<p><strong>tl;dr:</strong> The <a href="https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body" rel="nofollow noreferrer">orbital period </a>for a Keplerian orbit of a small body around a massive one with <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">standard gravitational parameter</a> <span class="math-container">$\mu=GM$</span> and semi-major axis <span class="math-container">$a$</span> is given by</p>
<p><span class="math-container">$$T = 2\pi\sqrt{\frac{a^3}{\mu}}.$$</span></p>
<p>I'm asking if for <em>near-circular orbits</em> the <em>rise times</em> of the object will <em>always come earlier than predicted</em> by a Keplerian model when the central body is oblate, i.e. has a positive <span class="math-container">$J_2$</span>.</p>
<hr />
<p>Assuming that we want to make a &quot;long-term&quot; prediction of the rise-time wrt any given observation point on a planet, of an object circling said planet, assuming that we want to make a rough estimate using the Kepler model. Is it correct to state that the said rough estimate is ALWAYS lagging because we ignore the oblateness of the planet? By &quot;lagging&quot;, I mean we should observe the rise somewhat BEFORE the predicted calculation, but always before?</p>
<p>A side question: Assuming we know the J2 zonal coefficient and the orbital elements at epoch (t0), is it correct to say that the prediction &quot;error&quot; at time t is simply linear with delta_t (t=t0+delta_t)? (the point is that if your answer is yes, then I can use the simplified model w/o making any error compared to a J2-aware model).</p>
<p>Note: For now, I am not interested in a real-life situation, where the orbit is perturbed by other forces than J2.</p>

