Doubt about Frame Dragging and planetary revolution

AI Thread Summary
The discussion centers on the effects of frame dragging on a satellite's orbit around the Sun, particularly if the satellite orbits in the opposite direction to the Sun's rotation. It is suggested that the satellite would not fall into the Sun, but would require a higher velocity for a stable orbit due to frame dragging effects. The conversation shifts to rotating black holes, where stable orbits are influenced by the ergosphere, complicating the dynamics of retrograde orbits. The mathematical framework provided, including the Kerr metric, outlines the necessary conditions for stable orbits around spinning objects. Overall, the complexities of orbital mechanics in the presence of rotation are highlighted, emphasizing the nuanced interactions between gravity and motion.
SreenathSkr
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Our sun rotates in counter clockwise direction and hence frame dragging will be also in that direction. Suppose if we put a satellite in sun's orbit (almost circular) to revolve in opposite direction to the sun's spin, what would happen to the orbit of the satellite? Would it fall into the sun?
 
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Hmmm... I don't think the satellite would fall into the Sun, but I'm not certain.
 
Of course, the effect for our Sun is negligible
So it is better to talk about the rotating BH. Black holes (not rotating) have zones where no stable orbits are possible. Rotating BHs have ergosphere where you can rotate only in the same direction with BH. I think the question "what is an area where stable orbits with a rotation opposite to the BH" is a good one and is quite complicated.
 
You'll probably find that the satellite orbiting against the suns rotation would need a slightly higher velocity. From Kerr metric, the velocity required for a stable orbit at the equator around a spinning object which takes into account frame dragging is-

v_\pm=\frac{r^2+a^2\mp 2a\sqrt{Mr}}{\sqrt{\Delta} \left[a\pm r\sqrt{r/M}\right]}

where \Delta= r^{2}+a^{2}-2Mr

\pm denotes prograde and retrograde orbit, M is the gravitational radius (M=Gm/c^2) and a is the spin parameter (a=J/mc) which for the sun is approx. a=0.185M.

Source- http://arxiv.org/abs/gr-qc/0407004 equation 25
 
Very interesting, Steve... Furthermore, a prograde equatorial orbiting clock and a retrograde orbiting clock would have different proper periods, t, ...
the difference given by ...
(t+) −(t−) ≈ 4πJ/(Mc^2).
And which, remarkably, is not dependent upon G or the orbital radius...as pointed out here: http://arxiv.org/abs/gr-qc/9912027
 
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