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If an object revolves around a circular path at the speed of light could it still generate gravitational waves and what would be the simulation in inner area?

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closed as unclear what you're asking by Rob Jeffries, John Rennie, user36790, CuriousOne, ACuriousMind Jul 22 '16 at 11:12

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  • $\begingroup$ It is unclear what you are asking. $\endgroup$ – Rob Jeffries Jul 22 '16 at 7:12
  • $\begingroup$ If earth revolved around sun at the speed of light without self rotation would it still generate gravitation al waves? $\endgroup$ – NOob94 Jul 22 '16 at 7:13
  • $\begingroup$ How do propose to make the Earth, or indeed anything else, rotate round the Sun at the speed of light? $\endgroup$ – John Rennie Jul 22 '16 at 7:18
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    $\begingroup$ The revolution of two objects around each other will generate gravitational waves, just not very strong ones, unless the motion is relativistic. The gravitational wave emissions of Earth revolving around the sun have been estimated at 200W: en.wikipedia.org/wiki/Gravitational_wave. This is simply not measurable with any known technology. $\endgroup$ – CuriousOne Jul 22 '16 at 7:21
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Object can't go with the speed of light on the Special Relativity, on which the General Relativity is based upon. The gravitational waves are predicted and calculated by the General Relativity, i.e. if you are talking about them, you are using the terminology of a theory which closes out the things going with the speed of the light.

But we can talk about an object near the speed of light.

It doesn't need to go with around $c$. If it accelerates, it produces gravitational waves and if it is on a circular orbit, it accelerates. (With some exception: for example an (around it centre) rotating planet doesn't emit GW.)

Note: in "normal" situations, this "gravitational radiation" is very small. For example, the orbit of the Earth around the Sun produces only 200W. The whole planet.

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  • $\begingroup$ what would the gravitional wave look like inside the orbit and outside ? would it be like a whirlpool on the inside? $\endgroup$ – NOob94 Jul 22 '16 at 10:23
  • $\begingroup$ Your third paragraph is wrong. The GW is produced by a time dependent quadrupole moment not by acceleration. The single mass rotating around the planet does have a time dependent quadrupole moment so it will emit GWs. However a continuous ring revolving at the same angulr velocity would not emit GWs. $\endgroup$ – John Rennie Jul 22 '16 at 10:48
  • $\begingroup$ @JohnRennie As I know, the third time derivate of the second moment counts, but I suspected it had been unclear for the OP. I extended the answer. $\endgroup$ – user259412 Jul 22 '16 at 11:19

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