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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

Deepak Patra
6 years ago

(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
Chitrangda dutta
28 Points
2 years ago
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) = 0
So, K total = 2 m v ^ 2