Question

Two stars each of mass $M$ and radius $R$ 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

• A. $v = \sqrt{GM\left(\frac{1}{R}-\frac{1}{r}\right)}$
• B. $v = \sqrt{GM\left(\frac{1}{2R}-\frac{1}{r}\right)}$
• C. $v = \sqrt{GM\left(\frac{1}{R}+\frac{1}{r}\right)}$
• D. $v = \sqrt{GM\left(\frac{1}{2R}+\frac{1}{r}\right)}$

Answer 1

Answer: (B)

Since the speeds of the stars are negligible when they are at a distance $r$, hence the initial kinetic energy of the system is zero. Therefore, the initial total energy of the system is $E _{ i }= KE + PE =0+\left(-\frac{ GMM }{ r }\right)=-\frac{ GM ^{2}}{ r }$ where M represents the mass of each star and $r$ is initial separation between them.
when two stars collide their centres will be at a distance twice the radius of a star i.e. $2 R$.

Let v be the speed with which two stars collide. Then total energy of the system at the instant of their collision is given by
$
E _{ f }=2 \times\left(\frac{1}{2} Mv ^{2}\right)+\left(-\frac{ GMM }{2 R }\right)= Mv ^{2}-\frac{ GM ^{2}}{2 R }
$
According to law of conservation of mechanical energy $E _{ f }= E _{ i }$
$
Mv ^{2}-\frac{ GM ^{2}}{2 R }=-\frac{ GM ^{2}}{ r } \text { or } v ^{2}= GM \left(\frac{1}{2 R }-\frac{1}{ r }\right)
$
or $v =\sqrt{ GM \left(\frac{1}{2 R }-\frac{1}{ r }\right)}$