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 speed at this separation is negligible, the speed 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

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

The speed of the stars at separation r is negligible. Therefore, their energy is entirely potential at this separation.
$E_1=-\frac{G{m}^2}{r}$
As the stars approach each other under gravitational attraction, they begin to acquire speed and hence kinetic energy at the expense of potential energy. When they eventually collide, the separation between their centres is r = R + R = 2R.
Now, the total energy is
$E_2=-\frac{G{m}^2}{2R}+\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=-\frac{G{m}^2}{2R}+m{v}^2$
From the principle of conservation of energy, $E_1=E_2$
$\Rightarrow -\frac{G{m}^2}{r}=-\frac{G{m}^2}{2R}+m{v}^2$
$\Rightarrow v=\sqrt{Gm(\frac{1}{2R}-\frac{1}{r})}$
Hence, the correct choice is (b).