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Mar 10, 2020 at 5:05 comment added A.V.S. Carlip's paper in my answer would do. But to see that EM and gravity are different in that regard one could also use post-Newtonian expansion: system of charges is conservative up to $1/c^2$ terms while system of gravitating bodies is conservative up to $1/c^4$. Radiative effects appear at $1/c^3$ for EM and at $1/c^5$ for gravity.
Mar 9, 2020 at 22:50 comment added J Thomas @A.V.S It sounds like there's something here I ought to learn about. Would you suggest a link?
Mar 9, 2020 at 22:48 history edited J Thomas CC BY-SA 4.0
Fixed errors
Mar 9, 2020 at 22:26 comment added J Thomas You're right the forces are equal. I should have said that the sun's response to that force is small, compared to the earth's response.
Mar 9, 2020 at 14:57 comment added Agnius Vasiliauskas The force on the sun from the earth -- kind of small Wrong. Have you ever checked Newton gravity law ? It states that gravity force is proportional to the product of masses : $F=G{\frac {m_{1}m_{2}}{r^{2}}}$. So force from sun to earth and from earth to sun are equal in magnitude.
Mar 9, 2020 at 14:43 comment added A.V.S. This is wrong. For EM field “effective position” is linearly extrapolated. For gravitational field this “effective position” is extrapolated quadratically, taking into account not only velocity but also acceleration.
Mar 9, 2020 at 9:12 history answered J Thomas CC BY-SA 4.0