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        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 (a) ½ mV2(b) mV2(c) 3/2 mV2(d) 2mV2
5 years ago

Deepak Patra
471 Points


(b)
V is the orbital velocity. If VC is the escape velocity then Ve
= v2 V . The kinetic energy at the time
of ejection

KE = ½ m V2e = ½ m (v2V)2 = mV2

5 years ago
Chitrangda dutta
28 Points
							We know that initial potential energy = - 2 Kinetic energy.$\inline U = 2. \frac{1}{2}(2m) v^{^{2}} = 2 mv^{^{2}}.$For escape velocity, U (initial) + K (total) = 0So, K total = 2 m v ^ 2

10 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions