Question

Two stars each of mass and radius Rare approaching
each other for a head-on collision. They start approaching
enh other when their separation is r>>R. If their spreads at
this separation are negligible, the speed y with which they
collide would be
I 1
(b) r= .6.11
1
DR
( 1= GI!
(J) V= /G/
21​

Answer 1

QUESTION:

Two stars each of same mass and radius are approaching each other for a head-on collision. They start approaching each other when their separation is r >> R. If their speeds at this separation are negligible, the speed v with which they collide would be

ANSWER:

The velocity $\sf v=\sqrt{GM\left(\dfrac{1}{2R}-\dfrac{1}{r}\right)}$

GIVEN:

• Two stars each of same mass and radius are approaching  each other for a head-on collision.

• They start approaching  each other when their separation is r >> R.

• Their speeds at  the separation are negligible.

TO FIND:

• The speed v with which they collide.

EXPLANATION:

Let the mass of stars be M and the radius of stars be R.

Initial separation between the stars = r

Diagram: [Before Collision]

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0) \put(0, 0) {\circle{4}} \put(5, 0){\circle{4}} \put(5, 0){\circle{4}}\put(0, 0){\circle{4}}\put(5, 0){\line( - 1,0){5}}\put(5, 0){\circle*{0.1}}\put(0, 0){\circle*{0.1}} \put( 0, - 0.1) { \underbrace{\qquad \qquad\qquad\qquad\qquad\qquad\qquad} }\put( 2.4, - 0.6) { \bf r } \put(0.2, 0.1){ \bf R }\put(4.6, 0.1){ \bf R }\end{picture}$

$\sf Let\ E_i\ be\ the\ initial\ Total\ Energy$

$\boxed{\large{\bold{\gray{E=K.E +P.E}}}}$

$\boxed{\large{\bold{\gray{E_i=(K.E)_i + (P.E)_i}}}}$

$\sf E_i=\dfrac{1}{2}Mv_i^2 + \dfrac{-GMM}{r}$

$\sf (v_i = 0)$

As given that the initial speed is negligible.

$\sf E_i=\dfrac{1}{2}M0^2 - \dfrac{GM^2}{r}$

$\sf E_i=0- \dfrac{GM^2}{r}$

$\sf E_i= - \dfrac{GM^2}{r}$

Diagram: [During Collision]

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0) \put(0, 0) {\circle{4}} \put(1.4, 0){\circle{4}} \put(1.4, 0){\circle{4}}\put(0, 0){\circle{4}}\put(0, 0){\line( 1,0){1.4}}\put(1.4, 0){\circle*{0.1}}\put(0, 0){\circle*{0.1}} \put( 0, - 0.1) { \underbrace{\qquad \qquad} }\put( 0.1, 0.1) { \bf R }\put( 0.9, 0.1) { \bf R }\put( 0.6, - 0.3){ \line( 0, - 1){0.5}} \thicklines\put( 0.58, - 0.8){ \vector( 1, 0){1}}\put( 1.8, - 0.9) { \bf 2R }\end{picture}$

$\sf Let\ E_f\ be\ the\ final\ Total\ Energy$

$\boxed{\large{\bold{\gray{E_f=(K.E)_f + (P.E)_f}}}}$

$\sf E_f=2\times\dfrac{1}{2}Mv^2 + \dfrac{-GMM}{2R}$

Here kinetic energy is multiplied by two because at collision the kinetic energies of two stars add up.

And also the distance between the two stars will be equal to two times of their radius as two stars will be in contact during collision.

$\sf E_f=Mv^2- \dfrac{GM^2}{2R}$

$\boxed{\large{\bold{\gray{E_f = E_i}}}}$

$\sf - \dfrac{GM^2}{r}=Mv^2- \dfrac{GM^2}{2R}$

Divide by M on both sides.

$\sf - \dfrac{GM}{r}=v^2- \dfrac{GM}{2R}$

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

$\sf v^2=\dfrac{GM}{2R}- \dfrac{GM}{r}$

Take GM as common.

$\sf v^2=GM\left(\dfrac{1}{2R}-\dfrac{1}{r}\right)$

$\sf v=\sqrt{GM\left(\dfrac{1}{2R}-\dfrac{1}{r}\right)}$

Hence the velocity with which they collide will be $\sf v=\sqrt{GM\left(\dfrac{1}{2R}-\dfrac{1}{r}\right)}$