# Integral of pressure over Newtonian star

Gold Member

## Homework Statement

Last part of MTW Gravitation exercise 23.7:

Calculate in Newtonian theory the energy one would gain from gravity if one were to construct a star by adding one spherical shell of matter on top of another, working from the inside outward. Use Laplace's equation d(r^2 (d phi/dr))/dr = 4 pi r^2 rho and the equation of hydrostatic equilibrium dp/dr = -rho d phi/dr to put the answer in the following equivalent forms:

(energy gained from gravity) is equivalent to -(gravitational potential energy)
= [various things which all come to Gm^2/2R]
= 3 times integral from 0 to R of 4 pi r^2 p dr.

Variables assumed:
R = total radius of star
m = total mass of star
p = pressure (function of r)
rho = density
phi = Newtonian gravitational potential (function of r)

2. Homework Equations :

d(r^2 (d phi/dr))/dr = 4 pi r^2 rho

Equation of hydrostatic equilibrium:
dp/dr = -rho d phi/dr

## The Attempt at a Solution

It's not clear from the context whether rho is assumed to be able to vary with radius, but if it works in that case it should also work when rho is constant, so I tried that first.

Assuming rho constant then by integrating the equation of hydrostatic equilibrium we have:

p = -rho phi(r) + k where k is constant of integration

I assume that the pressure must be zero at the surface of the star (and increases with decreasing radius), so

k = rho phi(R)

giving

p = -rho (phi(R) - phi(r))

(Convention in MTW is to use units where G=1 but I'll include it explicitly anyway)

phi(r) = -G/r * mass within radius r
= -G/r * mr^3/R^3
= -Gm/R * r^2/R^2

phi(R) = -Gm/R

This gives

p = -rho (-Gm/R (1 - r^2/R^2))

The quantity requested on the last line of the question is:

3 * integral from 0 to R of 4 pi r^2 p dr

Filling in the above expression for p, this is

3 * integral from 0 to R of 4 pi r^2 rho (Gm/R) (1-(r^2/R^2)) dr

= 3 * 4 pi Gm/R rho integral from 0 to R r^2 (1-(r^2/R^2)) dr
(moving the constants outside the integral)

= 12 pi rho Gm/R (r^3/3 - r^5/5R^2) between r = 0 and R
(integrating)

= 12 pi rho Gm/R R^3 (1/3 - 1/5)

= 12 pi rho Gm/R R^3 (2/15)

Using m = 4/3 * pi R^3 rho, this becomes

12 * 3/4 * (4/3 pi rho R^3) Gm/R (2/15)

= 9 * 2/15 * Gm^2/R

= 6/5 Gm^2/R

However, all of the other cases gave 1/2 Gm^2/R. Where did I go wrong, please? (I've tried various other ways of looking at it, and all of them give the same result, so at least I'm consistent). This isn't for homework - I'm just trying to understand the relationship to the pressure term in the Komar mass.