While video recording the star Tycho 1311-1818-1 in Taurus on a very cold Thursday evening last week (-4° F) in the hope that asteroid 126561 (2002 CF105) would pass in front of it (it didn’t), I was surprised and delighted to serendipitously record a very slow moving Earth-orbiting satellite crossing the field. Now, in order to see a satellite, it must be illuminated by sunlight. But to see any satellite during the first week of January only 10 minutes before local midnight, it must be very far from the Earth indeed (more on that later).
Here’s a video of the event showing its complete traversal of the field of view:
I’m hoping that one of the good people that frequent the satellite observers’ forum SeeSat-L will be able to identify this unusual object. Requisite to that, of course, are two precise positions at two precise times and the observer’s location.
A very useful online tool provided by the Department of Physics at Virginia Tech allows one to input the right ascension, declination, and x-y coordinates of between 4 and 10 known objects, and it does an astrometric solution across the field so you can determine the right ascension and declination of an unknown object.
Using Guide 9.1, Limovie, and this tool, I determined the following:
At 5 Jan 2018 5:42:58.122 UT, the satellite was located at:
5h45m48.14s +21°45’17.5″ (apparent coordinates, epoch of date).
At 5 Jan 2018 5:50:22.931 UT, the satellite was located at:
5h46m53.98s +21°48’06.3″ (apparent coordinates, epoch of date).
Observer Location: 42°57’36.9″N, 90°08’31.1″ W, 390 m.
Using the satellite coordinates above, and the angular separation calculator kindly provided by the Indian Institute of Astrophysics, we find that the satellite traversed just 0.2590° in 0.1236 hours. That’s 2.095° per hour, or only about four moon diameters in an hour!
Surely, this satellite must be way out there. How far? To determine that, I did a couple of what we used to call during my college physics days “back-of-the-envelope” (BOTEC) calculations. These are rough approximations—using simplifying assumptions—that should get you to an answer that is at least the right order of magnitude.
If we can estimate the orbital angular velocity of the satellite, we can determine its orbital period, and if we could determine that, we can calculate it orbital distance. Now, we don’t know yet if this satellite is in a near-circular or highly-elliptical orbit. If the satellite is an a highly-elliptical orbit and we observe it near apogee, its angular velocity will be somewhat slower than the angular velocity of a circular orbit at that same distance. If we observe it near perigee, then its angular velocity will be somewhat faster that the angular velocity of a circular orbit at that same distance. First simplifying assumption: let’s assume a circular orbit.
The next simplifying assumptions are that (1) the satellite passes through the observer’s zenith, and (2) the distance to the satellite is large in comparison to the radius of the Earth. At the time of observation, the satellite was at an altitude between 65° and 66° above the horizon. Not quite the zenith, but maybe close enough.
First, we need to compensate for the fact that the observer’s location on the surface of the Earth is moving in the same direction (along right ascension) as the satellite is orbiting (eastward) as the Earth rotates. We need to add the Earth’s rotational velocity to the right ascension component of the satellite’s velocity to get its true angular velocity relative to the center of the Earth. This of course assumes that the radius of the Earth is small compared to the distance to the satellite.
During the 0.1236 hours we observed the satellite, it moved 0.2743° eastward in right ascension and 0.0469° northward in declination. We now need to add a portion of the Earth’s angular velocity to the right ascension component of the satellite’s angular velocity. If the satellite were at the north celestial pole, the amount we would add would be zero. If, on the other hand, the satellite were on the celestial equator, we would add the full amount. Since cos 90° is 0 and cos 0° is 1, let’s add the Earth’s rotational angular velocity times the cosine of the satellite’s declination to the right ascension component of the satellite’s angular velocity.
