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 )