# Homework Help: Gravity of a Flat Disc

1. Jun 3, 2006

### speg

Hey guys, so I'm back in school after an 8-month break, and I'm feeling a bit rusty

So I've got a flat disc of radius A, with constant density p, in the z=0 plane. I want to calculate the gravitational field at any point up or down the z-axis.

I integrated the potential over the disc got the correct potential function (which is given) of :

$$\Phi(z)=-G\rho2\pi(\sqrt{a^2+z^2}-z)$$
So now I just take the negative derivative of this to get the Gravitational field, right?
$$G(z)=-\nabla\Phi(z)$$
$$G(z)=-G\rho2\pi(\frac{z}{\sqrt{a^2+z^2}}-1)$$

But this means there is a force at z=0 when I think there should not be...

$$How do I make a new line in Latex? \\ this doesn't seem to work? :@$$

Last edited: Jun 3, 2006
2. Jun 3, 2006

### Andrew Mason

What is the thickness of the disc? Where does that appear in your calculation?

The force/unit mass at (0,0,z) from a ring element of the disc of radius r thickness h and width dr would be:

$$dF = Gdm/s^2 = \frac{G\rho 2\pi r hdr}{r^2 + z^2}$$

assuming h to be small compared to z. Integrate that from r = 0 to r = A.

AM

Last edited: Jun 3, 2006
3. Jun 3, 2006

### arildno

The potential is not correct, since it is not symmetric about the disk.
The proper potential is:
$$\Phi(z)=-G2\pi\rho(\sqrt{a^{2}+z^{2}}-|z|)$$
yielding the proper force per unit mass along the z-axis (in the positive vertical direction) :
$$f(z)=2\pi\rho{G}(\frac{z}{\sqrt{a^{2}+z^{2}}}-\frac{z}{|z|})$$
The limiting values as z goes to zero,
$$\lim_{z\to{0}^{+}}f(z)=-2\pi\rho{G}, \lim_{z\to{0}^{-}}f(z)=2\pi\rho{G}$$
are the strengths of the force just outside the disk, on either side.
There is a leap of discontinuity across the disk, where AT the origin, the force is, indeed 0.

Last edited: Jun 3, 2006
4. Jun 3, 2006

### speg

$$\frac{d|z|}{dz}=\frac{z}{|z|}$$?

5. Jun 3, 2006

### arildno

Quite so.
The derivative of the absolute value function is not defined at z=0.

6. Jun 3, 2006

### speg

And so F cannot be defined at z=0? So we take that to mean there is no force there?

7. Jun 3, 2006

### arildno

No, it doesn't. It just means you have to consider the z=0 case separately.