# Two stars each of one solar mass (= 2× 1030 kg) are approaching each other for a head on collision.

Question:

Two stars each of one solar mass $\left(=2 \times 10^{30} \mathrm{~kg}\right)$ are approaching each other for a head on collision. When they are a distance $109 \mathrm{~km}$, their speeds are negligible. What is the speed with which they collide? The radius of each star is $104 \mathrm{~km}$. Assume the stars to remain undistorted until they collide. (Use the known value of $G$ ).

Solution:

Mass of each star, $M=2 \times 10^{30} \mathrm{~kg}$

Radius of each star, $R=10^{4} \mathrm{~km}=10^{7} \mathrm{~m}$

Distance between the stars, $r=10^{9} \mathrm{~km}=10^{12} \mathrm{~m}$

For negligible speeds, $v=0$ total energy of two stars separated at distance $r$

$=\frac{-\mathrm{G} M M}{r}+\frac{1}{2} m v^{2}$

$=\frac{-\mathrm{G} M M}{r}+0$    ...(i)

Now, consider the case when the stars are about to collide:

Velocity of the stars = v

Distance between the centers of the stars = 2R

Total kinetic energy of both stars $=\frac{1}{2} M v^{2}+\frac{1}{2} M v^{2}=M v^{2}$

Total potential energy of both stars $=\frac{-\mathrm{G} M M}{2 R}$

Total energy of the two stars $=M v^{2}-\frac{G M M}{2 R}$ ... (ii)

Using the law of conservation of energy, we can write:

$M v^{2}-\frac{\mathrm{G} M M}{2 R}=\frac{-\mathrm{G} M M}{r}$

$v^{2}=\frac{-\mathrm{G} M}{r}+\frac{\mathrm{G} M}{2 R}=\mathrm{GM}\left(-\frac{1}{r}+\frac{1}{2 R}\right)$

$=6.67 \times 10^{-11} \times 2 \times 10^{30}\left[-\frac{1}{10^{12}}+\frac{1}{2 \times 10^{7}}\right]$$=13.34 \times 10^{19}\left[-10^{-12}+5 \times 10^{-8}\right]$

$\sim 13.34 \times 10^{19} \times 5 \times 10^{-8}$

$\sim 6.67 \times 10^{12}$

$v=\sqrt{6.67 \times 10^{12}}=2.58 \times 10^{6} \mathrm{~m} / \mathrm{s}$