gR)
3. ve = v/√2
4. ve = v/(2√2)

Correct answer: ve = √(GM/R)

Solution

The escape velocity is derived from the condition that the total energy (kinetic + potential) of the body becomes zero at infinity. Using the given potential energy at height R, the escape velocity is ve = √(2GM/R). Since the potential energy is given as -GMm/(2R), the effective escape velocity simplifies to ve = √(GM/R).

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