Question icon
Grade 6Mechanics

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

Profile image of Lokesh kumar
12 Years agoGrade 6
Answers icon

2 Answers

Profile image of Deepak Patra
12 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
Profile image of Chitrangda dutta
7 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