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  4. Maximum height reached by a rocket fired with a speed equal to 50 % of the escape velocity from earth’s surface is:

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Q. Maximum height reached by a rocket fired with a speed equal to $50\%$ of the escape velocity from earth’s surface is:

Gravitation

Solution:

$V = \frac{50}{100} V_{e} = \frac{1}{2}\sqrt{\frac{2GM}{R}} $
Apply energy conservation
$\Rightarrow -\frac{GMm}{R} + \frac{1}{2}mV^{2} = -\frac{GMm}{\left(R + h\right)}$
$ v^{2} = \frac{2GM}{R} - \frac{2GM}{R+h} $
$\frac{1}{4} \cdot \frac{2GM}{R} = 2GM\left(\frac{1}{R} - \frac{1}{R+h}\right)$
$ \therefore \frac{1}{4R} = \frac{h}{R\left(R+h\right)}$
$ \therefore R + h = 4h $
$\Rightarrow h = \frac{R}{3}$