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Homework Help: Pressure in a neutron star

  1. Jan 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Assuming a neutron star is made from an incompressible material, what is the hydraulic pressure 50m below the surface?

    Mass = 1.98x10^30kg
    Radius = 10km

    2. The attempt at a solution

    g=GM/r^2 = (6.67x10^-11 x 1.98x10^30)/ 10000^2 = 1.32x10^12 m/s^2

    P=P[tex]_{0}[/tex] + [tex]\rho[/tex]gh Pa

    [tex]\rho[/tex] = m/v = 1.98x10^30 / (4/3 x pi x 10000^3) = 4.73x10^17 kg/m^3

    If pressure in space is 0 Pa, then P = 0 + 4.73x10^17 x 1.32x10^12 x 50

    = 3.12x10^31 Pa

    Is this correct? Thank for your time!

    Edit: I think it might be wrong because my equation assumes that space (as a vacuum) and the star are one material (?) :s.
     
    Last edited: Jan 1, 2010
  2. jcsd
  3. Jan 1, 2010 #2
    Use the differential equation of hydrostatic equilibrium:

    [tex]\frac{dP}{dh} = -\rho g[/tex]

    Remember that the gravitational acceleration [tex]g[/tex] changes with [tex]h[/tex]
    In your calculation, you assumed that it was constant and equal to the surface gravitation. That is incorrect.
    [tex]\rho[/tex], on the other hand, is constant as the star is incompressible.
     
  4. Jan 1, 2010 #3
    Thanks, for your reply. Here is my second attempt:

    dP = - [tex]\rho[/tex]gdh

    [tex]\frac{dP}{dh}[/tex] = -[tex]\rho[/tex]g

    [tex]\int[/tex][tex]\frac{dP}{dh}[/tex]dh = [tex]\int[/tex]-[tex]\rho[/tex]gdh

    P = -[tex]\rho[/tex]GM[tex]\int[/tex][tex]\frac{1}{r^{2}}[/tex]dh between 10000 and 9950

    P = -[tex]\rho[/tex]GM [[tex]\frac{1}{r}[/tex]] 10000...9950

    = -[tex]\rho[/tex]GM [[tex]\frac{1}{10000}[/tex]-[tex]\frac{1}{9950}[/tex]]

    =3.14x10[tex]^{31}[/tex] Pa

    I apologize for integrating r with respect to h, if that was confusing.
     
    Last edited: Jan 1, 2010
  5. Jan 1, 2010 #4
    That too is incorrect. ;) Though you're on the right track!

    Remember that a spherical shell contributes no [tex]\frac{1}{r^2}[/tex] field inwards. That is to say, when calculating the gravitational field inside the star at a radius [tex]r_0<R[/tex], you must only regard the contribution of the mass inside, at radii [tex]0<r<r_0[/tex], as all the mass on the outside contributes nothing to the gravitational field!
     
  6. Jan 1, 2010 #5
    OK, I've got it, thanks RoyalCat!
     
  7. Jan 1, 2010 #6
    You're welcome!

    As a hint, the gravitational field rises linearly with the radius inside the star, in case you happen to get anything different.
     
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