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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.

Option 1)
$\sqrt{2gR}$

Option 2)
$\sqrt{gR}$

Option 3)
$\sqrt{gR/2}$

Option 4)
$\sqrt{gR}(\sqrt{2}-1)$

Answers (1)

As we learnt in

Orbital velocity of satellite -

$V=\sqrt{\frac{GM}{r}}$

$r=R+h$

$r\rightarrow$ Position of satellite from the centre of earth

$V\rightarrow$ Orbital velocity

- wherein

The velocity required to put the satellite into its orbit around the earth.

and

Gravitational potential energy at height 'h' -

$U_{h}=-\frac{GMm}{R+h}$

$U_{h}=-\frac{gR^{2}m}{R+h}$

$U_{h}\rightarrow$ Potential energy at height $h$

$R\rightarrow$ Radius of earth

- wherein

$U_{h}=-\frac{mgR}{1+\frac{h}{R}}$

$v_0=\sqrt{\frac{Gm}{R+h}}$

$\frac{1}{2}mv_e^2 = {\frac{Gmm}{R+h}} \:\Rightarrow v_e=\sqrt{\frac{2Gm}{R+h}} =\sqrt{\frac{2Gm}{R}} (\because h< < R)$

required increment in the orbital velocity

$v_e-v_0=\sqrt{\frac{2Gm}{R}}-\sqrt{\frac{Gm}{R}} = \sqrt{\frac{Gm}{R}} \left ( \sqrt{2}-1 \right )$

$v_e-v_0=\sqrt{gR}\sqrt{2}-1$

Option 1)

$\sqrt{2gR}$

Incorrect

Option 2)

$\sqrt{gR}$

Incorrect

Option 3)

$\sqrt{gR/2}$

Incorrect

Option 4)

$\sqrt{gR}(\sqrt{2}-1)$

Correct

Posted by

Vakul

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