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Bouncing Ball Example

A particle is...

Question

A particle is projected from a point A that is at a distance of $4 R$ from the centre of the earth with speed $V_{1}$ in a direction making $30^{\circ}$ with the line joining the centre of the earth and point A as shown in figure. Consider the gravitational interaction only between these two bodies (use $\frac{G M}{R} = 6.4 \times 10^{7} m^{2} / s^{2}$ ). The speed $V_{1}$ if particle passes grazing the surface of the earth is:

2 \sqrt{2} \times 10^{3} m / s

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4 \sqrt{2} \times 10^{3} m / s

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4 \times 10^{3} m / s

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4 \sqrt{3} \times 10^{3} m / s

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Solution

The correct option is B $4 \sqrt{2} \times 10^{3} m / s$
Conserving angular momentum
$m V_{1} c o s 60^{\circ} \times 4 R = m V_{2} R$
$\frac{V_{2}}{V_{1}} = 2$
Conserving energy of the system
$\frac{- G M m}{4 R} + \frac{1}{2} m V_{1}^{2} = \frac{- G M m}{R} + \frac{1}{2} m V_{2}^{2}$
$\frac{1}{2} V_{2}^{2} - \frac{1}{2} V_{1}^{2} = \frac{3 G M}{4 R}$
or
$V_{1}^{2} = \frac{1}{2} \frac{G M}{R}$
$V_{1} = \frac{1}{\sqrt{2}} \sqrt{64 \times 10^{6}} = \frac{8000}{\sqrt{2}} m / s$

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