Integration problem deriving virial theorem

[Bried]

Hello there,

I'm reading a section of my textbook that is deriving the virial theorem from the hydrostatic equilibrium equation. In part of the derivation it states that
$$-\int_0^M\frac{Gm(r)}{r}dm(r)=E_{GR}=-\frac{GmM}{r}$$
When I perform this integral I get the wrong answer. Here's my working.
$$-\int_0^M\frac{Gm(r)}{r}dm(r)=-Gr^{-1}\int_0^Mm(r)dm(r)=-Gr^{-1}\left[\frac12m^2\right]_0^M=-Gr^{-1}\left(\frac12M^2-0\right)=-\frac{GM^2}{2r}$$
I could just move on and take the text for granted but I like to understand how everything is derived so that I don't get tripped up in the future.
I'm not used to seeing $$m(r)$$ in the $$dm(r)$$ but surely the $$(r)$$ could be omitted right? Isn't this just stating that $$m$$ is a function of $$r$$?
Sorry for being a bit of a simpleton but having not done any studying over the summer I seem to have lost my flow somewhat. If someone could point me in the right direction I would be most grateful.

Thanks

Regards

Brian

 

[Mark44]

Hello there,

I'm reading a section of my textbook that is deriving the virial theorem from the hydrostatic equilibrium equation. In part of the derivation it states that
$$-\int_0^M\frac{Gm(r)}{r}dm(r)=E_{GR}=-\frac{GmM}{r}$$

Is there any additional information on what m(r) represents? It would appear to be the mass at a distance of r units from the center, but that's just a guess. Is there an assumption that the mass increases in a linear fashion?

When I perform this integral I get the wrong answer. Here's my working.
$$-\int_0^M\frac{Gm(r)}{r}dm(r)=-Gr^{-1}\int_0^Mm(r)dm(r)=-Gr^{-1}\left[\frac12m^2\right]_0^M=-Gr^{-1}\left(\frac12M^2-0\right)=-\frac{GM^2}{2r}$$

I don't believe it's valid to treat r as if it were a constant, and bring it outside the integral as you did. dm(r) is the same thing as dm, but emphasizing that m is a function of r. As far as the integration is concerned, I believe that the "of r" part could be omitted, but as I said above, it doesn't seem valid to bring the r out, since m depends on it.

I could just move on and take the text for granted but I like to understand how everything is derived so that I don't get tripped up in the future.
I'm not used to seeing $$m(r)$$ in the $$dm(r)$$ but surely the $$(r)$$ could be omitted right? Isn't this just stating that $$m$$ is a function of $$r$$?
Sorry for being a bit of a simpleton but having not done any studying over the summer I seem to have lost my flow somewhat. If someone could point me in the right direction I would be most grateful.

Thanks

Regards

Brian

 

[Bried]

Hi Mark,

Thanks for getting back to me. Yes, m(r) is the mass of gas contained within a cloud of radius r. I don't believe the text actually states whether the mass increases in a linear fashion. The notation in this new module I'm doing is throwing quite a bit. I recently got a distinction in my Open University level 2 maths so I know I'm more than capable of doing it. I'm so used to evaluating integrals with dx, dt or d anything for that matter but the dm(r) has really confused me and my previous maths course hasn't prepared me for this. I'm feeling rather apprehensive about this module now.

I'm not sure if this will work but here's a dropbox link to the text in the book.

https://www.dropbox.com/s/rteojqv84gulfmd/Virial theorem derivation.jpg?dl=0

Thanks again for your help.

Regards

Brian

 

[Mark44]

In the text on dropbox, they don't evaluate this integral:
$-\int_0^M\frac{Gm(r)}{r}dm(r)=E_{GR}=-\frac{GmM}{r}$
They simply note that it is the gravitational potential energy of the cloud, E_GR. I'll have to take their word for it. What they do is to evaluate this integral:
$$3\int_{r = 0}^{r = R} V(r) dP(r)$$
using integration by parts. They result they get from integration is equal to E_GR.

 

[Bried]

Hi Mark,

Thanks for having a look at this again for me. Your response has put my mind at rest and gives me the confidence to carry on through the text without worrying too much about verifying every detail. I did manage to complete the integral on the left hand side, using integration by parts, without any problems.

Regards

Brian