A) \[v=\sqrt{GM\left( \frac{1}{R}-\frac{1}{r} \right)}\]
B) \[v=\sqrt{GM\left( \frac{1}{2R}-\frac{1}{r} \right)}\]
C) \[v=\sqrt{GM\left( \frac{1}{R}+\frac{1}{r} \right)}\]
D) \[v=\sqrt{GM\left( \frac{1}{2R}+\frac{1}{r} \right)}\]
Correct Answer: B
Solution :
| Since the speeds of the stars are negligible when they are at a distance r, hence the initial kinetic energy of the system is zero. Therefore, the initial total energy of the system is \[{{E}_{i}}=KE+PE=0+\left( -\frac{GMM}{r} \right)=-\frac{G{{M}^{2}}}{r}\] |
| where M represents the mass of each star and r is initial separation between them. |
| When two stars collide their centres will be at a distance twice the radius of a star i.e. 2R. |
| Let v be the speed with which two stars collide. |
| Then total energy of the system at the instant of their collision is given by |
| \[{{E}_{f}}=2\times \left( \frac{1}{2}M{{\text{v}}^{2}} \right)+\left( -\frac{GMM}{2R} \right)=M{{\text{v}}^{2}}-\frac{G{{M}^{2}}}{2R}\] |
| According to law of conservation of mechanical energy, \[{{E}_{f}}={{E}_{i}}\] |
| \[M{{\text{v}}^{2}}-\frac{G{{M}^{2}}}{2R}=-\frac{G{{M}^{2}}}{r}\] or \[{{\text{v}}^{2}}=GM\left( \frac{1}{2R}-\frac{1}{r} \right)\] |
| or \[\text{v=}\sqrt{GM\left( \frac{1}{2R}-\frac{1}{r} \right)}\] |
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