# What happens to gravity in one dimension of space?

1. Oct 8, 2013

### KeplerJunior

I recently heard that it was found how gravity would behave in different dimensions of space. Apparently, while in our 3 dimensions of space gravity is F= x/r^2, in 4 dimensions it would be F=x/r^3, so it would be weaker, and in 2 dimensions it would be F=x/r so it would be stronger.What it didn't say was what would happen in 1 dimension, it follows from the above that it would be F=x/r^-1, which simplifies to F=xr, which would indicate that gravity would get stronger as you went further away from the object. Is this what would happen? And if so does that mean gravity would become a repulsive force?

2. Oct 8, 2013

### D H

Staff Emeritus
You skipped over a key integer. Hint: What integer is between -1 and +1?

3. Oct 8, 2013

### KeplerJunior

Oh of course, so would it just be x/r^0 so x/1, so x. So then distance wouldn't change anything in 1 dimension

4. Oct 8, 2013

### Jarfi

Um, actually, the integer x/r dr is xln(r). Which means the equation of gravity in one dimension would be:

f(r) = gravitational force = xln(r). So gravity would according to this increase with increased distance, something I find hard to believe. It means the further you are away, the stronger it'd pull, well until you get to a distance of 0, then it starts to push you away again... because ln(r)<0 if r<1. This is something similar to strong force, it's repulsive at a certain closeness, but if they stray too for from each other it becomes a very strong attractive force.

Last edited: Oct 8, 2013
5. Oct 8, 2013

### Flatland

I thought that gravity did not change with distance in 1 dimensional space?

6. Oct 8, 2013

### WannabeNewton

It doesn't. Laplace's equation in one dimension for the gravitational potential exterior to a point mass is simply $\frac{\mathrm{d} ^2 \varphi}{\mathrm{d} x^2} = 0$ so $\varphi = Ax + B$ and $g = - \frac{\mathrm{d} \varphi}{\mathrm{d} x} = -A$.

7. Oct 8, 2013

### Jarfi

So I was wrong?