Question

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).

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Answer 1

Answer:

Given:-

• Mass of each star, (M) = 2×10³⁰ kg
• Radius of each star, (R) = 10⁴ km

In m:- 10⁴ km = 10⁷ m

• Distance between the Star's,(r) = 10⁹ km

In m:- 10⁹ km = 10¹² m

For negligible speeds, v = 0 total energy of two stars separated at distance r

$\mapsto \: - \frac{GMM}{r} + \frac{1}{2} {mv}^{2}$

$\mapsto \: - \frac{GMM}{r} + 0 \: \: \: ...(1)$

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Now, Consider the case when the stars are about to collide:

• Velocity of the star = v

Distance between the centres of the stars = 2R

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Total kinetic energy of both stars:

$\mapsto \: \frac{1}{2} {Mv}^{2} + \frac{1}{2} {Mv}^{2} = {Mv}^{2}$

Total potential energy of both stars:

$\mapsto \: \frac{ - GMM}{2R}$

★ Total energy of both stars:

$\mapsto \: {Mv}^{2} - \frac{GMM}{2R} \: ...(2)$

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Solution:

Using the law of conservation of energy, we can write :

$\mapsto \:{Mv}^{2} - \frac{GMM}{2R} = - \frac{GMM}{R}$

${v}^{2} = - \frac{GM}{r} + \frac{GM}{2R} = GM( - \frac{1}{2} + \frac{1}{2R} )$

$\mapsto \: 6.67 \times {10}^{ - 10} \times 2 \times {10}^{30} ( - \frac{1}{ {10}^{ - 12} } + \frac{1}{2 \times {10}^{7} } )$

$\mapsto \: 13.34 \times {10}^{19} ( { - 10}^{ - 12} + 5 \times {10}^{ - 8}$

$\mapsto \:13.34 \times {10}^{ 19} \times 5 \times {10}^{ - 8}$

$\mapsto \: 6.67 \times {10}^{12}$

$\longmapsto \: v = \sqrt{6.67 \times {10}^{12} } = \boxed{2.58 \times {10}^{6} m/s}$