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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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The value of eacape velocity on a certain planet is $ 2 km/s$ then the value of arbital speed for a satellite arbiting close to its surface is

$ a)\; 12 km/s \\b)\; 1 km/s \\c)\; 2\sqrt 2 km/s \\ d)\;\sqrt 2 km/s $
Can you answer this question?

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initial velocity $=\large\frac{escape\; velocity}{\sqrt 2}$
$v_0=\large\frac{2}{\sqrt 2}$$=\sqrt 2 \;km/s$
Hence d is the correct answer. 


answered Aug 9, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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