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 109 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Answer 1

Mass of each star, \(M\) = \(2\times 10^{30}\) \(kg\)
Radius of each star, \(R\) = \(10^4\) \(km\) = \(10^7 m\)
Distance between the stars,\(r\) = \(10^9\; km\) = \(10^{12}\;m\)
For negligible speeds, v = 0 total energy of two stars separated at distance r 

= \(\frac{-GMM}{r}+\frac{1}{2}mv^2\) 

= \(\frac{-GMM}{r}+0\) ...(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 = \(\frac{1}{2}Mv^2+\frac{1}{2}Mv^2\) = \(Mv^2\)

Total energy of both stars =\(Mv^2-\frac{GMM}{2R}\) ...(ii)

Total energy of the two stars = Using the law of conservation of energy, we can write:

\(MV^2-\frac{GMM}{2R}\) = \(\frac{-GMM}{r}\)

\(v^2 \)= \(\frac{-GM}{r}+\frac{GM}{2R}= GM\) \(\bigg(-\frac{1}{r}+\frac{1}{2R}\bigg)\) 

= \(6.67 \times 10^ {-11} \times2 \times10^{30} \bigg[-\frac{1}{ 10^{12}} + \frac{1}{2\times 10^7}\bigg ]\)

= \(13.34 \times 10^{19} [-10^{-12} + 5 \times 10^{-8}] ∼ 13.34 \times 10^{19} \times 5 \times10^{-8} ∼ 6.67 \times 10^{12}\)

\(v\) = \(\sqrt{6.67 \times 10^{12}}\) = \(2.58 \times 10^6 m/s\)