Two stars each of mass m are approaching each other for a head on collision
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Answer:
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 × 1030 kg
Radius of each star, R = 104 km = 107 m
Distance between the stars, r = 109 km = 1012m
For negligible speeds, v = 0 total energy of two stars separated at distance r
Or
Mass of each star, M = 2 \times 10^{30}\ kgM=2×10
30
kg
Radius of each star, R = 10^4\ km = 10^7\ mR=10
4
km=10
7
m
Distance between the stars,r=10^9\ km=10^{12}\ mr=10
9
km=10
12
m
Total energy of two stars separated at distance r =-GMM/r+(1/2)Mv^2=−GMM/r+(1/2)Mv
2
=-GMM/r ...........(1)=−GMM/r...........(1)
Let velocity of the stars when they are about to collide be vv
Distance between the centers of the stars =2R=2R
Total kinetic energy =(1/2)Mv^2+(1/2)Mv^2=Mv^2=(1/2)Mv
2
+(1/2)Mv
2
=Mv
2
Total potential energy =-GMM/2R=−GMM/2R
Total energy of the two stars just before collision=Mv^2-GMM/2R ........(2) =Mv
2
−GMM/2R........(2)
Using the law of conservation of energy, we can write:
Mv^2-GMM/2R = -GMM/rMv
2
−GMM/2R=−GMM/r
v^2 = GM/2R - GM/rv
2
=GM/2R−GM/r
=6.67 \times 10^{-11} \times 2 \times 10^{30}(\dfrac{1}{2 \times 10^7}-\dfrac{1}{10^{12}})=6.67×10
−11
×2×10
30
(
2×10
7
1
−
10
12
1
)
\approx 6.67 \times 10^{12}≈6.67×10
12
v=2.58 \times 10^6\ m/sv=2.58×10
6
m/s
I hope it's help you
Explanation: