Let's say it takes 5 minutes for the little "star" I observe at night to cross the sky from horizon to horizon. If one assumes a circular orbit and that the object in question is indeed in orbit around this planet; is it then possible to guesstimate orbital height?
How would the calculations be done?
Answer: Orbital altitude above the observer can be estimated from a satellite's apparent angular velocity.
The time it takes a satellite to cross the sky is very difficult to estimate with any accuracy, so I suggest you make your observation at the zenith instead.
The complete horizon is seldom visible, except at sea . The moment you first notice a satellite will seldom be the moment it appears at the horizon since you have no warning, so the “start” time will likely be inaccurate.
And, unlike on the Moon, the view of objects at the Earth's horizon are displaced and obscured. For instance, the sun is displaced an entire solar diameter when it is on the horizon. This is why low altitude sightings are not used in celestial navigation.
I suggest you rely on observing transits when the satellite is as high as possible, preferably within 30 degrees of the zenith.
The method described below will give a 2 significant figure estimate of a satellite’s orbital altitude (above the observer, not as measured from the center of the Earth). Its ease of use relies on some obviously simplistic assumptions:
Assumption: Satellite orbits, and the Earth, are flat. This is a reasonable assumption since the satellite altitude is an order of magnitude less than the earth’s radius.
Assumption: All observable satellites have the same speed. This is not true, but within tolerance for our purposes. The ISS speed is 7.66km/sec at 400km while Iridium satellites orbit with a speed of 7.45km/sec at an altitude of 780km. These examples cover the range of most visible satellites, so we can average it at 7.6km/sec.
Our observation datum is the time a satellite takes to cross the field of your binoculars. The field width is stamped on the binocular chassis, usually about 7.5 degrees. Since all LEO satellites travel at the same speed (for our purposes), the distance covered is proportional to the time taken to cross the field. The distance from the observer is a fixed multiple (1/sin7.5) of this distance.
To fine tune the result, correction can be made for Zenith angle (angle between the zenith and satellite at the time of observation). If the zenith has been previously identified (say, with a plumb bob or spirit level), the zenith angle can be estimated by the number of hand spans between satellite and zenith.
In practice, the observation is easiest from a supine position. Orient your feet towards the West since most satellites will be approaching from that quadrant. They will disappear as they reach the earth’s umbra before they get to the Eastern horizon. Have your assistant zero their stopwatch.
Pick a satellite which you judge will pass within 30 degrees of your zenith. Capture it in the field of view and follow it towards the zenith. Position the satellite against the trailing edge of the field of view and track it in that position. When the satellite is overhead, stop moving the binoculars to freeze the starfield in your view. Say “mark” so your assistant can start the stopwatch. When the satellite reaches the other edge of the field of view, say “mark” again to capture the transit time “t”. Say it is 8.5 sec in our example
The orbital distance “d” covered is
d=8.5sec(7.6km/sec)=64.6km.
Distance from the observer is
64.6/sin7.5=495km.
If the satellite passes off the zenith, a cosine correction can be applied using an estimated zenith angle. In our case, a zenith angle of 20* corrects the altitude to 465km
Consider this a supplementary answer for the sake of those readers who would opt for a bit more accuracy at the price of more work than a "guestimate".
Approximate orbital determination is frequently by Gauss's method, which is one of the earliest methods. This technique uses three positions and times of those observations. The positions have to be translated into radii from Earth's center with a little work. However, their accuracy is better is they are nearly overhead, instead of horizon to horizon because of parallax errors. See the Wikipedia article,
https://en.wikipedia.org/wiki/Gauss%27s_method It's simple enough one can make reasonable estimates using a calculator or spreadsheet. I'm sure if you look around, there is a canned method for reducing observations taken from a single location (latitude, longitude, time). Knowing the correct Universal Time (Zulu or GMT) is essential but this can be had from the exact local time and longitude. More accurate results are obtained from additional observations collected at other times and/or places.
Here's my own formula to make calculations of overhead pass time and orbital height:
$$t = \arccos \left(\frac{r}{r+x}\right) \frac{1}{\pi}\sqrt{k(r+x)^3}$$
where
t is time (in seconds) it takes to pass from horizon to horizon
r is Earth's (equatorial) radius (6378137) in meters
x is orbital height above Earth's surface, in meters
k = $4 \pi^2/(GM) \approx 9.90425 \times 10^{-14}$ where $GM$ is Earth's standard gravitational parameter
Trig and Kepler's third law did the trick! ChatGPT made the graph below. Thanks for all replies, BTW!
The formula seems to get me the correct answers.
