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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

Lokesh kumar , 10 Years ago
Grade 6
anser 2 Answers
Deepak Patra

Last Activity: 10 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

Last Activity: 5 Years ago

We know that initial potential energy = - 2 Kinetic energy.
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

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