# Gravitationnal potential of a sphere

1. Nov 18, 2004

### quasar987

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.

2. Nov 18, 2004

### Tide

The best approach would be to use Gauss' Law - MUCH easier!

3. Nov 19, 2004

### quasar987

Got it! Thanks a lot.

Last edited: Nov 19, 2004
4. Nov 19, 2004

### quasar987

I thought I had it but now I'm majorly confused. Gauss's law says that the field must satisfy

$$\vec{\nabla} \cdot \vec{g} = -4 \pi G \rho$$

And we know it must also satisfy

$$\vec{\nabla} \times \vec{g} = \vec{0}$$

and

$$\lim_{r \rightarrow \infty} \vec{g} = \vec{0}$$

I thought of the function

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

(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

$$\vec{g} = -\frac{MG}{r^3}\vec{r}$$

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

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

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

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

Last edited: Nov 19, 2004
5. Nov 19, 2004

### ZapperZ

Staff Emeritus
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. Nov 19, 2004

### Staff: Mentor

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, $\Phi = -4 \pi G M$ (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: Nov 19, 2004
7. Nov 19, 2004

### scifind

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. Nov 19, 2004

### quasar987

Yea, no I'm not very familiar with Gauss's law. But I'll give it a shot.

9. Nov 19, 2004

### quasar987

Yea that was easy.

10. Nov 19, 2004

### quasar987

(writing.....)

Last edited: Nov 19, 2004