<p>Due to the oblateness of the Earth (the Earth's equatorial bulge), a geocentric satellite's orbital plane will precess (rotate) relative to inertial space.
The rate at which the line of nodes moves owing to this bulge is given by
$$
\dot\Omega = -\frac{3}{2}J_2\left(\frac{r_E}{ℓ}\right)^2 n\cos\iota
$$<br>
where $J_2$ is the is the zonal harmonic coefficient ($1.08262668\times10^{-3}$ for Earth), $r_E$ is the body's equatorial radius ($6\,378\,137$ m for Earth), $ℓ$ is the orbit parameter (the semi-latus rectum), $n$ is the mean motion, and $\iota$ is the inclination of the orbit.</p>

<p>For a given orbit parameter ($ℓ$) and mean motion ($n$), the inclination of a geocentric satellite orbit can be selected to obtain, for example, a Sun-synchronous orbit ($\dot\Omega=360^\circ$ per $365.26$ days, or $0.9856$ degrees per day).</p>

<p><strong>Coordinate System</strong></p>

<p>This <a href="https://en.wikipedia.org/wiki/Orbital_elements" rel="noreferrer">Wikipedia article</a> (and diagram from that article) describes the coordinate system used.</p>

<p><a href="https://i.sstatic.net/tzuPy.png" rel="noreferrer"><img src="https://i.sstatic.net/tzuPy.png" alt="enter image description here"></a></p>

<p>The orbital plane (yellow) intersects a reference plane (gray). For Earth-orbiting satellites, the reference plane is usually the Earth's equatorial plane.  The intersection is called the line of nodes, as it connects the center of mass with the ascending and descending nodes.  This plane, together with the vernal point ($\gamma$), establishes a reference frame.</p>

<ul>
<li><p>Two elements that define the size and shape of the elliptical orbit
are the semi-major axis, $a$, and the eccentricity, $e$. The
semi-latus rectum is related to $a$ and $e$ by $ℓ=a(1-e^2)$.</p></li>
<li><p>Two elements define the orientation of the orbital plane in which the
ellipse is embedded: the inclination, $\iota$, and the longitude of the ascending node, $\Omega$.</p></li>
</ul>

CC BY-SA 3.0
2016-10-18T13:20:06.560Z
Wow - I think you introduced more "unknowns" in one sentence than I've ever seen :-) .   Can you provide the coordinate system orientation?  I'm trying to see how that equation yields $\dot\Omega = 0$ for an equatorial geosynchronous orbit.
Carl Witthoft
CC BY-SA 3.0
2016-10-18T13:42:42.207Z
@CarlWitthoft, for an equatorial geosynchronous orbit, the inclination, $\iota$, is zero so that the rate of change of the longitude of the ascending node, $\dot\Omega$, is also zero.  Anyway, if the inclination is zero, $\Omega$ is not defined.   For computation it is then, by convention, set equal to zero; that is, the ascending node is placed in the reference direction.

I will edit my answer to provide the coordinate system orientation.
Christo
CC BY-SA 3.0
2016-10-18T07:04:24.453Z
2016-10-18T14:12:54.483Z
2016-10-18T14:12:54.483Z
Christo
<p>I am trying to solve a problem where I need to calculate a satellite's orbit, but first I would like to ask for some clarifications from someone here that might know this stuff. I need to design an orbit so that no maneuvers (one part of the problem) are necessary to maintain it, and the hint is that there are J2 perturbation only and that the orbit is elliptical. I don't really understand this, so can someone explain what they mean with only J2 perturbations and how it is relevant here? (just asking for an explanation) </p>

<p>Thanks!  </p>

<orbital-mechanics>
J2 perturbations and orbits
Planet_hunter
# J2 perturbations and orbits

(Asked by: username_0 on 2016-10-17)

I am trying to solve a problem where I need to calculate a satellite's orbit, but first I would like to ask for some clarifications from someone here that might know this stuff. I need to design an orbit so that no maneuvers (one part of the problem) are necessary to maintain it, and the hint is that there are J2 perturbation only and that the orbit is elliptical. I don't really understand this, so can someone explain what they mean with only J2 perturbations and how it is relevant here? (just asking for an explanation)

Thanks!

**Tags:** orbital-mechanics


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## 1 answers

### Answer by username_1 on 2016-10-18. Score: 7

Due to the oblateness of the Earth (the Earth's equatorial bulge), a geocentric satellite's orbital plane will precess (rotate) relative to inertial space.
The rate at which the line of nodes moves owing to this bulge is given by
$$
\dot\Omega = -\frac{3}{2}J\_2\left(\frac{r\_E}{ℓ}\right)^2 n\cos\iota
$$  
where $J\_2$ is the is the zonal harmonic coefficient ($1.08262668\times10^{-3}$ for Earth), $r\_E$ is the body's equatorial radius ($6\,378\,137$ m for Earth), $ℓ$ is the orbit parameter (the semi-latus rectum), $n$ is the mean motion, and $\iota$ is the inclination of the orbit.

For a given orbit parameter ($ℓ$) and mean motion ($n$), the inclination of a geocentric satellite orbit can be selected to obtain, for example, a Sun-synchronous orbit ($\dot\Omega=360^\circ$ per $365.26$ days, or $0.9856$ degrees per day).

**Coordinate System**

This [Wikipedia article](https://en.wikipedia.org/wiki/Orbital_elements) (and diagram from that article) describes the coordinate system used.

[![enter image description here](https://i.sstatic.net/tzuPy.png)](https://i.sstatic.net/tzuPy.png)

The orbital plane (yellow) intersects a reference plane (gray). For Earth-orbiting satellites, the reference plane is usually the Earth's equatorial plane. The intersection is called the line of nodes, as it connects the center of mass with the ascending and descending nodes. This plane, together with the vernal point ($\gamma$), establishes a reference frame.

* Two elements that define the size and shape of the elliptical orbit
  are the semi-major axis, $a$, and the eccentricity, $e$. The
  semi-latus rectum is related to $a$ and $e$ by $ℓ=a(1-e^2)$.
* Two elements define the orientation of the orbital plane in which the
  ellipse is embedded: the inclination, $\iota$, and the longitude of the ascending node, $\Omega$.

> **Comments:**
> **username_2**: Wow - I think you introduced more "unknowns" in one sentence than I've ever seen :-) .   Can you provide the coordinate system orientation?  I'm trying to see how that equation yields $\dot\Omega = 0$ for an equatorial geosynchronous orbit.  
> **username_1**: @username_2, for an equatorial geosynchronous orbit, the inclination, $\iota$, is zero so that the rate of change of the longitude of the ascending node, $\dot\Omega$, is also zero.  Anyway, if the inclination is zero, $\Omega$ is not defined.   For computation it is then, by convention, set equal to zero; that is, the ascending node is placed in the reference direction.  I will edit my answer to provide the coordinate system orientation.  

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