# Gravitational potential energy of an isothermal sphere

I am trying to calculate the gravitational potential energy, W, defined as:

$$W = -\frac{1}{2}\int\rho(r)\Phi(r)d^{3}r$$

for an isothermal sphere. I am given that the density profile varies with r as:

$$\rho \propto r^{-2}$$

and that $$\Phi(r)$$ is defined as:

$$\Phi(r)=\Phi_{0} + \eta ln(r)$$

where $$\eta$$ is a constant. Given it is a spherical system I set up the following equation.

$$W = -\frac{1}{2}\int^{r_{0}}_{0}4\pi k(\Phi_{0} + \eta ln(r))dr$$

Where k is a constant of proportionality. I am not sure how to calculate this integral without ending up with infinities due to $$ln(0)$$ values. I am also told that the sphere can be considered to be truncated at $$r_{0}$$, does this mean I can somehow ignore the $$ln(r)$$ term in the integral?

• Doesn't a problem arise at the definition of $\Phi(r)$? And have you looked up the integral for ln(r)? Check out that integral and you should see a way out of the problem. – Bill N Apr 18 '19 at 17:27
• @BillN ah, so for a truncated sphere the definition of the definition of $\Phi(r)$ is $\Phi(r)=\Phi_{0}$? – Vinteuil Apr 18 '19 at 17:30
• No. Go find the integral of ln(r). What happens when $r\to 0$? – Bill N Apr 18 '19 at 17:31
• @BillN I get $r(1-ln(r))$, so as $r \rightarrow 0$, $r(1-ln(r)) \rightarrow 0$? – Vinteuil Apr 18 '19 at 17:46