Question 224: Gravitation

Question

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.)

Accepted Answer

Orbital velocity of satellite,
$${v_0} = \sqrt {{{GM} \over {R + h}}} $$
= $$\sqrt {{{GM} \over R}} $$ [ As $$h <  < R$$ then R + h = R ]
= $$\sqrt {gR} $$ [ As $$g = {{GM} \over {{R^2}}}$$ ]
Escape velocity
$${v_e} = \sqrt {{{2GM} \over R}} $$
= $$\sqrt {2gR} $$ [ As $$g = {{GM} \over {{R^2}}}$$ ]
$$	herefore$$ The minimum increase in its orbital velocity required to escape from the earth gravitational field
$$ = \sqrt {2gR}  - \sqrt {gR}  = \sqrt {gR} \left( {\sqrt 2  - 1} ight)$$

Answer

$$\sqrt{2 g R}$$

Answer

$$\sqrt{g R}$$

Answer

$$\sqrt{g R / 2}$$

Answer

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

Gravitation

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.)

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