Gravitationnal potential of a sphere

  • Thread starter Thread starter quasar987
  • Start date Start date
  • Tags Tags
    Potential Sphere
Click For Summary

Homework Help Overview

The discussion revolves around finding the gravitational potential function of a sphere with uniform density, characterized by its radius and mass. Participants explore various approaches to derive the potential, referencing known expressions for gravitational fields and the application of Gauss's Law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster considers integrating the potential from a shell but questions its validity. Some participants suggest using Gauss's Law as a simpler method. Others express confusion about the conditions that the gravitational field must satisfy and how they relate to the potential function.

Discussion Status

Participants are actively engaging with the problem, raising questions about the application of Gauss's Law and the conditions for gravitational fields. Some guidance has been offered regarding the use of spherical coordinates and the integral form of Gauss's Law, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is a mention of homework rules regarding the posting of such problems, and some participants express uncertainty about the assumptions being made in the problem setup.

quasar987
Science Advisor
Homework Helper
Gold Member
Messages
4,796
Reaction score
32
What would be the tactic to find the gravitationnal potential function of a sphere of uniform density with radius a and mass M ? I know the expression of the potential for a shell of radius R. I thought I could integrate from 0 to a and it would give the potential of the sphere, but apparently not.

Thanks for your help!
 
Physics news on Phys.org
The best approach would be to use Gauss' Law - MUCH easier!
 
Got it! Thanks a lot.
 
Last edited:
I thought I had it but now I'm majorly confused. Gauss's law says that the field must satisfy

[tex]\vec{\nabla} \cdot \vec{g} = -4 \pi G \rho[/tex]

And we know it must also satisfy

[tex]\vec{\nabla} \times \vec{g} = \vec{0}[/tex]

and

[tex]\lim_{r \rightarrow \infty} \vec{g} = \vec{0}[/tex]

I thought of the function

[tex]\vec{g}(x,y,z) = -\frac{4}{3} \pi G \rho (x\vec{i} + y\vec{j} + z\vec{k})[/tex]

(where the vectors i, j and k are unit vectors in the direction of the x, y and z axis respectivly) but this function, although it satisfy the two first conditions, doesn't meet the third.

But even more confusing is that the answer the book gives is

[tex]\vec{g} = -\frac{MG}{r^3}\vec{r}[/tex]

which certainly satisfy the limit condition but if we make the substitution [itex]M = \frac{4}{3} \pi a^3 \rho[/itex], we get

[tex]\vec{g}(x,y,z) = \frac{-\frac{4}{3} \pi a^3 G \rho}{(x^2+y^2+z^2)^{3/2}}(x\vec{i} + y\vec{j} + z\vec{k})[/tex]

and we see that the first condition isn't met!

However, we notice that if M were given by [itex]M = \frac{4}{3} \pi r^3 \rho[/itex], then the book's solution is the same as mine and all 3 conditions are met. But this equality is wrong, isn't it?

Please help me see what's the matter!
 
Last edited:
It appears that you somehow have never used Gauss's Law, or are aware of the integral form of Gauss's Law. Furthermore, why are you even using cartesian coordinates (which would make this problem WAY more complicated) and not spherical coordinates?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

It also appears that this is a homework problem, which should have been posted in the Homework zone section of PF.

Zz.
 
This all sounds overly complicated for such a simple problem. Why not use Gauss's law in its integral form: The gravitational flux through any closed surface, [itex]\Phi = -4 \pi G M[/itex] (where M is the mass enclosed by the surface and G is the gravitational constant).

Edit: Zapper beat me to it! And I will move this to the appropriate forum.
 
Last edited:
you seems to have a good guess for satsifying conservative condition.
but you would see that your guess doesn't satisfy our physical condition.
i.e. g will be infinity when r tends to infinity!
 
Yea, no I'm not very familiar with Gauss's law. But I'll give it a shot.
 
Yea that was easy. :smile:
 
  • #10
(writing...)
 
Last edited:

Similar threads

  • · Replies 43 ·
2
Replies
43
Views
5K
Replies
1
Views
1K
  • · Replies 29 ·
Replies
29
Views
2K
Replies
2
Views
670
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
2K
Replies
15
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
2K