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Standard X

Physics

Potential Energy of an Object at a Height

A projectile ...

Question

A projectile is fired vertically upwards from he surface of the earth with a velocity $K v_{e}$ , where $v_{e}$ is the escape velocity and $k < 1$ . If R is the radius of the earth, the maximum height to which it will rise, measured from the centre of the earth, will be (neglect air resistance)

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

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

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R (1 - k)^{2}

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

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Solution

The correct option is B $\frac{R}{1 - k^{2}}$
If the projectile is to escape from earth at $v_{e}$ , then
$\frac{1}{2} m v_{e}^{2} = \frac{G M m}{R}$
If the projectile is fired with $k v_{e}$ , conservation of energy would give,
$\frac{1}{2} m (k v_{e})^{2} - \frac{G M m}{R} = - \frac{G M m}{h}$
Using above value of $v_{e}$ ,
$h = \frac{R}{1 - k^{2}}$

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