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Standard XII

Physics

Escape Speed

A projectile ...

Question

A projectile is fired from the surface of earth of radius $R$ with a velocity $k v_{e}$ (where $v_{e}$ is the escape velocity from surface of earth and $k < 1)$ Neglecting air resistance, the maximum height of rise from the centre of earth is

\frac{R}{k^{2} - 1}

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k^{2} R

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\frac{R}{1 - k^{2}}

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k R

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Solution

The correct option is D $\frac{R}{1 - k^{2}}$
At maximum height h, KE of the projectile will be zero.
Using conservation of energy, $K E_{i} + P E_{i} = K E_{f} + P E_{f}$
or $\frac{1}{2} m (k v_{e})^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{h}$

or $\frac{1}{2} k^{2} (2 g R) - \frac{G M}{R} + \frac{G M}{h} = 0$ where escape velocity, $v_{e} = \sqrt{2 g R}$

or $g R k^{2} - g R + g R^{2} / h = 0$ where $g = \frac{G M}{R^{2}}$

or $k^{2} - 1 + R / h = 0 \Rightarrow h = \frac{R}{1 - k^{2}}$

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