Solution:
Using conservation of energy: \(-\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{r}\). Putting \(v = \frac{v_e}{2} = \sqrt{\frac{GM}{2R}}\) yields \(r = \frac{4R}{3}\), which is the distance from the earth's centre.
A rocket is fired vertically with a speed half of the escape speed from the earth’s surface. How far from the earth does the rocket go before returning to the earth? [Given mean radius of earth is \(R\)]
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