Deep in space, two neutron stars are separated (center-to-center) by a distance

Question

Deep in space, two neutron stars are separated (center-to-center) by a distance of 20 X 106 km apart. Neutron star A has a mass of 151 X 1028 kg and radius 50000 m while the neutron star B has a mass of 167 X 1028 kg and radius 70000 m. They are initially at rest with respect to each other.


a) With respect to that rest frame, how fast are the stars moving when their separation has decreased to one-half its initial value?
Va=?
Vb=?


b) How fast are each moving the instant before they collide?
Va=?
Vb=?

Explanation / Answer

Initial Energy = P.E. = - GMm/r = -6.67*10-11*151*167*1056/20*109

= - 8409.87 *1036 J

a) R = r/2

Final P.E. = -GMm/(r/2) = - 16819.74*1036 J

K.E. = I.E. - P.E. = 8409.87*1036 J

=>0.5* 151*1028*Va2 + 0.5*167*1028*Vb2 = 8409.87*1036

=> 75.5*Va2 + 83.5*Vb2 = 8409.87*108 ------------- eqn 1

Since Initial momentum = 0

Hence final Momentum = 0

=> 151*1028*Va + 167*1028*Vb = 0

=> Va = - 1.106Vb

Putting this in eqn 1

=> Vb2(83.5+1.1062*75.5) = 8409.87*108

=> Vb2 = 47.823*108

=> Vb = 6.92*104 m/s 69.2 km/s

& Va = - 7.65*104 m/s 76.5 km/s

b) R = 50000 + 70000 = 1.2*106 m

P.E. = - GMm/R = -6.67*10-11*151*167*1056/1.2*106

= - 140164.49*1039 J

K.E. = - 8409.87 *1036 J + 140164.49*1039 J

= 1.40156*1044 J

Since Initial momentum = 0

Hence final Momentum = 0

=> 151*1028*Va + 167*1028*Vb = 0

=> Va = - 1.106Vb

K.E. = 1.40156*1044

=>0.5* 151*1028*Va2 + 0.5*167*1028*Vb2 = 1.40156*1044

=> 75.5*Va2 + 83.5*Vb2 = 1.40156*1016

=> Vb2(83.5+1.1062*75.5) = 1.40156*1016

=> Vb2 = 7.97*1013

=> Vb = 8.9275*106 m/s = 8927.5 km/s

* Va = - 1.106Vb = 9.8738*106 m/s = 9873.8 km/s

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