8.20 Two stars each of one solar mass (= 2×10³⁰ 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).

6 Views | Posted 5 months ago

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Mass of each star, M = 2 * 10³⁰ kg, Radius of each star, R = 10⁴ km = $10^{7}$ m

Distance between stars, r = $10^{9}$ km = $10^{12}$ m

For negligible speed, v = 0

The total energy of two stars separated at a distance r

= $- \frac{G M M}{r} + \frac{1}{2} m v^{2}$ = $- \frac{G M M}{r}$ …….(i)

Now, consider the case when the stars are about to collide. Velocity of the stars = V, distance between the centres 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 = $- \frac{G M M}{2 R}$

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

Using the law of conservation of energy, we can write

M $V^{2} - \frac{G M M}{2 R}$ =

...more

Mass of each star, M = 2 * 10³⁰ kg, Radius of each star, R = 10⁴ km = $10^{7}$ m

Distance between stars, r = $10^{9}$ km = $10^{12}$ m

For negligible speed, v = 0

The total energy of two stars separated at a distance r

= $- \frac{G M M}{r} + \frac{1}{2} m v^{2}$ = $- \frac{G M M}{r}$ …….(i)

Now, consider the case when the stars are about to collide. Velocity of the stars = V, distance between the centres 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 = $- \frac{G M M}{2 R}$

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

Using the law of conservation of energy, we can write

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

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

$V^{2}$ = GM ( $\frac{1}{2 R}$ $-$ $\frac{1}{r})$ = 6.67 $* 10^{- 11} * 2 * 10^{30}$ ( $\frac{1}{2 * 10^{7}} - \frac{1}{10^{- 12}}$ )

= 1.334 $* 10^{20} *$ ( $\frac{1}{2 * 10^{7}} - \frac{1}{10^{- 12}}$ )

V = 2.6 $* 10^{6}$ m/s

less

1 Answer