Ncert Exercises & Solutions

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).

Mass of each star, M = 2 × $10^{30}$ 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 potential energy of both stars $= \frac{- GMM}{2 R}$

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

Applying the law of conservation of energy,

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

$v^{2} = \frac{- GM}{r} + \frac{GM}{2 R} = GM (- \frac{1}{r} + \frac{1}{2 R})$
$= 6.67 \times 10^{- 11} \times 2 \times 10^{30} [- \frac{1}{10^{12}} + \frac{1}{2 \times 10^{7}}]$
$= 13.34 \times 10^{19} [- 10^{12} + 5 \times 10^{- 8}]$
$~ 13.34 \times 10^{19} \times 5 \times 10^{- 8}$
$~ 6.67 \times 10^{12}$
$v = \sqrt{6.67 \times 10^{12}} = 2.58 \times 10^{6} m / s$

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