Potential energy of a plummer sphere

[ghetom]

Homework Statement

The Plummer sphere of total mass M and scale radius a is a simple if crude model for
star clusters and round galaxies. Its gravitational potential:

\phi(r) = -GM / (r^2 +a^2)^{1/2}

approaches that of a point mass for r >> a

Find the density of the sphere as a function of r, and calculate the potential energy of the distribution.

Homework Equations

\nabla^2 \phi = 4 \pi G \rho
U_i = \phi_i m_i
\nabla^2 F= (1/r^2) * d/dr(r^2 dF/dr)

The Attempt at a Solution

It's easy to show that \rho= \frac{3a^2*M^2 *G}{4 \pi (r^2 + a^2)^{5/2}}

but I can't calculate the potential;

I think that
U = \int {\phi \rho} d{Volume} (*)
thus
U = \int {\phi * \rho * 4 \pi r^2} dr
thus
U = A \int \frac{r^2}{(r^2 +a ^2)^3} dr
where A is a constant

but that integral is horrible (where as if it was r or r^3 I could do it ).
Is (*) correct? have I made a howler? or is what I've done so far correct?

 

[haene]

Hi ghetom

Thread is quite old but I had to do the same exercise so I would like to complete the thread.
The integral you was looking for is right and its solution is complicate. For such beasts i use
this page. Only in the limit R→\infty the integral has a elegant solution.

\int^{R}_{0} \frac{r^2}{(r^2+a^2)^3} dr = \frac{1}{8}\frac{1}{a^3}\frac{r^4}{(r^2+a^2)^2}tan^-1(\frac{r}{a}) = \frac{\pi}{2\times8} \frac{1}{a^3} ,limit R→∞

with that we got the total potential energy W

W = \frac{1}{2}\int \Phi \rho dV

integrated over the volume V at limit R→∞

W = \frac{3\pi G M^2}{32} \frac{1}{a}

took me quite a while to get there (and to type as well )