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Maximum angular velocity for a Neutron Star: Textbook - Physics for Scientists and Engineers

  1. Oct 23, 2014 #1
    1. The problem statement & given/known data
    A neutron star is a collection of neutrons bound together by their mutual gravitation. The density of such a star is comparable to that of an atomic nucleus (10^17 kg/m^3).

    a) Assuming the neutron star to be spherical in shape and of uniform mass density, show that the maximum frequency with which it may rotate, without mass flying off at the equator, is ω=(4πρG/3)^1/2, where ρ is the mass density.

    b)Calculate ω for ρ=10^17 kg/m^3.

    c) Determine the kinetic energy of such a star about its axis of rotation. The radius of the star is 10km.

    2. Relevant equations & information
    R=10km=10000m
    ρ=10^17 kg/m^3
    Vsphere=4/3πr^3
    Isphere=2/5mr^2
    M=Total Mass
    m=mass of particle at surface
    ω=vr
    ω=(4πρG/3)^1/2
    Fg=Gm1m2/r^2
    Fc=mv^2/r
    g=GM/r^2
    G=6.67x10^(-11)
    K=1/2Iω^2

    3. The attempt at a solution
    a) Solved
    Particle at surface experience two force: Fc (Centrifugal Force) & Fg (Gravitational Force).
    When the maximum angular velocity is reached, the two forces will be balanced (direction out & into center)
    Fg=Fc
    M=ρ(4/3πr^3)
    v=ω/r
    Fc=mv^2/r =m(ω/r)^2/r =mrω^2
    Fg=GMm/r^2
    ⇒ Fg=Fc → GMm/r^2=mrω^2 → GM=r^3ω^2 → ω^2=GM/r^3
    M=ρ(Vsphere)=4/3πr^3*p
    ⇒ Fg=Fc → ω^2=G(4/3πr^3*p)/r^3 → ω=(4/3πρG)^(1/2)

    b) Solved
    Alright. I spent at least an hour thinking about this problem and could not come up with the answer in the book. I couldn't stand the thought of leaving this thing unsolved, so I came here seeking help. Of course, now that I'm taking a fresh look at it, I figured out what I did wrong earlier! (I was thinking of G as the the force of gravity rather than the gravitational constant). Anyway, since I've gone to the trouble of solving this much, I might as well finish.
    From (a): ω=(4/3πρG)^(1/2)
    ⇒ ω=(4.19*10^17*G)^(1/2) → ω=(4.19e17G)^(1/2)
    G=6.67x10^(-11)
    ⇒ ω=(4.19x10^17(6.67x10^(-11))^(1/2) = 5.29x10^3 rad/s

    c) Solved
    K=1/2Iω^2 = 1/2*(2/5πr^2)*(5.29x10^3)^2 = 1/5π(10000^2)*(5.29x10^3)^2= 2.34x10^44 J
     
  2. jcsd
  3. Oct 24, 2014 #2

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    Hello Ken, welcome to PF

    Well posted, well done. Fresh looks do wonders indeed.
     
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