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

• A. $v=\sqrt{GM(\frac{1}{R}-\frac{1}{r})}$
• B. $v=\sqrt{GM(\frac{1}{2R}-\frac{1}{r})}$
• C. $v=\sqrt{GM(\frac{1}{R}+\frac{1}{r})}$
• D. $v=\sqrt{GM(\frac{1}{2R}+\frac{1}{r})}$

Answer 1

The correct option is B $v=\sqrt{GM(\frac{1}{2R}-\frac{1}{r})}$

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 $E_i=KE+PE=0+(-\frac{GMM}{r})=-\frac{G{M}^2}{r}$
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
$E_f=2\times \left(\frac{1}{2}M{v}^2\right)+(-\frac{GMM}{2R})=M{v}^2-\frac{G{M}^2}{2R}$
According to law of conservation of mechanical energy $E_f=E_i$
$M{v}^2-\frac{G{M}^2}{2R}=-\frac{G{M}^2}{r}$ or $v^2=GM(\frac{1}{2R}-\frac{1}{r})$
or $v=\sqrt{GM(\frac{1}{2R}-\frac{1}{r})}$