A satellite is moving with a constant speed $^{'} v^{'}$ in a circular orbit about 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 its ejection, the kinetic energy of the object is

Question

A satellite is moving with a constant speed

^{'} v^{'}

in a circular orbit about 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 its ejection, the kinetic energy of the object is

Toppr Admin · Accepted Answer

At height r from center of earth- orbital velocity-x221A-GMr-x2234- By energy conservationKE of -x2032-m-x2032-x2212-GMmr-0-0-At infinity- PE-KE-0-x21D2-KE of -x2032-m-x2032-GMmr-x221A-GMr-2m-mv2-

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 is32mv2mv22mv212mv2

At height r from center of earth. orbital velocity=√GMr∴ By energy conservationKE of ′m′+(−GMmr)=0+0(At infinity, PE=KE=0)⇒KE of ′m′=GMmr=(√GMr)2m=mv2.

At height $r$ from center of earth. orbital velocity
$= \sqrt{\frac{G M}{r}}$
$∴$ By energy conservation
$K E$ of $^{'} m^{'} + (- \frac{G M m}{r}) = 0 + 0$
(At infinity, $P E = K E = 0)$
$\Rightarrow K E$ of $^{'} m^{'} = \frac{G M m}{r} = {(\sqrt{\frac{G M}{r}})}^{2} m = m v^{2}$ .