JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 22)
Two stars of masses 3 $$ \times $$ 1031 kg each, and at distance 2 $$ \times $$ 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O 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 O is - (Take Gravitational constant; G = 6.67 $$ \times $$ 10–11 Nm2 kg–2)
2.4 $$ \times $$ 104 m/s
1.4 $$ \times $$ 105 m/s
3.8 $$ \times $$ 104 m/s
2.8 $$ \times $$ 105 m/s
Explanation
By energy convervation between 0 & $$\infty $$.
$$ - {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0$$
[M is mass of star m is mass of meteroite)
$$ \Rightarrow $$ v $$ = \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}$$m/s
$$ - {{GMm} \over r} + {{ - GMm} \over r} + {1 \over 2}m{V^2} = 0 + 0$$
[M is mass of star m is mass of meteroite)
$$ \Rightarrow $$ v $$ = \sqrt {{{4GM} \over r}} = 2.8 \times {10^5}$$m/s
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