Deriving the gravitational binding energy of the cluster

Click For Summary

Homework Help Overview

The original poster attempts to derive the gravitational binding energy of a cluster, expressed as $$U = -\alpha \frac{GM^2}{r}$$. They start from the differential form $$dU = -\frac{GM(r)dm}{r}$$ and express $$dm$$ in terms of density and volume elements, but encounter difficulties in progressing further with the derivation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants question the dependencies of $$M(r)$$ and $$\rho(r)$$, suggesting that without explicit definitions, the integral cannot be computed. Others inquire about the implications of spherical symmetry on the problem.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the assumptions regarding density and mass distribution. There is a recognition that a clearer statement of the problem could facilitate finding a solution, but no consensus has been reached yet.

Contextual Notes

Participants note the lack of explicit information regarding the dependencies of $$M(r)$$ and $$\rho(r)$$, which is critical for evaluating the integral. The original poster also expresses uncertainty about deriving the energy from an unknown matter density.

Arman777
Insights Author
Gold Member
Messages
2,163
Reaction score
191
I am trying to derive the gravitational binding energy of the cluster. Its given as

$$U = -\alpha \frac{GM^2}{r}$$

Now for the derivation I started from
$$dU = -\frac{GM(r)dm}{r}$$

I I tried to write ##dm = \rho(r)4 \pi r^2dr## and do it from there but I could not do much. Any ideas how can I proceed ?
$$dU = -\int_0^R \frac{GM(r)}{r}\rho(r)4\pi r^2dr$$

If there's a simpler way that's also fine.

[Moderator's note: Moved from a technical forum and thus no template.]
 
Last edited by a moderator:
Physics news on Phys.org
To tell you the truth I don't understand what is exactly your problem?
I mean you want to get ##U=-\alpha GM^2/r## from ##dU = -\int_0^R \frac{GM(r)}{r}\rho(r)4\pi r^2dr##, but you didn't state what are the explicit dependencies of ##M(r)## and ##\rho(r)##, so how can you calculate this integral?!
I mean if they are constants with respect to r then it's a simple matter to calculate this integral, but there is insufficient data to compute this integral as I see it.
 
MathematicalPhysicist said:
To tell you the truth I don't understand what is exactly your problem?
I mean you want to get ##U=-\alpha GM^2/r## from ##dU = -\int_0^R \frac{GM(r)}{r}\rho(r)4\pi r^2dr##, but you didn't state what are the explicit dependencies of ##M(r)## and ##\rho(r)##, so how can you calculate this integral?!
I mean if they are constants with respect to r then it's a simple matter to calculate this integral, but there is insufficient data to compute this integral as I see it.
If its spherically symmetric ?
 
You mean the density ##\rho(r)## is proportional to ##r^{-3}##? or is it ##M(r)##?
Anyway, if you state your problem with all the assumptions as clear as possible then I believe a solution is possible.
 
MathematicalPhysicist said:
You mean the density ##\rho(r)## is proportional to ##r^{-3}##? or is it ##M(r)##?
Anyway, if you state your problem with all the assumptions as clear as possible then I believe a solution is possible.
I seen some solution here but it does not look like my thing...

https://physics.stackexchange.com/q...otential-energy-of-any-spherical-distribution

Thats the problem. I mean I thought I mentioned the spherically symmetry thing. Which I noticed I did not so I said it. I just wondered is it derivable from a unknown matter density. If its not that's also okay.
 

Similar threads

Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
29
Views
2K
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K