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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?

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    $\begingroup$ 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. $\endgroup$ – Bill N Apr 18 at 17:27
  • $\begingroup$ @BillN ah, so for a truncated sphere the definition of the definition of $\Phi(r)$ is $\Phi(r)=\Phi_{0}$? $\endgroup$ – Vinteuil Apr 18 at 17:30
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    $\begingroup$ No. Go find the integral of ln(r). What happens when $r\to 0$? $\endgroup$ – Bill N Apr 18 at 17:31
  • $\begingroup$ @BillN I get $r(1-ln(r))$, so as $r \rightarrow 0 $, $r(1-ln(r)) \rightarrow 0$? $\endgroup$ – Vinteuil Apr 18 at 17:46

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