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  4. Two stars each of mass m and radius R approach each other to collide head-on. Initially the stars are at a distance r(>>R) . Assuming their speeds to be negligible at this distance of separation, the speed with which the stars collide is

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Q. Two stars each of mass $m$ and radius $R$ approach each other to collide head-on. Initially the stars are at a distance $r(>>R) .$ Assuming their speeds to be negligible at this distance of separation, the speed with which the stars collide is

Gravitation

Solution:

Since the speeds of the stars are negligible when they are at a distance of $r$, the initial kinetic energy of the system is zero. Therefore, initial total energy of the system is
$E_{i}= KE + PE =0+\left(-\frac{G m m}{r}\right)$...(i)
where $m$ represents the mass of each star and $r$ the initial separation between them.
When two stars collide, their centres will be at a distance twice the radius of a star $i . e .2 R .$ Let $v$ be the speed with which two stars collide. Then total energy of the system at the instant of their collision is
$E_{f}=\left(\frac{1}{2} m v^{2}\right) \times 2+\left(-\frac{G m m}{2 R}\right)=m v^{2}-\frac{G M m}{2 R}$
According to law of conservation of energy, $E_{f}=E_{i}$
$\therefore m v^{2}-\frac{G m m}{2 R}=-\frac{G m m}{r}$
or $ v^{2}=G m\left(\frac{1}{2 R}-\frac{1}{r}\right)$
or $v=\sqrt{G m\left(\frac{1}{2 R}-\frac{1}{r}\right)}$