Last modified on 7 August 2013, at 19:42

Astrodynamics/Classical Orbit Elements

Orbit ParametersEdit

It is possible to specify an orbit entirely using a set of 5 parameters. With these 5 parameters, we can specify precisely where an orbit is, how it is oriented in 3-D space, and what size it is. If we have an optional sixth parameter, we can determine exactly where the satellite is in its orbit at any arbitrary time t. These 6 parameters are called the Keplerian Elements.

However, before we discuss the 6 orbit parameters, we need to introduce a few new terms.

Ascending NodeEdit

If the Earth's equator is a plane, the fundamental plane (as it is in the geocentric-equatorial coordinate system, which we will be using in this section), and the orbit forms a plane (the "orbital plane") that contains the vectors r and v, then the line formed by the intersection of these two planes is known as the line of nodes.

We can find the vector n by taking the cross product of the angular momentum vector, h, and the unit vector K:

\bold{n} = \bold{K} \times \bold{h}

n is a unit vector on the line of nodes that points in the direction of the ascending node. The ascending node is the spot where the satellite crosses the equatorial plane in a northerly direction. Likewise, the descending node is the point where the satellite crosses the equatorial plane in the southerly direction.

In an equatorial orbit, n is undefined, and there are no nodes.

Direct and RetrogradeEdit

Direct means the satellite is traveling around the earth from west to east. Retrograde means the satellite is traveling from east to west.

Visual RepresentationEdit


This image shows two planes, the orbital plane (pale yellow) and the ecliptic (gray). The earth, or prime focus is considered to be at the dot at the center of the two planes. Notice how the orbit crosses the ecliptic at two points: once going northward (the ascending node), and once going southward (the descending node). The descending node is not marked in this picture, but its location should be obvious. We will explain some of the other terms below.

Orbital elements.svg

Classical ElementsEdit

Now that we have the node vector, along with the vectors r, v, and h, we can find our 5 classical orbital elements.

Semi-Major AxisEdit

The semi-major axis, a can be found via the Vis-Viva equation:



The eccentricity e can be determined from the magnitude of the eccentricity vector:


Alternately, it can be expressed as:


though the eccentricity vector will be needed in the calculation of later elements.


The inclination, i is the angle between the K unit vector and the angular momentum vector h. It can be calculated using:

[Orbital Inclination]

i = \arccos\frac{h_K}{h}

The angle of inclination is always less than 180°.

Longitude of the Ascending NodeEdit

The longitude of the ascending node, Ω, is the angle between the ascending node and the I unit vector. It can be calculated as:

[Longitude of the Ascending Node]

\Omega = \arccos\frac{n_I}{n}

If the nJ component is greater than zero, then Ω is less than 180°. Otherwise, Ω is greater than 180°.

Argument of PeriapsisEdit

The argument of periapsis is the angle in the orbital plane between the ascending node and the periapsis. We can calculate it as:

[Argument of Periapsis]

\omega = \arccos \frac{\bold{n} \cdot \bold{e}}{ne}

If eK is greater than 0, then ω is less than 180°. Otherwise, ω is greater than 180°.

Time of Periapsis PassageEdit

The time of periapsis passage is the time when the satellite reaches periapsis. We define this time as Tp. It can be determined from observation. This is the sixth orbital parameter that allows calculating the position of a celestial object in any time.

Optional ElementsEdit

Longitude of PeriapsisEdit

Longitude of the Periapsis is the sum of the Longitude of the Ascending Node and the Argument of Perigee

[Longitude of Periapsis]

\Pi = \Omega  + \omega

True Anomaly at EpochEdit

Argument of Latitude at EpochEdit

True Longitude at EpochEdit

Finding r and vEdit

From the five classical orbital elements, we can determine r and v.

\bold{r} = r\cos(\nu)\bold{P} + r\sin(\nu)\bold{Q}
\bold{v} = \sqrt{\frac{\mu}{p}}[-sin(\nu)\bold{P} + e\bold{Q} + \cos(\nu)\bold{Q}]

Where P and Q are unit vectors in the perifocal coordinate system. Using our coordinate transformations, we can convert these into geocentric-equatorial coordinates.