Two neutron stars are separated by a distance of 1.0 x 10 10 m. They each have a

Question

Two neutron stars are separated by a distance of 1.0x1010 m. They each have a mass of 1.0x1030 kg and a radius of 500000 m. They are initially at rest with respect to each other. As measured from the rest frame, how fast are they moving when their separation has decreased to one-half its initial value?

___________  m/s (The correct answer for this is 81663.94319)

How fast are they moving when they collide?
___________  m/s (I can't figure out this answer.)

Please use my numbers to work out the solution. Thanks!

Explanation / Answer

m = 1*10^30 kg

r = 500000 m

distance of separation, d = 1*10^10 m

Initially the Potential energy of the two stars,

P.Ei = - G*m*m/r = -Gm^2/r

where G = Gravitational cosntant = 6.67*10^-11

So, P.Ei = -6.67*10^-11*(1*10^30)^2/(1*10^10)

So, P.Ei = -6.67*10^39 J

Now, after their separtion has been halved,

d' = 1*10^10/2 = 5*10^9 m

So, P.Ef = -G*m^2/d'

So, P.Ef = 6.67*10^-11*(1*10^30)^2/(5*10^9) = -1.33*10^40 J

So, By conservation of energy the P.E lost must be converted to Kinetic energy of the masses

So, P.Ei - P.Ef = K.E

So,  -6.67*10^39 - (-1.33*10^40) = 2*(0.5*m*v^2) <-------- v =speed of the masses

So, 6.63*10^39 = 1*10^30*v^2

So, v = 8.15*10^4 m/s <------------answer

2)

when they collide,

d' = 2*r = 2*500000 = 1*10^6 m

So, PEf = -Gm^2/d'

So, PEf = 6.67*10^-11*(1*10^30)^2/(1*10^6)

So, PEf = 6.67*10^43 J

So, By conservation of energy,

-6.67*10^39 - (-6.67*10^43) = 2*(0.5*1*10^30*v^2)

So, v = 8.17*10^6 m/s <-----------answer

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