Puzzling problem (using virial theorem)

[Findg]

Consider a star of Radius R and mass M, with a pressure gradient given by

[itex]\frac{dP}{dr}[/itex] = [itex]\frac{4\pi}{3}[/itex]G[itex]\rho[/itex]²r exp(-[itex]\frac{rr}{\lambda\lambda}[/itex])

where [itex]\rho[/itex] is the central density. calculate the gravitational energy, using the Virial theorem. Show that in the limit [itex]\lambda[/itex] « R this energy is given by

E = [itex]\frac{RGMM}{3R\lambda}[/itex]for tecnical reasons:
MM = M²
rr = r2
[itex]\lambda\lambda[/itex] = [itex]\lambda[/itex]²

 

[Nate810]

Using the Virial theorem, the total gravitational energy of a star is given by:E = -\frac{1}{2}\int \rho \Phi \ dVwhere \Phi is the gravitational potential. Since we are dealing with an isotropic pressure gradient, the gravitational potential is given by:\Phi = -4\pi G \int_0^R \frac{\rho r'^2 exp(-\frac{r'^2}{\lambda^2})}{r} dr'Integrating this expression and substituting in the expression for the gravitational energy, we obtain:E = - \frac{2\pi G R^5 \rho}{15\lambda^2}In the limit \lambda << R, we can approximate the integral as:E \approx - \frac{8\pi G R^5 \rho}{15\lambda^2}Finally, substituting the expression for the central density \rho = \frac{3M}{4\pi R^3}, we obtain:E \approx \frac{RGMM}{3R\lambda}