Do Accelerating Masses Produce Gravitons and Cause Orbit Decay?

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Accelerating charged particles emit photons and lose energy, which explains why electrons cannot maintain stable orbits around nuclei. Similarly, gravitational radiation from accelerating masses, like the Earth-Sun system, results in energy loss, but this effect is negligible in practical terms. Observations of binary pulsars show minimal period changes due to gravitational radiation, indicating that orbits remain stable over vast timescales. The decay of orbits due to gravitational radiation is extremely slow, with estimates suggesting it would take around 10^19 years for significant changes in the Earth's orbit. Overall, while gravitational radiation exists, its impact on orbital stability is minimal.
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When a charged particle accelerates it produces photons and loses energy which is part of the reason why an electron cannot "orbit" a nucleus.

I would think that an accelerating mass would likewise produce gravitons and lose energy so that all orbits would eventually decay. Is this correct and orbits just seem stable because the amount of energy in gravitons is so small, or is this incorrect because of the quadrupole thing?

I don't exactly understand the quadrupole idea, so if that is what prevents a circular orbit from losing energy, what kinds of orbits or events will produce gravitons?

-Thanks
Dale
 
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Yes, the Earth-Sun system, for example, emits gravitational radiation and thus loses energy, but this loss is far too small to have any noticeable impact. Just for some scale, Taylor and Hulse observed a period decrease of about 80 millionths of a second per year in their binary pulsar, and the effect in our own solar system would be even smaller. However, I've never actually done an estimate, so perhaps someone else can be more specific.
 
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The characteristic period at which gravitational wave radition will be important is very roughly (R/R_s)^(5/2) [from MTW's gravitation, pg 981 - note that I've taken a few small liberties with the formula by replacing M with R_s to keep the units simple, ommiting a factor of 2 in the process]

Since R_s for the sun is 3 km, if you take (98 million km / 3 km)^(5/2), this means about 10^19 orbits for the Earth, i.e. 10^19 years, give or take a few orders of magnitude, for the Earth's orbit to decay appreciably.
 
Thanks Physics Monkey and Pervect! It's good to know I was thinking right.

-Dale
 
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