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

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