# Sun-synchronous orbit

Diagram showing the orientation of a Sun-synchronous orbit (green) in four points of the year. A non-sun-synchronous orbit (magenta) is also shown for reference. Dates are shown (in white): 21 March (right), 21 June (top), 23 September (left) and 22 December (bottom).

A Sun-synchronous orbit (SSO, also called a heliosynchronous orbit[1]) is a geocentric orbit that combines altitude and inclination in such a way that the satellite passes over any given point of the planet's surface at the same local solar time. Such an orbit can place a satellite in constant sunlight and is useful for imaging, spy, and weather satellites. More technically, it is an orbit arranged in such a way that it precesses once a year. The surface illumination angle will be nearly the same every time that the satellite is overhead. This consistent lighting is a useful characteristic for satellites that image the Earth's surface in visible or infrared wavelengths (e.g. weather and spy satellites) and for other remote sensing satellites (e.g. those carrying ocean and atmospheric remote sensing instruments that require sunlight). For example, a satellite in sun-synchronous orbit might ascend across the equator twelve times a day each time at approximately 15:00 mean local time. This is achieved by having the osculating orbital plane precess (rotate) approximately one degree each day with respect to the celestial sphere, eastward, to keep pace with the Earth's movement around the Sun.[2]

The uniformity of Sun angle is achieved by tuning the inclination to the altitude of the orbit (details in section "Technical details") such that the extra mass near the equator causes the orbital plane of the spacecraft to precess with the desired rate: the plane of the orbit is not fixed in space relative to the distant stars, but rotates slowly about the Earth's axis. Typical sun-synchronous orbits are about 600–800 km in altitude, with periods in the 96–100 minute range, and inclinations of around 98° (i.e. slightly retrograde compared to the direction of Earth's rotation: 0° represents an equatorial orbit and 90° represents a polar orbit).[2]

Special cases of the sun-synchronous orbit are the noon/midnight orbit, where the local mean solar time of passage for equatorial longitudes is around noon or midnight, and the dawn/dusk orbit, where the local mean solar time of passage for equatorial longitudes is around sunrise or sunset, so that the satellite rides the terminator between day and night. Riding the terminator is useful for active radar satellites as the satellites' solar panels can always see the Sun, without being shadowed by the Earth. It is also useful for some satellites with passive instruments that need to limit the Sun's influence on the measurements, as it is possible to always point the instruments towards the night side of the Earth. The dawn/dusk orbit has been used for solar observing scientific satellites such as Yohkoh, TRACE, Hinode and PROBA2, affording them a nearly continuous view of the Sun.

Sun-synchronous orbits can happen around other oblate planets, such as Mars. A satellite around the almost spherical Venus, for example, will need an outside push to be in a sun-synchronous orbit.

## Technical details

The angular precession per orbit for an orbit around an oblate planet is (Equation (24) of the article Orbital perturbation analysis (spacecraft)) given by

${\displaystyle \Delta \Omega =-2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ {\frac {3}{2}}\ \cos i\,}$

where

${\displaystyle J_{2}\,}$ is the coefficient for the second zonal term (1.7555 · 1010 km5 / s2) related to the oblateness of the earth (see Geopotential model),
${\displaystyle \mu \,}$ is the Standard gravitational parameter of the planet (398600.440 km3 / s2 for Earth)
${\displaystyle p}$ is the semi-latus rectum of the orbit,
${\displaystyle i}$ is the inclination of the orbit to the equator.

An orbit will be Sun-synchronous when the precession rate, ${\displaystyle \rho }$, equals the mean motion of the Earth about the Sun which is 360° per sidereal year (1.99096871 · 10−7 radians / s) so we must set ${\displaystyle \Delta \Omega /P=\rho }$ where P is the orbital period.

