JEE Advance - Physics (2011 - Paper 2 Offline - No. 5)
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
$${1 \over 2}m{V^2}$$
$$m{V^2}$$
$${3 \over 2}m{V^2}$$
$$2m{V^2}$$
คำอธิบาย
A particle escapes from the gravitational pull if its total energy (T) i.e., sum of kinetic energy (K) and potential energy (U), is greater than or equal to zero. The condition for just escape T = K + U = 0 i.e.,
K = $$-$$ U. .............. (1)
In a circular orbit of radius r, gravitational attraction provides the centripetal acceleration, mV2/r = GMm/r2, which gives
$$r = {{GM} \over {{V^2}}}$$ ........ (2)
From equations (1) and (2), the kinetic energy of the particle at the time of injection is given by
$$K = - U = - \left( { - {{GMm} \over r}} \right) = {{GMm} \over {GM/{V^2}}} = m{V^2}$$.