question_answer

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 0. A meteorite passes through 0 moving perpendicular to the star/s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at 0 is-  (Take       Gravitational      constant;             \[G=6.67\times {{10}^{-11}}\,N{{m}^{2}}k{{g}^{-}}^{2})\] [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

A) \[2.4\times {{10}^{4}}m/s\]     

B) \[1.4\times {{10}^{5}}m/s\]

C) \[3.8\times {{10}^{4}}m/s\]

D) \[2.8\times {{10}^{5}}m/s\]

Correct Answer: D

Solution :

When total mechanical energy of meteorite become 0 then meteorite will escape out \[\frac{1}{2}m{{v}^{2}}+m\,\left[ -\frac{GM}{r}-\frac{GM}{r} \right]\,\,=\,\,0\] \[{{v}^{2}}\,=\,\frac{4GM}{r}\,=\,\frac{4\times 6.67\times 3\times {{10}^{-11}}\times {{10}^{31}}}{{{10}^{11}}}\] \[v\,\,=\,\,2.8\times {{10}^{5}}\,m/s\]