CC BY-SA 4.0
2021-07-13T16:40:58.777Z
@uhoh, please check my calculus, but this is what I obtain by putting the eccentricity to zero and isolating the term involving the inclination: Tnodal=TKepler[ 1 - K(2-sin²(i)) ]. K being a positive constant involving J2,a,R. Whichever i, I conclude that 2-sin²(i)>0
Ng Ph
CC BY-SA 4.0
2021-07-14T00:43:20.620Z
Somehow I got the link wrong in that last comment so deleted it, but you figured out what I meant anyway. I think there is a niche at high eccentricity ($\epsilon >0.8, \omega < 10°$ where the expression shows a *longer* period, but double-checking now (I'm [prone to errors](https://math.stackexchange.com/questions/4181503/ray-tracing-in-nonuniform-media-did-i-write-this-second-order-differential-equa#comment8667501_4181503)). Also, this expression is an approximation only, it may not even be correct for extreme eccentricity; I'll look for an answer about that as well and update here.
uhoh
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2021-05-25T22:03:10.603Z
*Great question!* I've added a short orbital-mechanics summary of your question at the beginning to avoid "unclear what you're asking" close votes, please check to see if I've got it right and feel free to edit further.
uhoh
CC BY-SA 4.0
2021-05-25T22:11:16.417Z
By the way, if the orbit is low and near zero inclination (equatorial) then if the reduced $J_2$ is say +0.001 the gravity field near the planet in the orbital plane will always be 0.001 stronger and so the orbit will still be circular with a period will be 0.9995 $T_{Kepler}$. In this condition the small moon will just think the planet is one part per thousand more massive. It of course gets much more complicated for higher inclinations, and the orbit will *precess around the planet* which Kepler doesn't predict; after a while the Keplerian predictions will not resemble anything like reality.
uhoh
CC BY-SA 4.0
2021-05-25T22:14:51.040Z
So I think a better question *might be* more like your different but related one in Space SE about the period defined by subsequent [ascending node crossings](https://en.wikipedia.org/wiki/Nodal_period) rather than tied to rise times a a specific location on the planet's surface, since [nodal precession](https://en.wikipedia.org/wiki/Nodal_precession) (of the orbital plane) adds a complication, *unless that really is the ultimate question you need addresses* in which case it's fine!
uhoh
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2021-05-26T08:25:24.540Z
@uhoh, Great edit. Agreed.
Ng Ph
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2021-05-26T08:30:06.930Z
@uhoh, Intuitively, when there is symmetry and a single central force, then the Kepler model must apply. Hence, Kepler must apply to an equatorial orbit around an oblate planet, imho.
Ng Ph
CC BY-SA 4.0
2021-05-26T08:45:20.257Z
@uhoh, I would say that my related question in Space SE is to validate (or invalidate) a reasoning step. This one here is about the practical use. Indeed, the effect of J2 on the orbit plane brings an additional complication to the effect on the mean motion. However, this could be bounded by clarifying the magnitude of "long-term" in the statement of problem (I hope). If not the simple prediction could be a little bit more sophisticated (which may defeat the ultimate goal, on the other hand).
Ng Ph
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2021-05-26T08:53:01.553Z
The equatorial orbit would be *Keplerian* because it would be a circle, but it would not have the Keplerian period if you just use the mass of the planet, it would be a little shorter. That's because if you just use the mass you have to use [Newton's shell theorem](https://en.wikipedia.org/wiki/Shell_theorem) to replace any spherically symmetric mass distribution with a single point mass at the center. Since an oblate spheroid is not spherically symmetric, then you can't replace it with *the same mass* at the center. Thus *Keplerian in shape* but not in period.
uhoh
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2021-05-26T21:01:05.233Z
@uhoh, Interesting theorem, but I can't follow your reasoning. My guess is that scientists measure the J2 of a planet by observing how orbiting objects around it deviate from Kepler's model. However, I can't see how an object orbiting in the Equator plane can "sense" the presence of a J2 perturbation? Intuitively, I would postulate that, with Equatorial orbits, you cannot differentiate even between the 2 extremes: a sphere and a disk.
Ng Ph
CC BY-SA 4.0
2021-05-26T21:06:45.270Z
Oh, I'm only saying that if one *already has a known numerical value for the mass* $M$ of the Earth, then the circular equatorial orbit would have a slightly shorter period than $2\pi \sqrt{a^3 / \mu}$ predicts. That doesn't disagree at all with what you just wrote. If folks on the surface observed the equatorial moon's motion and analyzed it, they would determine that the planet's mass was slightly larger than it really was and believe the orbit was Keplerian. Once they launched an artificial satellite at some other inclination though, they'd get a big surprise!
uhoh
<orbital-mechanics><orbital-elements>
When an object revolves around an oblate planet, in near-circular orbit, does the oblateness ALWAYS make it moves FASTER?
Ng Ph
# When an object revolves around an oblate planet, in near-circular orbit, does the oblateness ALWAYS make it moves FASTER?

(Asked by: username_0 on 2021-05-25)

**tl;dr:** The [orbital period](https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body) for a Keplerian orbit of a small body around a massive one with [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) $\mu=GM$ and semi-major axis $a$ is given by

$$T = 2\pi\sqrt{\frac{a^3}{\mu}}.$$

I'm asking if for *near-circular orbits* the *rise times* of the object will *always come earlier than predicted* by a Keplerian model when the central body is oblate, i.e. has a positive $J\_2$.

---

Assuming that we want to make a "long-term" prediction of the rise-time wrt any given observation point on a planet, of an object circling said planet, assuming that we want to make a rough estimate using the Kepler model. Is it correct to state that the said rough estimate is ALWAYS lagging because we ignore the oblateness of the planet? By "lagging", I mean we should observe the rise somewhat BEFORE the predicted calculation, but always before?

A side question: Assuming we know the J2 zonal coefficient and the orbital elements at epoch (t0), is it correct to say that the prediction "error" at time t is simply linear with delta\_t (t=t0+delta\_t)? (the point is that if your answer is yes, then I can use the simplified model w/o making any error compared to a J2-aware model).

Note: For now, I am not interested in a real-life situation, where the orbit is perturbed by other forces than J2.

