What is the gravitational field of a flat disc?

[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? :@

 

[Andrew Mason]

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...

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

 

[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.

 

[speg]

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

 

[arildno]

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

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

 

[speg]

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

 

[arildno]

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

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