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For GEO satellite, 24h period, it has four equilibrium points (two stable point and two unstable point). At the equilibrium point, the transverse acceleration is zero, so semimajor axis changes slowly. At stable point, satellite will oscillate around it. But at the unstable point, satellite will drift and move to the closest one of a stable point. My question is, why does satellite at the stable point just oscillate around it while at the unstable point the satellite will drift to the closest one of a stable?

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    $\begingroup$ By way of analogy, think of an undamped pendulum that starts at zero angular velocity at initial deflection angle with respect to straight down. With two exceptions, the pendulum will oscillate about straight down. The two exceptions are when the pendulum is initially in the straight down orientation or initially in the straight up (perfectly inverted) orientation. In both cases, the pendulum just stays there, forever. A tiny deviation from straight down results in tiny oscillations about straight down, but a tiny deviation from straight up results in huge oscillations about straight down. $\endgroup$ Jun 15 '20 at 15:14
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    $\begingroup$ The straight down orientation is the pendulum's stable equilibrium point. The straight up orientation (an inverted pendulum) is the pendulum's unstable equilibrium point. The pendulum doesn't oscillate about the unstable equilibrium point. $\endgroup$ Jun 15 '20 at 15:15
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Let's step away from satellites and just look at stable vs. unstable equilibrium. (I'm making use of standard examples from 1st-semester Calculus textbooks). The top of a hill is unstable equilibrium, because an object there won't move on its own. The only force is gravity, pointing straight down. But the tiniest push in any horizontal direction moves the object to a slope, and then it moves away from the top of the hill, never to return.
A stable equilibrium is a valley, or better a bowl. Again, at the very bottom, the only force is gravity, straight down. But now a small horizontal push causes the object to move uphill, and when that push is terminated, gravity pulls the object back towards the bottom of the bowl
In the absence of friction (as with the satellite), the object will continue to move up and back down in a sinusoidal motion.

But be warned - the satellite's "stable equilibrium" is a local minimum , so a large enough push will move it into an unstable non-equilibrium orbit.

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  • $\begingroup$ I got it, thank you so much $\endgroup$ Sep 12 '20 at 14:54
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As you have written, an object (infinitesimal near) near an unstable point will drift to a stable point, but it will not stop there but drift further. So an object (starting infinitesimal near) an unstable point will also oscillate around a stable point. But while the "amplitude" near a stable point is very small, an object at an unstable point has an amplitude of nearly 180 deg. it will accelerate to the stable point, "overshoot" it, be decelerated till it nearly reaches the other unstable point and go back again and again...

..::EDIT::.. (I started this as a comment but its to long and actually part of a valid answer:) @ElisaFitri the amplitude near a stable point is SMALL. So basically all objects oscillate around stable points. But for example: there is a stable point at about 108 deg. If you start at 107deg, you drift to 109 deg, and back to 107 and so on. Starting at 8 deg lets you drift to 208 deg and back. All this comes from the shape of the earth.

Looking from top earth is more elliptical than circular. So somewhere on orbit (not at the four equilibrum points at the axis) where is always "more" earth right or left from the zenit, so you will be ac-/de-celerated towards. For LEO, this compensates in one tour around earth.

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  • $\begingroup$ mhh, I try to rewrite it $\endgroup$
    – CallMeTom
    Jun 15 '20 at 4:21
  • $\begingroup$ Hope it is clearer now $\endgroup$
    – CallMeTom
    Jun 15 '20 at 4:24
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    $\begingroup$ Thank you very much for your answer @CallMeTom. Would you explain why does the amplitude of oscillation near a stable point is very much, while at an unstable point has an amplitude of nearly 180 deg? $\endgroup$ Jun 15 '20 at 4:41
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    $\begingroup$ Also, thank you to @uhoh for the correction. $\endgroup$ Jun 15 '20 at 4:41
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    $\begingroup$ I added an explanation to my answer @ElisaFitri $\endgroup$
    – CallMeTom
    Jun 15 '20 at 4:59

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