As the orbital period of a spacecraft is ${\displaystyle 2\pi \ a{\sqrt {\frac {a}{\mu }}}\,}$ (where a is the semi-major axis of the orbit) and as ${\displaystyle p\approx a\,}$ for a circular or almost circular orbit it follows that

${\displaystyle \rho \approx -{\frac {3J_{2}\cos i}{2a^{7/2}\mu ^{1/2}}}=-(360{\text{° per year}})\times (a/12352{\text{ km}})^{-7/2}\cos i=-(360{\text{° per year}})\times (P/3.795{\text{ hrs}})^{-7/3}\cos i}$

or when ${\displaystyle \rho }$ is 360° per year,

${\displaystyle \cos i\ \approx \ -{\frac {\rho \ {\sqrt {\mu }}}{{\frac {3}{2}}\ J_{2}}}\ a^{\frac {7}{2}}=-(a/12352{\text{ km}})^{7/2}=-(P/3.795{\text{ hrs}})^{7/3},}$

As an example, for a=7200 km (the spacecraft about 800 km over the Earth surface) one gets with this formula a Sun-synchronous inclination of 98.696 deg.

Note that according to this approximation cos i equals −1 when the semi-major axis equals 12 352 km, which means that only smaller orbits can be Sun-synchronous. The period can be in the range from 88 minutes for a very low orbit (a=6554 km, i=96°) to 3.8 hours (a=12 352 km, but this orbit would be equatorial with i=180°). (A period longer than 3.8 hours may be possible by using an eccentric orbit with p<12 352 km but a>12 352 km.)

If one wants a satellite to fly over some given spot on Earth every day at the same hour, it can do between 7 and 16 orbits per day, as shown in the following table. (The table has been calculated assuming the periods given. The orbital period that should be used is actually slightly longer. For instance, a retrograde equatorial orbit that passes over the same spot after 24 hours has a true period about 365/364 ≈ 1.0027 times longer than the time between overpasses. For non-equatorial orbits the factor is closer to 1.)

Orbits per day Period (hrs) Height above
Earth's surface
(km)
Maximum latitude
16 ${\displaystyle 1{\tfrac {1}{2}}}$ = 1 hr 30 min 282 83.4°
15 ${\displaystyle 1{\tfrac {3}{5}}}$ = 1 hr 36 min 574 82.3°
14 ${\displaystyle 1{\tfrac {5}{7}}}$ ≈ 1 hr 43 min 901 81.0°
13 ${\displaystyle 1{\tfrac {11}{13}}}$ ≈ 1 hr 51 min 1269 79.3°
12 ${\displaystyle 2}$ 1688 77.0°
11 ${\displaystyle 2{\tfrac {2}{11}}}$ ≈ 2 hrs 11 min 2169 74.0°
10 ${\displaystyle 2{\tfrac {2}{5}}}$ = 2 hrs 24 min 2730 69.9°
9 ${\displaystyle 2{\tfrac {2}{3}}}$ = 2 hrs 40 min 3392 64.0°
8 ${\displaystyle 3}$ 4189 54.7°
7 ${\displaystyle 3{\tfrac {3}{7}}}$ ≈ 3 hrs 26 min 5172 37.9°

When one says that a Sun-synchronous orbit goes over a spot on the earth at the same local time each time, this refers to mean solar time, not to apparent solar time. The Sun will not be in exactly the same position in the sky during the course of the year. (See Equation of time and Analemma.)

The Sun-synchronous orbit is mostly selected for Earth observation satellites that should be operated at a relatively constant altitude suitable for its Earth observation instruments, this altitude typically being between 600 km and 1000 km over the Earth surface. Because of the deviations of the gravitational field of the Earth from that of a homogeneous sphere that are quite significant at such relatively low altitudes a strictly circular orbit is not possible for these satellites. Very often a frozen orbit is therefore selected that is slightly higher over the Southern hemisphere than over the Northern hemisphere. ERS-1, ERS-2 and Envisat of European Space Agency as well as the MetOp spacecraft of EUMETSAT are all operated in Sun-synchronous, "frozen" orbits.[citation needed]

## References

1. ^ Tscherbakova, N. N.; Beletskii, V. V.; Sazonov, V. V. (1999). "Stabilization of heliosynchronous orbits of an Earth's artificial satellite by solar pressure.". Cosmic Research. 37 (4): 393–403. Bibcode:1999KosIs..37..417S.
2. ^ a b Rosengren, M. (November 1992). "ERS-1 - An Earth Observer that exactly follows its Chosen Path". ESA Bulletin (72). Bibcode:1992ESABu..72...76R.