# Semi-major axis

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Semi-major axis
The semi-major axis of an 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; essentially, it is the measure of the radius of an orbit taken at the orbit's two most distant points. For the special case of a circle, the semi-major axis is the radius. One can think of the semi-major axis as an ellipse's long radius.

The length of the semi-major axis a of an ellipse is related to the semi-minor axis' length b through the eccentricity e and the semi-latus rectum , as follows:

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

The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches. Thus it is the distance from the center to either vertex (turning point) of the hyperbola.

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 fixed. Thus $a\,\!$ and $b\,\!$ tend to infinity, a faster than b.

## Ellipse

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,

$r(1-e\cos\theta)=\ell.\,$

The mean value of $r={\ell\over{1+e}}\,\!$ and $r={\ell\over{1-e}}\,\!$, (for $\theta = \pi \, \text{and} \, \theta = 0$) is

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

In an ellipse, the semimajor axis is the geometric mean of the distance from the center to either focus and the distance from the center to either directrix.

## Hyperbola

The semi-major axis of a hyperbola is, depending on the convention, plus or minus 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

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

The transverse axis of a hyperbola coincides with the semi-major axis.[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:

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

where:

a is the length of the orbit's semi-major axis
μ 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.

The angular momentum H of a small body orbiting a central body in a circular or elliptical orbit is:

$H = \sqrt{a \cdot \mu \over (1-e^2)}$

where:

a and μ are as defined above
e is the eccentricity of the orbit

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),

$T^2 \propto a^3 \,$

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.

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 focus of the ellipse and the orbiting body. This is not quite precise, as it depends on what the average is taken over.

• 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
$a \left(1 + \frac{e^2}{2}\right).\,$

The time-average of the inverse of the radius[clarification needed], 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:

$a = { - \mu \over {2\varepsilon}}\,$

for an elliptical orbit and, depending on the convention, the same or

$a = {\mu \over {2\varepsilon}}\,$

and

$\varepsilon = { v^2 \over {2} } - {\mu \over \left | \mathbf{r} \right |}$

and

$\mu = G(M+m ) \,$

(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 and m are the masses of the bodies.

Note that for a given amount of total mass, the specific energy and the semi-major axis are always the same, regardless of eccentricity or the ratio of the masses. Conversely, for a given total mass and semi-major axis, the total specific energy is always the same. This statement will always be true under any given conditions.

## References

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