- #1

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For some shell at a given radius away from the center of the star, we can write its mass as:

[tex]dm=4\pi r^2 \rho (r) dr[/tex]

Since the entire shell is at the same radius from the center, we can essentially treat it as a sheet and write the equation for the differential force as:

[tex]dF=\frac{G M_{enc} dm}{r^2}[/tex]

Where,

[tex]M_{enc}=\int_{0}^{r}dm=\int_{0}^{r}4\pi r^2 \rho (r) dr[/tex]

If we take,

[tex]\rho (r)=\frac{\rho_0}{r}[/tex]

Then we can rewrite this as:

[tex]dF=\frac{G (\int_{0}^{r}4\pi r \rho_0 dr) 4\pi r \rho_0 dr}{r^2}[/tex]

[tex]dF=\frac{16 \pi ^{2} \rho_0^2(r^2/2)r dr}{r^2}[/tex]

[tex]dF=8 \pi^2 \rho_0^2 r dr[/tex]

If we call the radius to some shell in question r_s, and integrate from r_s to some larger radius, r_0,

[tex]F=\int_{r_s}^{r_{\odot}}8 \pi^2 \rho_0^2 r dr[/tex]

[tex]F=4\pi^2\rho_0^2(r_{\odot}^2-r_s^2)[/tex]

Which seemed all fine and good until I realized this gives a weight of zero for the outermost shell. My thoughts are that maybe there is some way to split the weight into the sum of the layers above the layer in question (effectively what I've found) and the weight of the layer itself, but when I write the layer as an infinitesimal it has a weight zero.

Anyone have any thoughts, or if I'm on the right track?