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What happens to gravity in one dimension of space?

  1. Oct 8, 2013 #1
    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. jcsd
  3. Oct 8, 2013 #2

    D H

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    You skipped over a key integer. Hint: What integer is between -1 and +1?
     
  4. Oct 8, 2013 #3
    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
     
  5. Oct 8, 2013 #4
    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
  6. Oct 8, 2013 #5
    I thought that gravity did not change with distance in 1 dimensional space?
     
  7. Oct 8, 2013 #6

    WannabeNewton

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    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##.
     
  8. Oct 8, 2013 #7
    So I was wrong?
     
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