# Gravitational field of a disc

1. Aug 15, 2004

### LENIN

What if we had a large and massive object shaped like a disc (look at the attachment) and we would walk on its edge. If we would step on the flat part what would happen?

P.S. I am going on holidays for a fe days so i won't be abel to check out your answers very soon.

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2. Aug 15, 2004

### Gonzolo

Just how big is your disk? Are you considering the G field OF the disc? If it's big enough to create a G field sticking you together, it'll probably fall apart.

If the disk is small enough to be on earth, and you're talking about the Earth's G, you can simply try this on a merry-go-round!

3. Aug 15, 2004

### Nenad

I dont understand you diagram, could you elaborate.

4. Aug 15, 2004

### pmb_phy

You would probably be drawn toward the center of the disk, i.e. there might be a component of the gravitational force parallel to the surface of the disk and directed towards the center. But to really determine this you'd have to calculate it. This is just an educated guess.

Pete

5. Aug 15, 2004

### pervect

Staff Emeritus
I'm not sure what your diagram means either, but it sounds like you may be describing an Alderson disk?

http://en.wikipedia.org/wiki/Alderson_disk

If we put the disk in the x-y plane, gravity will point roughly in the z direction if one is not too close to the edge of the disk.

The Newtonian surface gravity at the center of an infinite thin disk, with a density $$\rho$$ kg/m^2 will be $$2*Pi*G*\rho$$. For an infinite disk, the gravity will not change with height, for a finite disk the gravity will start to drop off eventually.

6. Aug 18, 2004

### LENIN

I will try to explain more precisaly. The basic idea is that you are standing on the edg of a large disk. Something the saize of a planet. Than at some moment you jump from the edg to the surfec. The question I am asking is will you fall towerds the canter, or will you be abel to stand on the sufec of the disk close to the edg.

P.S I hope this diagrem will be easyer to understand.

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7. Aug 18, 2004

### Metallicbeing

Common sense would say that you would fall to the center since there is no "edge" to stop you.

Is this a simplified model of what you think gravity is? A 2D "Earth" vs. 3D "gravity" is what I imagine here.

8. Aug 18, 2004

### pervect

Staff Emeritus
At the center of the disk, there will be some local acceleration pulling you straight down. Call this value 'g'.

At the edge of the disk, this force will be less, but will rapidly increase as you move towards the center. Modelling the situation as being some height h above an infinitely dense disk, the downwards force will be approximated by (and greater than)

$$g \left( 1-{\frac {1}{\sqrt {1+{\frac {{d}^{2}}{{h}^{2}}}}}} \right)$$

where you've moved a distance d towards the center of the disk

The force pulling you towards the center of the disk will be approximated by (and less than)

$$\frac {g}{\pi}\ln \left( {\frac {D}{d}} \right)$$

where d is as before your distance from the edge, and D is the total diameter of the disk.

So this means you'll fall some distance, then stop, I'd need some more details to tell you exactly how far you'd fall.

Last edited: Aug 18, 2004
9. Aug 18, 2004

### pervect

Staff Emeritus
The last equation should be
$$\frac {g}{\pi}\ln \left( {\frac {D}{d}} \right)$$

I can't seem to get the latex to edit. (If a moderator fixes this problem, this post can be deleted).

10. Aug 19, 2004

### LENIN

Thanks pervect

11. Aug 21, 2004

### zynko

how do you guys know what equations to us? it's like the concept and math changes according to the shape of an object.

12. Aug 21, 2004

### pervect

Staff Emeritus
Here's what I did - note the analysis is only approximate.

First off, I modelled the disk as being infinitely thin and infinitely dense, to get rid of the third dimension.

For the first part of the problem, I assumed that the person was a height h above the infinitely dense disk, this would represent half the thickness of a more realistic disk of finite thickness. To get an estimate of the downwards force for a person standing on the disk, I integrated the force for a radius 'd=kh' around the person. This component will provide a pure downward force. It sets a lower bound on the downward force.

$$\int_{r=0}^{r=kh} 2\,{\frac {G\rho\,\pi \,r{\it dr}\,h}{ \left({h}^{2}+{r}^{2} \right) ^{3/2}}}$$

Motivation: the mass of a circle element of radius r is $$2 \pi \rho r dr$$. The distance of the person from this circular mass element is $$\sqrt{r^2+h^2}$$. F = G*m / r^2. The downward component gets multiplied by a factor of $$\frac{h}{\sqrt{r^2+h^2}}$$, this factor is 1 when r = 0, and goes towards zero as r increases, it is the sine of the angle if you draw the diagram (opposite side over hypotenuse). The lateral component cancels out.

The value of the integral is

$$2\,{\frac {G\rho\,\pi \, \left( -1+\sqrt {1+{k}^{2}} \right) }{\sqrt {1+{k}^{2}}}}$$

The rapidity with which this force approaches a constant value $$2 \pi G \rho$$ argues that the ignored components are probably not too significant.

The above is a "good force" from the point of view of someone wanting to stand up and not slide along the disk.

To evaluate the force pulling a person sideways, towards the center of the disk, I ignored the height above the disk and evaluated the integrall

$$\int_{r=d}^{r=D}\int_{\theta=-\frac{pi}{2}}^{\frac{pi}{2}} {\frac {G\rho\,{\it dr}\,{\it d \theta}\,\cos \left( \theta \right) }{r}$$

This is a rather crude approximation, but it sets an upper bounds on the sidewards force. It could probably be improved. It is an integral around a half-circe of radius r>d, r<D. The result is

$$2\,G\rho\,\ln \left( {\it \frac{D}{d}} \right)$$

The integrals were all done via computer, which makes it both fairly easy to do, and less error-prone - except for setting them up, of course.

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