JEE Advance - Physics (2013 - Paper 2 Offline - No. 2)
Two bodies, each of mass M, are kept fixed with a separation $$2L$$. A particle of mass m is projected from
the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The
correct statement(s) is (are)
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $$4\sqrt {{{GM} \over L}} $$
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $$2\sqrt {{{GM} \over L}} $$
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $$\sqrt {{{2GM} \over L}} $$
The energy of the mass m remains constant.
คำอธิบาย
Here we will use the law of conservation of energy as there are no non-conservative force involved so the total energy of mass m remains constant.
For minimum initial velocity to escape the gravitational field, the velocity at infinity will be zero.
And we know PE at infinity is zero.
By energy conservation,
$$P{E_i} + K{E_i} = P{E_f} + K{E_f}$$
$$ \Rightarrow {{ - GMm} \over L} - {{GMm} \over L} + {1 \over 2}m{v^2} = 0 + {1 \over 2}m{(0)^2}$$
$$ \Rightarrow - {{2Gm} \over L}M + {1 \over 2}m{v^2} = 0$$
$$ \Rightarrow {1 \over 2}{V^2} = {{2Gm} \over L}$$
$$ \Rightarrow V = 2\sqrt {{{GM} \over L}} $$
Hence, options (B) and (D) are correct.