A satellite is revolving in a circular orbit at a height $^{'} h^{'}$ from the earth's surface (radius of earth $R$ ; $h < < R$ ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to:(Neglect the effect of atmosphere.)

\sqrt{2 g R}

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\sqrt{g R}

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\sqrt{g R / 2}

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\sqrt{g R} (\sqrt{2} - 1)

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Solution

The correct option is D $\sqrt{g R} (\sqrt{2} - 1)$
height $= h$
No, the given formula,
Orbital velocity
$v = \frac{G M}{R + h} = \frac{G M}{R}$ as $h << R$
Velocity required to escape
$\frac{1}{2} m v^{' 2} = \frac{G M m}{R + h} v^{'} = \frac{2 G M}{R + h} = \frac{2 G M}{R} (h << R)$
$∴$ Increase in velocity
$v^{'} = v = \sqrt{\frac{2 G M}{R}} - \sqrt{\frac{G M}{R}} = \sqrt{2 g R} - \sqrt{g R} = \sqrt{g R} (\sqrt{2} - 1)$

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A satellite is revolving in a circular orbit at a height ′h′ from the earth's surface (radius of earth R;h < < R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to:(Neglect the effect of atmosphere.)

The correct option is D √gR(√2−1)height=hNo, the given formula,Orbital velocityv=GMR+h=GMR as h<<RVelocity required to escape12mv′2=GMmR+hv′=2GMR+h=2GMR(h<<R)∴ Increase in velocityv′=v=√2GMR−√GMR=√2gR−√gR=√gR(√2−1)