Question

Two stars each of one solar mass ($=2\times {10}^{30}kg$) are approaching each other for a head on collision. When they are a distance ${10}^{9}km$, their speeds are negligible. What is the speed with which they collide? The radius of each star is ${10}^{4}km$. Assume the stars to remain undistorted until they collide. (Use the known value of $G$)

Answer 1

For negligible speeds,

v = 0 total energy of two stars separated at distance r

= $\left[\frac{-GMM}{r}\right]+\left(\frac{1}{2}\right)m{v}^2$

= $\left[\frac{-GMM}{r}\right]$ + 0 . . .. ( i )

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

Velocity of the start = v

Distance between the centers of the start = 2R

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

Total potential energy of both stars = $\frac{-GMM}{2R}$

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

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

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

$v^2=-\frac{GM}{r}+\frac{GM}{2R}$

= $GM\left[\right(-\frac{1}{r})+(\frac{1}{2R}\left)\right]$

= $6.67\times {10}^{-11}\times 2\times {10}^{30}\left[\right(-\frac{1}{{10}^{12}})+(\frac{1}{2}\times {10}^7\left)\right]-6.67\times {10}^{12}$

v = $(6.67\times {10}^{12}{)}^{\frac{1}{2}}$

v = $2.58\times {10}^6$ m/s.