Question 214: Gravitation

Question

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 -

Accepted Answer

At height r from center of earth. orbital velocity
= $$\sqrt {{{GM} \over r}} $$
$$ 	herefore $$  By energy conservation
KE of 'm' + $$\left( { - {{GMm} \over r}} ight)$$ = 0 + 0
(At infinity, PE = KE = 0)
$$ \Rightarrow $$  KE of 'm' = $${{{GMm} \over r}}$$ = $${\left( {\sqrt {{{GM} \over r}} } ight)^2}$$ m = mv
2

Answer

mv2

Answer

$${1 \over 2}$$ mv2

Answer

$${3 \over 2}$$ mv2

Answer

2 mv2

Gravitation

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 -

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