# Determining the distance of an orbiting satellite

If you see a satellite moving overhead (at night obviously), what’s a better indicator of its orbit distance; its brightness or the speed its moving at?

Its velocity. Brightness is a combination of reflectivity, distance and size and as such, it’s not a reliable indicator of any individual property. Velocity, on the other hand, is directly related to orbital distance, so that if you know the one, you can calculate the other.

I’d say its speed (arc subtended over time) although this won’t tell you the orbital eccentricity. You need a few measurements some time apart to pin that down.

The observed brightness can vary depending on its size, shape, angle wrt the observer and the sun and its albedo. Too many variables there.

On preview what Q.E.D. said.

Most artificial satellites are in a pretty circular orbit. In such a case the distance can be estimated as:

d[sup]3[/sup] = 10[sup]13[/sup]t[sup]2[/sup] meters[sup]3[/sup] t is time in seconds for one circuit.

Take the cube root and that’s the distance in meters from the center of the earth. Subtract the earth’s radius of 6.367*10[sup]6[/sup] meters and you’ve got it.

Even simpler: If you see it moving through the sky, it’s almost certainly in low-earth orbit, about 200-300 km up. That piece of information alone will give you the height more precisely than anything else you could do without instruments.

Actually many satellites are in non-circular orbits. Some, like the GPS constellation satellites, are indeed in circular (and non-geosynchronous) orbits, which is allowed because they are sufficiently numerous to overlap coverage (typcially a minimum of 4-5 are in view above the horizon at most navigatiable locations on the planet. Ditto for some weather satellites. However, many surveillance and communication satellites are in highly elliptical orbits, which allows them to lay a complex meandering path that dwells over a certain area of the globe–say, a particular nation of Northeast Asia that you’d like to keep a regular eye on–and makes them available for use for a greater amount of their total orbital period than a circular geosynchronous orbit would allow.

If you assume a circular orbit, the equation given by David Simmons will give you a good estimation. If you assume a more eccentric elliptical orbit then the situation becomes more complex. Its not that the ballistics are that much more difficult–Kepler’s Laws will give you an exact solution given sufficiently accurate data–but the problem is making good observations. You can mark angular position with respect to a rotating ground reference frame, as has already been indicated, estimating radial distance via size or brightness is guesswork at best. In order to plot an elliptical orbit, you need a minimum of two positions in their radial, normal angular, and inclination measurements plus velocity at each position; since all you can measure for any individual point is it angular and (with two data points) inclination, you need to take a series of observations, fit a path through them, correct for errors in measurement, and account for any orbital eccentricities like lunar and solar tides (though with an eyeball/astrolabe observation, the eccentricities are going to be lost in the natural error of observation).

The most visible artificial satellite you are going see in the sky is the ISS; according to NASA, its orbit is at an altitude of 250 statute miles with an inclination of 51.6 degrees. Try making some measurements on that first and see how well they match up to the designated orbit.

Stranger

I stand corrected. Of course if the orbit has a lot of eccentricity then asking for “the distance” isn’t exactly meaningful.

“Brightness” is dependent on a lot of variables, most importantly the reflectivity of the satellite. The Iridium satellites are notoriously reflective, so much so that they light up a swath of the earth as they pass. Other satellites are tumbling, and have highly variable reflectivity as different surfaces reflect.

My favorite site for satellite watching is Heavens Above (known affectionately by my friends as heavens dash above dot com). You can look up the visible satellites for your location, their magnitude, altitude, track, and distance, among other details.

Looking at the chart for today, for my location, I see that the Lacrosse 3 passes at 5:30, magnitude 2.6, distance 420 miles; while Landsat 4 comes by at 6:06, magnitude 4.3 (a lot dimmer), distance 360 miles. Just from comparing those two examples one can see that magnitude has no correlation with distance.

Meanwhile, I’ll see a -5 magnitude (brighter than Venus) Iridium flare at 5:40, from Iridium 80, distance (probably) 480 miles.

Well, you’ll have the instantaneous distance, but as Q.E.D. notes, that’s very difficult to measure–even through a powerful telescope, a large satellite is barely more than a dot, and with atmospheric distortion and changing albedo with orientation, you’re not going to be able to scale it to figure distance. With elliptical orbits they are generally defined by an apogee and perigee and either orbital period or eccentricity, plus inclination.

But most satellites that are in highly elliptical orbits aren’t going to be readily visible to the unaided eye, and may not be visible to amateur observers at apogee. As Chronos says, if you can see it, it’s clearly in a Low Earth Orbit, and assuming that it is circular (i.e. you don’t see it speeding up or slowing down dramatically near perigee) that will give you an order of magnitude estimate.

Stranger