Edmund Weiss (1837-1917) and many astronomers since have called asteroids “vermin of the sky”, but on October 4, 1957 another “species” of sky vermin made its debut: artificial satellites. In the process of video recording stars for possible asteroid occultations, I frequently see satellites passing through my 17 × 11 arcminute field of view.
I’ve put together a video montage of satellites I serendipitously recorded during the first half of 2020. Many of the satellites move across the field as “dashes” because of the longer integration times I need to use for some of my asteroid occultation work. A table of these events is shown below the video. The range is the distance between observer and satellite at the time of observation. North is up and east is to the left.
Interestingly, three of the above satellites (7,9,18) are in retrograde orbits, that is their orbital inclination is > 90˚ and their east-west component of motion is towards the west instead of the east. However, I was surprised to find that two of the prograde orbiting satellites (5,6) appear to be orbiting retrograde! Both have orbital inclinations close to 90˚ (82.6˚ and 87.5˚, respectively), and both were in the western sky at northern declinations at the time of observation. But another satellite (8) with an orbital inclination of 82.5˚ at a southern declination in the southern sky at the time of observation exhibited the expected “barely” prograde motion. I suspect the ~0.5 km/s rotation of the Earth towards the east might have something to do with this “apparent retrograde” motion, but I was unable to find any reference that describes this situation.
Satellite #12 has an interesting story. It is the Inertial Upper Stage (IUS) used to launch USA-48 (Magnum), a classified DoD payload, from the Space Shuttle Discovery (STS-33).
In addition to these 18 satellites, I observed 7 geosynchronous satellites, shown below.
This non-operational Soviet communications satellite is a “tumbler”, meaning its changing orientation causes variation in its brightness, as shown below.
This non-operational communications satellite is also a tumbler, as seen in this light curve from a portion of the video.
There were four satellites I was unable to identify, shown in the video below. They were either classified satellites or, more likely, small pieces of space debris that only government agencies or contractors are keeping track of.
Occasionally, I record other phenomena of interest. Meteors during this period are described here, and you will find a couple of other curiosities below.
References Hughes, D. W. & Marsden, B. G. 2007, J. Astron. Hist. Heritage, 10, 21
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
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
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.
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
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
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.
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.
Gérard Maral, Michel Bousquet, Zhili Sun. Satellite Communications Systems: Systems, Techniques and Technology, Fifth Edition. Wiley, 2009. See section 188.8.131.52 Polar mounting.