At height r from the centre of the Earth, orbital velocity is given as:
orbitalvelocity=rGM
Ve=2v
K.E=21m(2v)2=mv2
JEE Main 2019 — Physics Mechanics
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:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
23mv2
mv2
21mv2
2mv2
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Two masses $m$ and 2 m are connected by a light string going over a pulley (disc) of mass 30 m with radius $r=0.1 \mathrm{~m}$. The pulley is mounted in a vertical plane and it is free to rotate about its axis. The 2 m mass is released from rest and its speed when it has descended through a height of 3.6 m is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. (Assume string does not slip and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)
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