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 closed to(Neglect the effect of atmosphere.)

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Answer 1

For a satellite in orbit, the velocity can be evaluated from the following equation,
$\frac{G M m}{r ^{2}} = \frac{m v _{o}^{2}}{r}$
$v_{o} = \frac{G M}{r}$
Here, $r = R + h$

For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be $v_{e}$ .
$E = P E + K E = 0$
$\frac{- G M m}{r} + \frac{1}{2} m v_{e}^{2} = 0$
$v_{e} = \frac{2 G M}{r}$

Difference in velocities is given by
$Δ v = v_{e} - v_{o}$
$Δ v = \frac{G M}{r} (2 - 1)$
$Δ v \approx \frac{G M}{R} (2 - 1)$ as $h < < R$

But $g = \frac{G M}{R ^{2}}$

$∴ Δ v \approx g R (2 - 1)$

Answer 2

For a satellite in orbit, the velocity can be evaluated from the following equation,

\frac{G M m}{r ^{2}} = \frac{m v _{o}^{2}}{r}

v_{o} = \frac{G M}{r}

Here,

r = R + h

For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be

v_{e}

.

E = P E + K E = 0

\frac{- G M m}{r} + \frac{1}{2} m v_{e}^{2} = 0

v_{e} = \frac{2 G M}{r}

Difference in velocities is given by

Δ v = v_{e} - v_{o}

Δ v = \frac{G M}{r} (2 - 1)

Δ v \approx \frac{G M}{R} (2 - 1)

as

h < < R

But

g = \frac{G M}{R ^{2}}

∴ Δ v \approx g R (2 - 1)