# Semi-major axis

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In geometry, the term semi-major axis (also semimajor axis) is used to describe the dimensions of ellipses and hyperbolae.

## Ellipse

The major axis of an ellipse is its longest diameter, a line that runs through the centre and both foci, its ends being at the widest points of the shape. The semi-major axis is one half of the major axis, and thus runs from the centre, through a focus, and to the edge of the ellipse.

It is related to the semi-minor axis $b\,\!$ through the eccentricity $e\,\!$ and the semi-latus rectum $\ell\,\!$, as follows:

$b = a \sqrt{1-e^2}\,\!$
$\ell=a(1-e^2)\,\!$.
$a\ell=b^2\,\!$.

A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping $\ell\,\!$ fixed. Thus $a\,\!$ and $b\,\!$ tend to infinity, $a\,\!$ faster than $b\,\!$.

The semi-major axis is the mean value of the smallest and largest distances from one focus to the points on the ellipse. Now consider the equation in polar coordinates, with one focus at the origin and the other on the positive x-axis,

$r(1-e\cos\theta)=l\,\!$

The mean value of $r={\ell\over{1+e}}\,\!$ and $r={\ell\over{1-e}}\,\!$, is $a={\ell\over{1-e^2}}\,\!$.

## Hyperbola

The semi-major axis of a hyperbola is one half of the distance between the two branches; if this is a in the x-direction the equation is:

$\frac{\left( x-h \right)^2}{a^2} - \frac{\left( y-k \right)^2}{b^2} = 1$

In terms of the semi-latus rectum and the eccentricity we have

$a={\ell\over e^2-1 }$

## Astronomy

### Orbital period

In astrodynamics the orbital period $T\,$ of a small body orbiting a central body in a circular or elliptical orbit is:

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

where:

$a\,$ is the length of the orbit's semi-major axis
$\mu$ is the standard gravitational parameter

Note that for all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity.

In astronomy, the semi-major axis is one of the most important orbital elements of an orbit, along with its orbital period. For solar system objects, the semi-major axis is related to the period of the orbit by Kepler's third law (originally empirically derived),

$T^2=a^3\,$

where T is the period in years, and a is the semimajor axis in astronomical units. This form turns out to be a simplification of the general form for the two-body problem, as determined by Newton:

$T^2= \frac{4\pi^2}{G(M+m)}a^3\,$

where G is the gravitational constant, and M is the mass of the central body, and m is the mass of the orbiting body. Typically, the central body's mass is so much greater than the orbiting body's, that m may be ignored. Making that assumption and using typical astronomy units results in the simpler form Kepler discovered.

Remarkably, the orbiting body's path around the barycentre and its path relative to its primary are both ellipses. The semi-major axis used in astronomy is always the primary-to-secondary distance; thus, the orbital parameters of the planets are given in heliocentric terms. The difference between the primocentric and "absolute" orbits may best be illustrated by looking at the Earth-Moon system. The mass ratio in this case is 81.30059. The Earth-Moon characteristic distance, the semi-major axis of the geocentric lunar orbit, is 384,400 km. The barycentric lunar orbit, on the other hand, has a semi-major axis of 379,700 km, the Earth's counter-orbit taking up the difference, 4,700 km. The Moon's average barycentric orbital speed is 1.010 km/s, whilst the Earth's is 0.012 km/s. The total of these speeds gives the geocentric lunar average orbital speed, 1.022 km/s; the same value may be obtained by considering just the geocentric semi-major axis value.

### Average distance

It is often said that the semi-major axis is the "average" distance between the primary (the focus of the ellipse) and the orbiting body. This is not quite accurate, as it depends over what the average is taken.

• averaging the distance over the eccentric anomaly (q.v.) indeed results in the semi-major axis.
• averaging over the true anomaly (the true orbital angle, measured at the focus) results, oddly enough, in the semi-minor axis $b = a \sqrt{1-e^2}\,\!$.
• averaging over the mean anomaly (the fraction of the orbital period that has elapsed since pericentre, expressed as an angle), finally, gives the time-average (which is what "average" usually means to the layman): $a (1 + \frac{e^2}{2})\,\!$.

The time-average of the inverse of the radius, $r^{-1}\,\!$, is $a^{-1}\,\!$.

### Energy; calculation of semi-major axis from state vectors

In astrodynamics semi-major axis $a \,$ can be calculated from orbital state vectors:

$a = { - \mu \over {2\epsilon}}\,$ for an elliptical orbit and $a = {\mu \over {2\epsilon}}\,$ for a hyperbolic trajectory

and

$\epsilon = { v^2 \over {2} } - {\mu \over \left | \mathbf{r} \right |}$ (specific orbital energy)

and

$\mu = GM \,$ (standard gravitational parameter),

where:

• $v\,$ is orbital velocity from velocity vector of an orbiting object,
• $\mathbf{r }\,$ is cartesian position vector of an orbiting object in coordinates of a reference frame with respect to which the elements of the orbit are to be calculated (e.g. geocentric equatorial for an orbit around Earth, or heliocentric ecliptic for an orbit around the Sun),
• $G \,$ is the gravitational constant,
• $M \,$ the mass of the central body.

Note that for a given central body and total specific energy, the semi-major axis is always the same, regardless of eccentricity. Conversely, for a given central body and semi-major axis, the total specific energy is always the same.

## Example

The International Space Station has an orbital period of 91.74 minutes, hence the semi-major axis is 6738 km [1]. Every minute more corresponds to ca. 50 km more: the extra 300 km of orbit length takes 40 seconds, the lower speed accounts for an additional 20 seconds.