The Earth turns through 360° in one mean sidereal day (23h 56m 04s = 86,164s). That’s 1.8591° during the 0.1236 hours we observed the satellite. Taking that times the average declination of the satellite during the observation time, we get 1.8591° cos 21.7783° =1.7264°. Adding this to the 0.2743° the satellite moved in right ascension, we get new components for the satellite’s angular displacement of 0.2743° + 1.7264° = 2.0007° in right ascension and 0.0469° in declination. This gives us the “true” angular displacement for the satellite of
This is a motion of about 16.19° per hour, giving us a rough orbital period of 22.235 hours or 80,045 seconds.
Using Newton’s form of Kepler’s Third Law to calculate the orbital semi-major axis, we get (as a very rough estimate):
where G is the gravitational constant, M is the mass of the Earth in kg, and P is the satellite’s orbital period in seconds.
Geosynchronous satellites have an orbital radius of 42,164 km, so our mystery satellite is almost as far out as the geosynchronous satellites. If it were further, the satellite would have been moving westward across our field of view, not eastward.
Admittedly, this is a lot of hand waving and is almost certainly wrong, but perhaps it gets us reasonably close to the right answer.
Now, let’s consider the shadow of the Earth to give us another estimate of the satellite’s distance.
At the time of observation, the Sun was located at 19h04m23s -22°36’40”. The anti-solar point, which is the center of the Earth’s shadow cone, was then located at 7h04m23s +22°36’40”. That is only 18.1° from the satellite. The Sun’s angular diameter at that time was 32.5 arcminutes. In order for the satellite to not be shadowed by the Earth, the angular diameter of the Earth as seen from the satellite must be less thanThe distance from the center of the Earth at which the Earth subtends an angle of 18.6° is given bySo, using this method, the satellite must be at an orbital radius of at least 38,905 km to be outside the Earth’s umbral shadow cone.
Now, on to something less speculative: the varying brightness of the satellite. I used Limovie to track the satellite across most of the field and got the following light curve.
At first blush, it appears the satellite is tumbling with a period of around 51.2s. But a closer inspection reveals that a larger amplitude is followed by a smaller amplitude is followed by a larger amplitude, and so on. So the tumbling period looks to me to be more like 102.4s. The mean (unfiltered) magnitude of the satellite looks to be around 11.8m, but ranging between 10.7m and 13.0m. Thus the amplitude is around 2.3 magnitudes. You will find the raw data here.
Update January 10, 2018
Alain Figer, French astronomer and satellite enthusiast, was kind enough to identify this object for me. Alain writes, “At first glance I noticed, using Calsky, that Falcon 9 rocket, 2017-025B, #42699, might be your satellite…From the MMT data (astroguard russian site) 2017-025B rotation period was measured at 89.55s on 13 OCT 2017. That figure seems to me in rather good agreement with yours at 102.4s, since the rotation period of this rocket might be quickly lengthening, a rather classical behaviour for such newly launched rockets.” Alain goes on to say, “For estimating the satellite altitude from your own observations you have to consider its highly eccentric elliptical orbit.” Thank you, Alain!
After I got home from work this evening, I began thinking, “Hmm, Guide is such an amazing program, maybe it can show me accurate satellite positions as well.” Turns out, it can! After downloading the current orbital elements for all satellites and turning on the satellite display, I was able to confirm Alain’s determination that this object is indeed Falcon 9 rocket body 2017-025B.
SpaceX launched the Inmarsat-5 F4 commercial communications satellite from historic Launch Complex 39A at NASA’s Kennedy Space Center in Florida using a Falcon 9 rocket on May 15, 2017. Here are some pictures and a video of that launch.
The Falcon 9 rocket body currently orbits the Earth once every 23h21m19s in a highly-elliptical orbit (e=0.8358) that ranges from a perigee height of 432.4 km to an apogee height of 69,783 km. During the time of observation, its range (i.e. distance from me, the observer) went from 64,388 km to 64,028 km. The semi-major axis of its orbit is 41,481 km which is 3.3% higher than my (lucky) estimate above. The shadow criterion of > 38,905 km is met as well.
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Updated January 10, 2018