Question 145 of 223

Two stars of masses $3\times {10}^{31}kg$ each, and at distance $2\times {10}^{11}m$ rotate in a plane about their common centre of mass $O.$ A meteorite passes through $O$ moving perpendicular to the stars,s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at $O$ is ( Take Gravitational constant $G=6.67\times {10}^{-11}N{m}^{2}{kg}^{-2}$)

A$2.4\times {10}^{4}m{s}^{-1}$B$3.8\times {10}^{4}m{s}^{-1}$C$2.8\times {10}^{5}m{s}^{-1}$D$1.4\times {10}^{5}m{s}^{-1}$

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Answer: (C) $2.8\times {10}^{5}m{s}^{-1}$

To escape, the total energy of small particle must be zero.



$TE=KE+U$

$0=\dfrac{1}{2}m{V}^{2}+\left(\dfrac{-GMm}{\dfrac{d}{2}}\right)$$\times 2$

$\Rightarrow V=\sqrt{\dfrac{8GM}{d}}$

$=\sqrt{\dfrac{8\times 6.67\times {10}^{-11}\times 3\times {10}^{31}}{2\times {10}^{11}}}$

$=2.8\times {10}^{5}m{s}^{-1}$