JEE Main & Advanced
JEE Main Paper (Held On 10-Jan-2019 Evening)
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 \[\frac{1}{2}m{{v}^{2}}+m\,\left[ -\frac{GM}{r}-\frac{GM}{r} \right]\,\,=\,\,0\] \[{{v}^{2}}\,=\,\frac{4GM}{r}\,=\,\frac{4\times 6.67\times 3\times {{10}^{-11}}\times {{10}^{31}}}{{{10}^{11}}}\] \[v\,\,=\,\,2.8\times {{10}^{5}}\,m/s\]