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

Home » Two stars each of one solar mass are approaching each other for a headon collision. When they are a distance , their speeds are negligible. What is the speed with which they collide? The radius of each star is Assume the stars to remain undistorted until they collide. (Use the known value of G).

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

CBSE | Class 11 | Gravitation | NCERT | Physics

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

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

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

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

Initial potential energy of the system will begiven as,

$-\mathrm{GM} \mathrm{m} / \mathrm{r}=(1)$

So, Total kinetic energy of the stars will be

$1 / 2 M v^{2}+1 / 2 M v^{2}=M v^{2}$

where $v$ is the speed at which the stars collide.
r=2R is the distance between the centres of the stars when they collide.

Hence, the final potential energy of two stars when they are close to each other wll be $-\mathrm{GMm} / 2 \mathrm{R}$

Thus, the total energy of the two stars just before collision can be calculated as

$E=M v^{2}-G M m / 2 R$

(1)

$M v^{2}-G M m / 2 R=-G M m / r$
$v^{2}=(G m / 2 R)-(G m / r)$
$=G m(1 / 2 R-1 / r)$
$=6.67 \times 10^{-11} \times 2 \times 10^{30}\left[\left(1 / 2 \times 10^{7}\right)-\left(1 / 10^{12}\right)\right]$
$v^{2}=6.67 \times 10^{12}$
$v=2.58 \times 10^{6} \mathrm{~m} / \mathrm{s}$