Gravitationnal potential of a sphere

In summary, using Gauss's Law, the gravitational potential function of a sphere of uniform density with radius a and mass M can be found as :\vec{\nabla} \cdot \vec{g}= -4 \pi G \rho\vec{r}
  • #1
quasar987
Science Advisor
Homework Helper
Gold Member
4,807
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
  • #2
The best approach would be to use Gauss' Law - MUCH easier!
 
  • #3
Got it! Thanks a lot.
 
Last edited:
  • #4
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:
  • #5
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.
 
  • #6
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:
  • #7
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!
 
  • #8
Yea, no I'm not very familiar with Gauss's law. But I'll give it a shot.
 
  • #9
Yea that was easy. :smile:
 
  • #10
(writing...)
 
Last edited:

What is the formula for calculating the gravitational potential of a sphere?

The formula for calculating the gravitational potential of a sphere is V = -GM/r, where V is the potential, G is the gravitational constant, M is the mass of the sphere, and r is the distance from the center of the sphere.

How does the size of a sphere affect its gravitational potential?

The gravitational potential of a sphere is directly proportional to its mass. This means that the larger the sphere, the greater its gravitational potential will be.

What is the relationship between the distance from a sphere and its gravitational potential?

The gravitational potential of a sphere is inversely proportional to the distance from its center. This means that as the distance from the center of the sphere increases, the gravitational potential decreases.

Can the gravitational potential of a sphere be negative?

Yes, the gravitational potential of a sphere can be negative. This occurs when the distance from the center of the sphere is greater than the radius, resulting in a negative potential value.

How does the gravitational potential of a sphere affect objects around it?

The gravitational potential of a sphere determines the strength of its gravitational field, which can affect the motion of objects around it. Objects will experience a greater force of attraction towards the sphere if its gravitational potential is higher.

Similar threads

  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
548
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
775
  • Introductory Physics Homework Help
Replies
21
Views
2K
Replies
10
Views
428
  • Introductory Physics Homework Help
Replies
6
Views
668
  • Introductory Physics Homework Help
Replies
23
Views
349
  • Introductory Physics Homework Help
Replies
17
Views
404
Back
Top