Most stars maintain an equilibrium size by balancing two forces - an inward grav

Question

Most stars maintain an equilibrium size by balancing two forces - an inward gravitational force and an outward force resulting from the star's nuclear reactions. When the star's fuel is spent, there is no counterbalance to the gravitational force. Whatever material is remaining collapses in on itself. Stars about the same size as the sun become white dwarfs, which glow from leftover heat. Stars that have about three times the mass of the Sun compact into neutron stars. And a star with a mass greater than three times the Sun's mass collapses into a single point, called a black hole. In most cases, protons and electrons are fused together to form neutrons - this is the reason for the name neutron star. Neutron stars rotate very fast because of the conservation of angular momentum. Imagine a star of mass 4.63

Explanation / Answer

This is a conservation of angular momentum problem. Since there are no external torques the angular momentum "L" is constant in time.

Li = Iw =(2/5)M(Ri^2)(2pi/Ti) , where Ri & Ti are the initial radius & period

Lf = Iw =(2/5)M(Rf^2)(2pi/Tf)

Equate and solve for Rf

Rf^2 = (Ri^2)(Tf/Ti)

where Ti = (27)(24)(3600) = 2.3 x 10^-6

Rf ^2/Ri^2= Tf/(2.3 x 10^-6)

I guess the initial radius

Ri=9.29*10^8

Rf= 13.5 km= 13.5 *10^3 m

Tf=4.85*10^-16 s

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