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Gravity decreases as 1/r

by ObsessiveMathsFreak
Tags: 1 or r, decreases, gravity
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ObsessiveMathsFreak
#19
Apr15-03, 05:10 PM
P: 406
Sorry, probes, satellites, whatever.

The point of this post was that the sun's gravitational field will be surpassed by the gravitational effect of the sphere of light surrounding it at very large distances.
damgo
#20
Apr15-03, 06:21 PM
P: n/a
Aha! I knew there was something fishy here, and I figured it out. :)

OMF, your derivation is valid... for a "freeze frame" in time. But, it won't have any actual effect on motion. The basic reason you get the 1/r effect is because as r increases, more and more light is subsumed into the sphere of radius r behind you, which acts like all its mass was at the origin.

*But*, in actuality, the energy of this light is coming from the star itself, and moving outward at c. So, if you were moving away from the star at c, you would be just keeping up with your current shell of light. More energy and hence gravitational pull would show up in the sphere of light -- but this would be exactly cancelled by the energy&gravitational pull loss from the star's mass decrease. So you get the usual 1/r^2 effect.

If you're moving away from the star slower than c, gravitational pull will actually decrease faster than 1/r^2 , as more of the mass-energy of the star passes you. In the case where you are standing still, the effective mass will drop at a rate of Luminosity/c^2 and so gravitation force by a rate G*L/(c^2*r^2).

For v<<c, this will dominate, and the outgoing object will see a force of G*(M/r^2 - L/(c^2*r*v))
ObsessiveMathsFreak
#21
Apr17-03, 06:27 AM
P: 406
The derivation only deals with the instantaneous gravity at a point from the star.

The key point here is that the usual 1/r^2 effect won't appear to take place.

This is because the further you move away from the star, the more mass is within the sphere surrounding the star.

Due to the density of this "mass" of light, gravity decreases as 1/r. If you instead took a black hole, emmitting no light, its gravity would decrease as 1/r^2.

The formula find the total "mass" of the light spheres, compresses them at a point, the star, and then uses the 1/r^2 rule to find the gravity.
However due to the variable density in each sphere, proportial to 1/r^2 itself, the total gravity will "appear" to diminish as 1/r.

The gravity isn't actually obeying a 1/r rule. It will just appear to do so.

Isn't this what the MOND people hold to be true?


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