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If you're looking down at an infinite plane, then the plane occupies half your field of view (assuming you have a 360 degree field of view) no matter how far yo

If you're looking down at an infinite plane, then the plane occupies half your field of view (assuming you have a 360 degree field of view) no matter how far you are from it. Given a 2d object of uniform density, the component of the gravitational force towards that object in the direction perpendicular to the plane it lies in is proportional to the solid angle of your field of view it occupies. For an infinite plane, the components in other directions cancel out. I think this is related to how a sphere acts equivalently to a point at its center - an infinite plane is just a sphere of infinite radius. The situation is similar for an opaque light source (apart from the lack of atmosphere in the way, I don't think the sun would be much brighter if you were much closer to it and looking through a pinhole), but I'm not sure it applies to gamma radiation which is emitted uniformly in all directions from each point and treats most stuff as pretty transparent - unless there's some weird interference you don't get to cancel out the parts coming from different directions.

tomxor 4y ago
Yes, I actually kind of get the infinite part, but thanks for the explanation it's interesting. I was wondering what the intensity looks like for a distance field around a finite plane and how that relates. This might be harder to describe. What I'm suspecting is that under a certain distance for a given finite plane it's almost constant similar to the infinite version, but outside of that distance there must be some non uniform falloff function.
- penteract 4y ago
  For a finite circle, where you're on a line perpendicular to it going through its center, the gravitational force you experience towards it is exactly proportional to the amount of your field of view it occupies - if I'm right, that should be proportional to `1-cos(atan(r/d))` where r is the radius of the circle and d is the distance between you and the center of the circle.
- gus_massa 4y ago
  You are correct. For a short distance it's a almost a constant 1 like an infinite wall with the same charge density. For a long distance is almost 1/r^2 like punctual source. For an intermediate distance, it's a mess and you hope nobody ask for an exact solution. A rectangle is a mess. It's easier with a circle. The approximations at short and long distance are the same, but a circle has an "easy" formula in between. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.h... 1-x/sqrt(x^2+R^2) where R is the radio of the circle. Let's use R=10 to keep it simple. For a short distance 0<x<R/10, then it's almost a constant https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29++and+0+for+x+from+0+to+1 https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... For a long distance, x>4R it's almost like R^2/(2*x^2) https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29+and+10%5E2%2F%282x%5E2%29+for+x+from+0+to+100&lang=es https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... There are some trick to add more term to the approximations to reduce the middle part where both approximations are bad, like using A+Bx or A+Bx+Cx^3 for short distance and D/x^2+E/x^4 or D/x^2+E/x^4+F/x^4 for long distance. It depends on how much you care about the precision and how many calculations you want to do. In some particular cases like a circle or the shell of a sphere there are closed formulas for the intermediate distances. In other cases there is no nice formula.

tomxor 4y ago

Yes, I actually kind of get the infinite part, but thanks for the explanation it's interesting. I was wondering what the intensity looks like for a distance field around a finite plane and how that relates. This might be harder to describe. What I'm suspecting is that under a certain distance for a given finite plane it's almost constant similar to the infinite version, but outside of that distance there must be some non uniform falloff function.

penteract 4y ago
For a finite circle, where you're on a line perpendicular to it going through its center, the gravitational force you experience towards it is exactly proportional to the amount of your field of view it occupies - if I'm right, that should be proportional to `1-cos(atan(r/d))` where r is the radius of the circle and d is the distance between you and the center of the circle.
gus_massa 4y ago
You are correct. For a short distance it's a almost a constant 1 like an infinite wall with the same charge density. For a long distance is almost 1/r^2 like punctual source. For an intermediate distance, it's a mess and you hope nobody ask for an exact solution. A rectangle is a mess. It's easier with a circle. The approximations at short and long distance are the same, but a circle has an "easy" formula in between. http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.h... 1-x/sqrt(x^2+R^2) where R is the radio of the circle. Let's use R=10 to keep it simple. For a short distance 0<x<R/10, then it's almost a constant https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29++and+0+for+x+from+0+to+1 https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... For a long distance, x>4R it's almost like R^2/(2*x^2) https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%5E2%29+and+10%5E2%2F%282x%5E2%29+for+x+from+0+to+100&lang=es https://www.wolframalpha.com/input?i=1-x%2Fsqr%28x%5E2%2B10%... There are some trick to add more term to the approximations to reduce the middle part where both approximations are bad, like using A+Bx or A+Bx+Cx^3 for short distance and D/x^2+E/x^4 or D/x^2+E/x^4+F/x^4 for long distance. It depends on how much you care about the precision and how many calculations you want to do. In some particular cases like a circle or the shell of a sphere there are closed formulas for the intermediate distances. In other cases there is no nice formula.