JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 23)
A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the
orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape
velocity from the planet is:
2
1
$$\sqrt 2 $$
$${1 \over {\sqrt 2 }}$$
Explanation
$${V_0} = \sqrt {{{GM} \over r}} $$
$${V_e} = \sqrt {{{2GM} \over r}} $$
$$ \therefore $$ $${{{V_0}} \over {{V_e}}} = \sqrt {{{GM} \over r} \times {{2GM} \over r}} = {1 \over {\sqrt 2 }}$$
$${V_e} = \sqrt {{{2GM} \over r}} $$
$$ \therefore $$ $${{{V_0}} \over {{V_e}}} = \sqrt {{{GM} \over r} \times {{2GM} \over r}} = {1 \over {\sqrt 2 }}$$
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