<p>I can explain it for rings.</p>

<p>The phenomenon you're looking for is called nodal precession. Meaning, when the planet has an equatorial bulge, all non-equatorial orbits will precess, will wobble around the spin axis.</p>

<p><a href="https://en.wikipedia.org/wiki/Nodal_precession" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Nodal_precession</a></p>

<blockquote>
  <p>Because of the bulge the gravitational force on the satellite is not
  directly toward the center of the central body, but is offset toward
  the equator. Whichever hemisphere the satellite is in it is
  preferentially pulled slightly toward the equator. This creates a
  torque on the orbit. This torque does not reduce the inclination;
  rather, it causes a torque-induced gyroscopic precession, which causes
  the orbital nodes to drift with time.</p>
</blockquote>

<p>For rings, any such precessing orbits would be unstable - ring particles would collide with each other. The only non-precessing, non-colliding orbits are the equatorial ones. So that's where the rings persist.</p>

<p>As for satellites, I would venture a guess that their orbits actually do not migrate towards the equatorial plane, they stay where they are, just slowly precessing. Someone please correct me if your orbital mechanics is in better shape than mine.</p>

<p>EDIT: There's a ton of perturbations from an oblate planet, and some do apply to the inclination of the orbit. So it's not just precession. Some terms are periodic. Google "inclination of orbit oblate" without the quotes. It's... complex. I'll edit this answer tomorrow if I find something relevant.</p>

<p>EDIT2: This is far more complex than I thought. A thorough treatment is given in Kozai, Y. The motion of a close earth satellite:</p>

<p><a href="http://adsabs.harvard.edu/abs/1959AJ.....64..367K" rel="nofollow noreferrer">http://adsabs.harvard.edu/abs/1959AJ.....64..367K</a></p>

<p>Chapter 11.15 in J.M.A. Danby, Fundamentals Of Celestial Mechanics, is entirely dedicated to this topic. He talks about critical values for initial parameters where his results are not applicable, so obviously there isn't a simple conclusion that applies in all cases.</p>

<p>I am not going to interpret the math, sorry. That's all I had to say on this topic.</p>

CC BY-SA 4.0
2019-04-18T08:53:43.890Z
2019-04-18T20:08:01.997Z
2019-04-18T20:08:01.997Z
Florin Andrei
<p>I tried looking this up without success and I imagine there's a formula for it, with some ridiculous numbers like it drops off to the 5th power of the distance or maybe even more.</p>

<p>Imagine a planet like Jupiter, large equatorial bulge and imagine an artificial satellite orbiting in a polar orbit, not an equatorial orbit.    Jupiter's enormous equatorial bulge should disrupt the satellite orbit and in time (perhaps a very long time), turn it into an equatorial orbit.</p>

<p>Is that correct or incorrect thinking on my part and is there a formula that applies?    I imagine that some kind of measure for equatorial bulge would be needed as well and I'm not sure how that would be measured in a gravitational field from orbit.</p>

CC BY-SA 4.0
2019-04-15T16:43:56.183Z
Certain Earth-orbiting satellites exploit this, see [Wikipedia:Sun-synchronous orbit](https://en.wikipedia.org/wiki/Sun-synchronous_orbit)
Mike G
CC BY-SA 4.0
2019-04-15T17:26:11.037Z
Keep in mind that you need not only an equitorial bulge in volume, but in total mass.
Carl Witthoft
CC BY-SA 4.0
2019-04-15T20:51:02.077Z
@CarlWitthoft   Yes, in my mind, I imagine mass variation due to equatorial bulge largely disappears when you leave the surface and get equidistant from the center of mass.    I want to say that perhaps there is close to zero equatorial mass concentration from orbit, or at least, it's greatly reduced, but I'd only be guessing.   I've tried to google this but nearly all the answers involve standing on the surface and taking into account rotation, not a uniform distance from orbit.
userLTK
CC BY-SA 4.0
2019-04-16T01:48:56.177Z
@MikeG No they don't. The question asks about *a dramatic change in inclination, form polar to equatorial*. Nodal precession used for Sun-synchronous orbits is a totally different thing.
uhoh
CC BY-SA 4.0
2019-04-16T02:40:01.900Z
@uhoh  Title edited.
userLTK
<orbital-mechanics>
What is the force that changes a satellite’s inclination when orbiting a planet with an equatorial bulge
userLTK
# What is the force that changes a satellite’s inclination when orbiting a planet with an equatorial bulge

