## Geostationary Satellite Declinations

A few years ago, I was doing some telescope sweeping of the meridian sky around declination -6˚ when, to my surprise and delight, a 10th- or 11th-magnitude slow-moving object entered my field of view. As it slowly traversed eastward through the field, I remembered the declination I was pointed to and realized that it must be a geostationary, or at least a geosynchronous, satellite. Centering the moving object and then turning off the telescope’s clock drive confirmed my suspicions. The object was a geosynchronous satellite because it appeared to lay motionless while all the stars in the field drifted toward the west. Serendipity is the spice of life!

Satellites stationed in orbits that are always directly above the Earth’s equator and that have an orbital period of 23h 56m 04.0905s (one sidereal day) have the interesting property of remaining stationary as seen from any point on the surface of the Earth. This property of geostationary satellites, as they are called, is used to great advantage by many communications and weather satellites. There are currently at least 554 satellites in geosynchronous orbits. They are stationed all around the Earth at various longitudes.

At what altitude do geostationary satellites orbit the Earth? It is well above human-occupied spacecraft like the International Space Station which currently orbits 260 miles above the Earth’s surface. Geosynchronous orbit lies some 22,236 miles above the Earth’s equator. This is quite a ways out, as the entire Earth subtends an angle of only 17° 12′ at this distance—about the same as the angular distance between Capella (α Aur) and Elnath (β Tau).

Looking at it another way, geostationary satellites orbit at an altitude that is 2.8 Earth diameters above the equator. Since the Moon orbits at a distance that ranges between 27.4 and 31.4 Earth diameters above the Earth’s surface, geosynchronous orbit is about 1/10 of the way to the Moon.

If you have a telescope, know where to point it, and turn tracking off, you can see a geostationary satellite as a stationary point of light while the stars drift by due to the Earth’s rotation. At our latitude here in southern Wisconsin (43° N), the area where you want to search for geostationary satellites (near the meridian) is around declination -6° 37′. Remember, declination tells you how many degrees above or below the celestial equator an object is, and the numbers range from -90° to +90°, the south celestial pole and north celestial pole, respectively. The celestial equator has a declination of 0°.

For any latitude1, the declination you want to search is given by $\delta _{gs}=\textup{tan}^{-1}\left [ 6.611\textup{ csc }\phi - \textup{cot }\phi \right ]-90^{\circ}$

where δgs is the declination of the geostationary satellite in degrees
and ϕ is your latitude in degrees

Since most calculators don’t have the cosecant (csc) or cotangent (cot) functions, this formula can be rewritten in a slightly more complicated form as $\delta _{gs}=\textup{tan}^{-1}\left [ \frac{6.611}{\textup{sin }\phi }-\frac{1}{\textup{tan }\phi } \right ]-90^{\circ}$

Why aren’t the satellites right on the celestial equator (δ = 0°)? They would be if they were millions of miles away or if we were located on the Earth’s equator, but at our northern latitude trigonometric parallax causes us to see the satellites somewhat below the celestial equator, relative to the distant stars.

What if the geostationary satellite is situated east or west of your meridian? How do you calculate its declination then? As you might expect, because the range (observer-to-satellite distance) is greater the further from the meridian the satellite is, the less the parallax is, and therefore the closer the declination is to the equator, though not by a lot. The declination is also symmetric about the meridian, east and west: a geostationary satellite one hour east of the meridian will have the same declination as another geostationary satellite one hour west of the meridian.

If you know the longitude of the geostationary satellite (for example, the GOES-16 weather satellite is stationed above 75.2˚ W longitude), you can calculate its declination (and right ascension) using the following two-step process. $\textup{h}=\textup{tan}^{-1}\left [ \frac{\textup{sin }\Delta\lambda }{\textup{cos }\Delta \lambda-0.15126\textup{ cos }\phi } \right ]$

where h is the hour angle in degrees
and Δλ = λsat − λobs , the difference between the satellite and observer
longitudes, in degrees
and ϕ is the latitude of the observer in degrees $\delta _{gs}=\textup{tan}^{-1}\left [ \frac{-0.15126\textup{ sin }\phi \textup{ sin h}}{\textup{sin }\Delta \lambda } \right ]$

To determine the right ascension of the geostationary satellite, add the value of h to your local sidereal time (the right ascension of objects on your meridian). Make sure you convert h to hours before adding it to your LST.

