# Formulae for gravitatitional equilibrium

I am trying to calculate at which point gravitational equilibrium sets in for various bodies (planets stars neutron stars etc.) assuming they are perfect spheres. However, the radius I get is not equal to what it should be according to Wikipedia (I tried it for the Sun, a planet and a neutron star, but all of them were off by quite a bit).

Below is the example of the neutron star radius I'm trying to get:

My result (factor) is 1 at:

Radius: $2228588\mathrm{\ m}$

Density: $1.2768744892680034 \times 10^{11} \mathrm{\ kg/m^3}$

The Wikipedia article about neutron stars says the radius of a neutron star is about 12 km and the density is about 10 times higher than mine. My result is 2228km, which is quite a bit off, so I have a second question:

1. Am I calculating all the forces I need?
2. Are the formulae I'm using correct?

I took them from Wikipedia and university slides and I found multiple variations of almost all of them (most of which I actually didn't list below), so I'm quite confused as to what is correct.

I know that I'm using iron as the element and I should split it up and convert the protons to neutrons, but that would only increase the degeneracy pressure and result in an even bigger radius, if I'm not mistaken.

Here's how I calculated it:

Mass of an electron: $M_e$

Standard atomic weight of hydrogen: $M_h = 1.00782503223$

Mass of a neutron: $M_n$

Atomic mass of ${\mathrm{Fe}^{56}_{26}}$: $55.934936 \mathrm{\ u}$

Number of atoms: $7.938 \times 10^{31}$

Number of neutrons per atom: $(56-26) \times \text{number of atoms}$

Number of electrons per atom: $26 \times \text{number of atoms}$

Total number of particles: $\text{number of atoms} + \text{number of electrons}$

Mass: $\text{atomic mass} \times \text{number of atoms} = 4.4400513610370694 \times 10^{30} \mathrm{\ kg} \approx 2.3M_☉$

Radius: $2,228,588 \mathrm{\ m}$

Temperature: $9.452 \times 10^{-7}° K$

Force of gravity: $\frac{GM^2}{r}$

Volume: $\frac{4\pi r^3}{3}$

Particle density: $\frac{\text{number of particles}}{\text{volume}}$

a: $\frac{4\sigma}{c}$

Radiation pressure: $\frac{a T^4}{3}$

Gas pressure: $\text{particle density} \times k_B T$

Electron degeneracy pressure: $\frac{\pi^3 \hbar^2} {15 M_e} \times \frac{3 \times \text{number of electrons}} {V \pi}^{\frac{5}{3}}$

Electron degeneracy pressure 2: $\frac{\pi^2 \hbar^2} {5 M_e \times M_h^{\frac{5}{3}}} \times \frac{3}{\pi}^{\frac{2}{3}} \times \frac{\frac{M}{V}}{1}^ {\frac{5}{3}}$

Neutron degeneracy pressure: $\frac{\pi^3 \hbar^2 }{15 M_e} \times \frac{3 \times \text{number of neutrons}}{V \pi}^{\frac{5}{3}}$

Neutron degeneracy pressure 2: $\frac{3^{\frac{10}{3}} \hbar^2 }{ 15 \pi^{\frac{1}{3}} \times M_n^{\frac{8}{3}} \times r^5} \times \frac{M}{4}^{\frac{5} {3}}$

Total pressure: $\text{gas pressure + radiation pressure + neutron degeneracy pressure + electron degeneracy pressure}$

Total pressure force: $\text{total pressure} \times \text{volume}$

If total pressure force = force of gravity, then the body is in equilibrium.

• Can you improve the formatting and grammar here? At the moment, it's a bit unclear. Also, Mathjax is enabled on the site. Here's a good tutorial. – HDE 226868 Jan 20 '16 at 0:37
• Additionally, the density of a neutron star - like that of any celestial body - is not constant. It varies with radius. Plus, you'll need to take into account relativistic effects. – HDE 226868 Jan 20 '16 at 0:38
• the inputs are the amount and the element(and thus the mass) the temperature and the radius everything else is calculated based on those values – asdf Jan 20 '16 at 2:08
• I can't give you a complete answer, but at first glance you're calculating with the electron pressure. But actually in the core of the neutron star, protons and electrons fuse to neutrons. So you get less particles with higher mass. You might eventually get a quark-gluon plasma, which might again need to be treated differently. – Gerald Jan 20 '16 at 12:56
• Force is not pressure times volume; among many errors. – ProfRob Oct 4 '16 at 20:50

Mass = $\text{atomic mass} \times \text{number of atoms} = 4.4400513610370694 \times 10^{30} \mathrm{\ kg} \approx 2.3M_☉$
Herein lies the problem: your units and exponents are messed up. The mass of your iron isotope is $55.934936 \mathrm{\ u}$, and there are $7.938⋅10^{31}$ atoms. Do the math, and you'd get a total mass of $4.4400513610370694 \times 10^{33}\mathrm{\ u}$. That converts to approximately $7372878 \mathrm{\ kg}$. Nowhere near the mass of the Sun!