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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

$a)\; 12 km/s \\b)\; 1 km/s \\c)\; 2\sqrt 2 km/s \\ d)\;\sqrt 2 km/s$

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.

edited Feb 17, 2014 by meena.p