(Asked by: username_0 on 2019-04-15)

I tried looking this up without success and I imagine there's a formula for it, with some ridiculous numbers like it drops off to the 5th power of the distance or maybe even more.

Imagine a planet like Jupiter, large equatorial bulge and imagine an artificial satellite orbiting in a polar orbit, not an equatorial orbit. Jupiter's enormous equatorial bulge should disrupt the satellite orbit and in time (perhaps a very long time), turn it into an equatorial orbit.

Is that correct or incorrect thinking on my part and is there a formula that applies? I imagine that some kind of measure for equatorial bulge would be needed as well and I'm not sure how that would be measured in a gravitational field from orbit.

**Tags:** orbital-mechanics

> **Comments:**
> **username_1**: Certain Earth-orbiting satellites exploit this, see [Wikipedia:Sun-synchronous orbit](https://en.wikipedia.org/wiki/Sun-synchronous_orbit)  
> **username_2**: Keep in mind that you need not only an equitorial bulge in volume, but in total mass.  
> **username_0**: @CarlWitthoft   Yes, in my mind, I imagine mass variation due to equatorial bulge largely disappears when you leave the surface and get equidistant from the center of mass.    I want to say that perhaps there is close to zero equatorial mass concentration from orbit, or at least, it's greatly reduced, but I'd only be guessing.   I've tried to google this but nearly all the answers involve standing on the surface and taking into account rotation, not a uniform distance from orbit.  
> **username_3**: @MikeG No they don't. The question asks about *a dramatic change in inclination, form polar to equatorial*. Nodal precession used for Sun-synchronous orbits is a totally different thing.  
> **username_0**: @username_3  Title edited.  

---

## 1 answers

### Answer by username_4 on 2019-04-18. Score: 3

I can explain it for rings.

The phenomenon you're looking for is called nodal precession. Meaning, when the planet has an equatorial bulge, all non-equatorial orbits will precess, will wobble around the spin axis.

<https://en.wikipedia.org/wiki/Nodal_precession>

> Because of the bulge the gravitational force on the satellite is not
> directly toward the center of the central body, but is offset toward
> the equator. Whichever hemisphere the satellite is in it is
> preferentially pulled slightly toward the equator. This creates a
> torque on the orbit. This torque does not reduce the inclination;
> rather, it causes a torque-induced gyroscopic precession, which causes
> the orbital nodes to drift with time.

For rings, any such precessing orbits would be unstable - ring particles would collide with each other. The only non-precessing, non-colliding orbits are the equatorial ones. So that's where the rings persist.

As for satellites, I would venture a guess that their orbits actually do not migrate towards the equatorial plane, they stay where they are, just slowly precessing. Someone please correct me if your orbital mechanics is in better shape than mine.

EDIT: There's a ton of perturbations from an oblate planet, and some do apply to the inclination of the orbit. So it's not just precession. Some terms are periodic. Google "inclination of orbit oblate" without the quotes. It's... complex. I'll edit this answer tomorrow if I find something relevant.

EDIT2: This is far more complex than I thought. A thorough treatment is given in Kozai, Y. The motion of a close earth satellite:

<http://adsabs.harvard.edu/abs/1959AJ.....64..367K>

Chapter 11.15 in J.M.A. Danby, Fundamentals Of Celestial Mechanics, is entirely dedicated to this topic. He talks about critical values for initial parameters where his results are not applicable, so obviously there isn't a simple conclusion that applies in all cases.

I am not going to interpret the math, sorry. That's all I had to say on this topic.


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