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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A satellite revolving roung the earth with orbital speed $v_0$, If it stops suddenly, the speed with which it will strike the surface of the earth would be (-ve escape velocity of the particle on earth's surface)

\[(a)\;\frac{v_e^2}{v_0} \quad (b)\; v_0 \quad (c)\;\sqrt {v_e^2-v_0^2} \quad (d)\;\sqrt {v_e^2-2 v_0^2}\]
Can you answer this question?
 
 

1 Answer

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Let r be the radius of satellite then,
$v_0^2 =\large\frac{GM}{r}$
Applying conservation of Medical energy between points A (the satellite position when its suddenly stops ) to point B on the surface of earth where it strikes with velocity v
$KE$=change in potential energy
$\large\frac{1}{2}$$mv^2 =\large\frac{-GMm}{r}-\bigg(\large\frac{-GMm}{R}\bigg)$
$v^2=\large\frac{2GM}{R}-\frac{2GM}{r}$
$\qquad= v_e^2 -2 v_0^2$
$v=\sqrt {v_e^2-2v_0^2}$
Hence d is the correct answer.

 

answered Aug 30, 2013 by meena.p
edited Feb 17, 2014 by meena.p
 

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