Two neutron stars are separated by a distance of 1.0 times 10^11 m. They each ha

Question

Two neutron stars are separated by a distance of 1.0 times 10^11 m. They each have a mass of 1.0 times 10^30 kg and a radius of 2.0 times 10^5 are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial value and (b) they are about to collide? In a certain binary-star system, each star has the same mass as our Sun, and they revolve about their center of mass. The distance between them is the same as the distance between Earth and the Sun. What is their period of revolution in years? Miniature black holes. Left over from the big-bang beginning of the universe, tiny black holes might still wander through the

Explanation / Answer

Q-2)

Initially, U = - Gm^2 /R, and since at rest, their moments are zero.

Thus, by symmetry, they each must be moving with the same speed in opposite directions at any subsequent time.

At half that distance, U’ = 2U = - 2Gm^2 /R

dU = - Gm^2 /R

dKE = -dU

so for each star

K = Gm^2/2R = 0.5*mv^2

v = (Gm/R)^0.5

= [(6.67*10^-11)*(1*10^30) / (1*10^11) ]^0.5

v = 25826.34 m/sec

B. With a final separation distance of 2r = 4.0*10^5 m

dU = Gm^2 (1/R - 1/2r)

dKE = Gm^2 (1/2r - 1/R) = mv^2

v = [Gm(1/2r - 1/R)]^0.5

= [(6.67*10^-11)*(1*10^30) { 1/(4*10^5) - 1/(1*10^11) } ]^0.5

v = 12913145.74 m/s = 1.3*10^7 m/sec

Q-3) Kepler's third law:

(M1 + M2) = a^3 / P^2

where M is the mass in Solar masses, a is the distance in AU, and P is the period in years.

In the problem
M1 = M2 = Msol
a = 1 AU

so

P = (a^3 / (M1 + M2))^(1/2)

P = (1^3 / (1 + 1))^(1/2)

P= (1/2)^(1/2)

P = 0.707 years

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.