What happens to gravity in one dimension of space?

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SUMMARY

This discussion explores the behavior of gravity in various dimensions of space, specifically focusing on one-dimensional space. In three dimensions, gravity follows the formula F = x/r², while in four dimensions, it becomes F = x/r³, indicating a weaker force. The conversation suggests that in one dimension, gravity could be represented as F = xln(r), leading to the conclusion that gravity would increase with distance until a certain point, where it would then become repulsive. The participants clarify that Laplace's equation in one dimension shows that gravitational potential does not change with distance, resulting in a constant gravitational force.

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  • Understanding of gravitational force equations, specifically F = x/r²
  • Familiarity with Laplace's equation and its implications in physics
  • Basic knowledge of logarithmic functions and their properties
  • Concepts of dimensional analysis in physics
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KeplerJunior
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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?
 
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You skipped over a key integer. Hint: What integer is between -1 and +1?
 
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
 
KeplerJunior said:
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?

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.
 
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I thought that gravity did not change with distance in 1 dimensional space?
 
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##.
 
WannabeNewton said:
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##.

So I was wrong?
 

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