What if you want to calculate the geostationary declination at a particular hour angle? That is a bit trickier. I could not figure out how to manipulate the equation for h above so that Δλ = f (h,φ). Instead, I rewrote the equation as $\sin \Delta \lambda =\tan h\cdot \left ( \cos \Delta \lambda -0.15126\cos \phi \right )$

and using h as a starting value for Δλ, substituted it into the cos Δλ expression, calculated sin Δλ, took the arcsine to get a new value of Δλ, then substituted that back into the cos Δλ expression, and iterated. Fortunately, the value of Δλ converges very fast. Once you have Δλ, you can use the two-step process we used earlier to determine the declination of the geostationary satellite for a particular hour angle.

Please note that the value of the hour angle h we use here is positive east of the meridian and negative west of the meridian. This is opposite from the normal astronomical sense.

Here is a simple SAS program illustrating how to do all these calculations using a computer.

And here is the output from that program.

1 For latitudes south of the equator, add 180° to get your meridian geostationary declination. The equation goes singular at the equator (φ=0°) and at the poles (φ=90° N and 90° S) since we’re dividing by sin φ = 0 at the equator and tan φ is undefined at the poles. However, as you asymptotically get closer and closer to latitude 0° (0.0001° and -0.0001°, for example) you find that the meridian geostationary declination approaches δ = 0°. Likewise, as you asymptotically approach latitude 90° N and 90° S, you’ll find that the meridian geostationary declination approaches -8°36′ and +8°36′, respectively. Of course, in both cases the geostationary satellites always remain below your horizon. How far north or south in latitude would you have to go, then, to find that geostationary satellites on your meridian are on your horizon due south or due north, respectively? Through a little algebraic manipulation of the first equation above and utilizing some simple trigonometric identities, one finds that at latitudes 81°18′ N and 81°18′ S, geostationary satellites on your meridian would be on the horizon. North or south of there, respectively, you would not be able to see them because the Earth would be in the way.

References

Gérard Maral, Michel Bousquet, Zhili Sun. Satellite Communications Systems: Systems, Techniques and Technology, Fifth Edition. Wiley, 2009. See section 8.3.6.3 Polar mounting.

## Thank the Sumerians

Over five thousand years ago, the Sumerians in the area now known as southern Iraq appear to have been the first to develop a penchant for the numbers 12, 24, 60 and 360.

It is easy to see why. 12 is the first number that is evenly divisible by six smaller numbers:

12 = 1×12, 2×6, 3×4 .

24 is the first number that is evenly divisible by eight smaller numbers:

24 = 1×24, 2×12, 3×8, 4×6 .

60 is the first number than is evenly divisible by twelve smaller numbers:

60 = 1×60, 2×30, 3×20, 4×15, 5×12, 6×10 .

And 360 is the first number that is evenly divisible by twenty-four smaller numbers:

360 = 1×360, 2×180, 3×120, 4×90, 5×72, 6×60, 8×45, 9×40, 10×36, 12×30, 15×24, 18×20 .

And 360 in a happy coincidence is just 1.4% short of the number of days in a year.

We have 12 hours in the morning, 12 hours in the evening.

We have 24 hours in a day.

We have 60 seconds in a minute, and 60 minutes in an hour.

We have 60 arcseconds in an arcminute, 60 arcminutes in a degree, and 360 degrees in a circle.

The current equatorial coordinates for the star Vega are

α2019.1 = 18h 37m 33s
δ2019.1 = +38° 47′ 58″

Due to precession, the right ascension (α) of Vega is currently increasing by 1s (one second of time) every 37 days, and its declination (δ) is currently decreasing by 1″ (one arcsecond) every 5 days.

With right ascension, the 360° in a circle is divided into 24 hours, therefore 1h is equal to (360°/24h) = 15°. Since there are 60 minutes in an hour and 60 seconds in a minute, and 60 arcminutes in a degree and 60 arcseconds in an arcminute, it follows that 1m = 15′ and 1s = 15″.

Increasingly, you will see right ascension and declination given in decimal, rather than sexagesimal, units. For Vega, currently, this would be

α2019.1 = 18.62583h
δ2019.1 = +38.7994°

Or, both in degrees

α2019.1 = 279.3875°
δ2019.1 = +38.7994°