question_answer

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 - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

A) \[m{{v}^{2}}\]                                               

B) \[\frac{1}{2}m{{v}^{2}}\]

C) \[\frac{3}{2}m{{v}^{2}}\]                  

D)   \[2\,m{{v}^{2}}\]

Correct Answer: A

Solution :

KE of revolving particle = \[\frac{1}{2}\,m{{v}^{2}}\] Potential energy =- 2KE \[=\,\,-m{{v}^{2}}\] for escape, out, total mechanical energy of particle should become zero. \[KE+PE=0\] \[KE-m{{v}^{2}}=0\] \[KE=m{{v}^{2}}\]