JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 19)
A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -
mv2
$${1 \over 2}$$ mv2
$${3 \over 2}$$ mv2
2 mv2
Explanation
At height r from center of earth. orbital velocity
= $$\sqrt {{{GM} \over r}} $$
$$ \therefore $$ By energy conservation
KE of 'm' + $$\left( { - {{GMm} \over r}} \right)$$ = 0 + 0
(At infinity, PE = KE = 0)
$$ \Rightarrow $$ KE of 'm' = $${{{GMm} \over r}}$$ = $${\left( {\sqrt {{{GM} \over r}} } \right)^2}$$ m = mv2
= $$\sqrt {{{GM} \over r}} $$
$$ \therefore $$ By energy conservation
KE of 'm' + $$\left( { - {{GMm} \over r}} \right)$$ = 0 + 0
(At infinity, PE = KE = 0)
$$ \Rightarrow $$ KE of 'm' = $${{{GMm} \over r}}$$ = $${\left( {\sqrt {{{GM} \over r}} } \right)^2}$$ m = mv2
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