Calculating Hydraulic Pressure in a Neutron Star

[j-e_c]

Homework Statement

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_{0} + \rhogh Pa

\rho = 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.

 

[RoyalCat]

Use the differential equation of hydrostatic equilibrium:

\frac{dP}{dh} = -\rho g

Remember that the gravitational acceleration g changes with h
In your calculation, you assumed that it was constant and equal to the surface gravitation. That is incorrect.
\rho, on the other hand, is constant as the star is incompressible.

 

[j-e_c]

Thanks, for your reply. Here is my second attempt:

dP = - \rhogdh

\frac{dP}{dh} = -\rhog

\int\frac{dP}{dh}dh = \int-\rhogdh

P = -\rhoGM\int\frac{1}{r^{2}}dh between 10000 and 9950

P = -\rhoGM [\frac{1}{r}] 10000...9950

= -\rhoGM [\frac{1}{10000}-\frac{1}{9950}]

=3.14x10^{31} Pa

I apologize for integrating r with respect to h, if that was confusing.

 

[RoyalCat]

That too is incorrect. ;) Though you're on the right track!

Remember that a spherical shell contributes no \frac{1}{r^2} field inwards. That is to say, when calculating the gravitational field inside the star at a radius r_0<R, you must only regard the contribution of the mass inside, at radii 0<r<r_0, as all the mass on the outside contributes nothing to the gravitational field!

 

[j-e_c]

OK, I've got it, thanks RoyalCat!

 

[RoyalCat]

OK, I've got it, thanks RoyalCat!

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.