First slide

Gravitaional potential energy

Question

Two stars each of mass M and radius R are approaching each other for a head-on collision. They start approaching each other when their separation is r >> R. If their speeds at this separation are negligible, the speed v with which they collide would be

Moderate

A

v = $\sqrt{GM (\frac{1}{R} - \frac{1}{r})}$
B

v = $\sqrt{GM (\frac{1}{2 R} - \frac{1}{r})}$
C

v = $\sqrt{GM (\frac{1}{R} + \frac{1}{r})}$
D

v = $\sqrt{GM (\frac{1}{2 R} + \frac{1}{r})}$

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
$E_{i} = KE + PE = 0 + (- \frac{GMM}{r}) = - \frac{{GM}^{2}}{r}$
where M represents the mass of each star and r is initial separation between them; twice the radius of 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
$E_{f} = 2 \times (\frac{1}{2} {Mv}^{2}) + (- \frac{GMM}{2 R}) = {Mv}^{2} - \frac{{GM}^{2}}{2 R}$
According to law of conservation of mechanical energy
$E_{f} = E_{i}$
${Mv}^{2} - \frac{{GM}^{2}}{2 R} = - \frac{{GM}^{2}}{r} or v^{2} = GM (\frac{1}{2 R} - \frac{1}{r})$
or v = $\sqrt{GM (\frac{1}{2 R} - \frac{1}{r})}$

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