**Tags:** orbital-mechanics, orbital-elements
> **Comments:**
> **username_1**: *Great question!* I've added a short orbital-mechanics summary of your question at the beginning to avoid "unclear what you're asking" close votes, please check to see if I've got it right and feel free to edit further.  
> **username_1**: By the way, if the orbit is low and near zero inclination (equatorial) then if the reduced $J_2$ is say +0.001 the gravity field near the planet in the orbital plane will always be 0.001 stronger and so the orbit will still be circular with a period will be 0.9995 $T_{Kepler}$. In this condition the small moon will just think the planet is one part per thousand more massive. It of course gets much more complicated for higher inclinations, and the orbit will *precess around the planet* which Kepler doesn't predict; after a while the Keplerian predictions will not resemble anything like reality.  
> **username_1**: So I think a better question *might be* more like your different but related one in Space SE about the period defined by subsequent [ascending node crossings](https://en.wikipedia.org/wiki/Nodal_period) rather than tied to rise times a a specific location on the planet's surface, since [nodal precession](https://en.wikipedia.org/wiki/Nodal_precession) (of the orbital plane) adds a complication, *unless that really is the ultimate question you need addresses* in which case it's fine!  
> **username_0**: @uhoh, Great edit. Agreed.  
> **username_0**: @uhoh, Intuitively, when there is symmetry and a single central force, then the Kepler model must apply. Hence, Kepler must apply to an equatorial orbit around an oblate planet, imho.  
> **username_0**: @uhoh, I would say that my related question in Space SE is to validate (or invalidate) a reasoning step. This one here is about the practical use. Indeed, the effect of J2 on the orbit plane brings an additional complication to the effect on the mean motion. However, this could be bounded by clarifying the magnitude of "long-term" in the statement of problem (I hope). If not the simple prediction could be a little bit more sophisticated (which may defeat the ultimate goal, on the other hand).  
> **username_1**: The equatorial orbit would be *Keplerian* because it would be a circle, but it would not have the Keplerian period if you just use the mass of the planet, it would be a little shorter. That's because if you just use the mass you have to use [Newton's shell theorem](https://en.wikipedia.org/wiki/Shell_theorem) to replace any spherically symmetric mass distribution with a single point mass at the center. Since an oblate spheroid is not spherically symmetric, then you can't replace it with *the same mass* at the center. Thus *Keplerian in shape* but not in period.  
> **username_0**: @uhoh, Interesting theorem, but I can't follow your reasoning. My guess is that scientists measure the J2 of a planet by observing how orbiting objects around it deviate from Kepler's model. However, I can't see how an object orbiting in the Equator plane can "sense" the presence of a J2 perturbation? Intuitively, I would postulate that, with Equatorial orbits, you cannot differentiate even between the 2 extremes: a sphere and a disk.  
> **username_1**: Oh, I'm only saying that if one *already has a known numerical value for the mass* $M$ of the Earth, then the circular equatorial orbit would have a slightly shorter period than $2\pi \sqrt{a^3 / \mu}$ predicts. That doesn't disagree at all with what you just wrote. If folks on the surface observed the equatorial moon's motion and analyzed it, they would determine that the planet's mass was slightly larger than it really was and believe the orbit was Keplerian. Once they launched an artificial satellite at some other inclination though, they'd get a big surprise!  
> **username_0**: @uhoh, please check my calculus, but this is what I obtain by putting the eccentricity to zero and isolating the term involving the inclination: Tnodal=TKepler[ 1 - K(2-sin²(i)) ]. K being a positive constant involving J2,a,R. Whichever i, I conclude that 2-sin²(i)>0  
> **username_1**: Somehow I got the link wrong in that last comment so deleted it, but you figured out what I meant anyway. I think there is a niche at high eccentricity ($\epsilon >0.8, \omega < 10°$ where the expression shows a *longer* period, but double-checking now (I'm [prone to errors](https://math.stackexchange.com/questions/4181503/ray-tracing-in-nonuniform-media-did-i-write-this-second-order-differential-equa#comment8667501_4181503)). Also, this expression is an approximation only, it may not even be correct for extreme eccentricity; I'll look for an answer about that as well and update here.  

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## 0 answers

