A natural, tiny (meters-size, maybe 10.000kg mass) natural satellite could be trapped in a geostationary orbit. I wondered for quite some time:

When and how are we able to detect these satellites?

I assume the distance of 35.700km of way too far for naked eye detection. So the next realistic opportunity would have been Galilei who first used telescopes for scientific observations of the sky. Could he have detected such an object?

Now even if he had enough resolution, he certainly didn't systematically cover the whole 4$\pi$ of the sky (especially as the satellite could be syncronous with the other side of the earth, and he could never see it).

Then, would it be at the times of Hubble (because of the Mount Wilson Observatory and other similarly powerful telescopes)? Would it at the times when quality equipment for hobby astronomers became cheap and therefore widespread enough (to cover huge areas of the sky)? Or would we - until today - not be able to detect such objects?

To answer this question, one needs to consider both the technical ability as well as the area of the sky that is covered.

  • $\begingroup$ Keep in mind that geostationary orbits only exist over limited latitudes. $\endgroup$ Commented Mar 12, 2019 at 18:18
  • $\begingroup$ Thanks for the edit, it looks much better now! $\endgroup$
    – uhoh
    Commented Mar 13, 2019 at 4:43

2 Answers 2


tl;dr: At distances far enough from Earth that the motion with respect to the stars was slow, a serendipitous survey photographic plate from a large enough telescope might catch a trail, and in a doubly serendipitous situation it might have been a short exposure, duplicated on the next night, an Earth orbit suspected, and a hunt for a second Earth satellite begun.

However starting in the 1960's and 1970's radar and visual scans for artificial satellites in Earth orbit would have found this natural satellite in Earth orbit if it were low enough.

I'll start with @CarlWitthoft's 5 meter asteroid refer to this answer and especially this answer. Two equivalent equations to get the absolute magnitude of an asteroid are:

$$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$

where $H$ is absolute magnitude, $p_V$ is albedo, D is in km, and $C$ = 15.618, and

$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right).$$

A 5 meter diameter asteroid with an albedo of 0.1 has an absolute magnitude of +29.6.

The apparent magnitude from this answer:

Knowing the absolute magnitude of an object, you calculate the apparent magnitude $m$ using:

$$ m = M_{Abs} + 5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right), $$

where $d_{SR}$ and $d_{RE}$ are the Sun-Roadster and Roadster-Earth Sun-satellite and satellite-Earth distances, each normalized by 1 AU, and the factor $O(1)$ is the phase integral, of order unity, taking into account the angular difference between the direction of illumination and the direction of viewing. In an order of magnitude calculation, this only becomes really significant when the body moves between the Sun and the viewer. See https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H).

Let's pick two distances. One is the geostationary distance where the satellite would appear to hover above the observer, probably drifting up and down in a roughly annelema shape because the Earth's oblanteness will eventually tilt the orbit. See Geostationary orbit; Orbital stability. It will have a distance to Earth as close as 36,000 km.

The other is a low Earth orbit, but high enough that it will not decay due to drag too soon. Call it 1000 km altitude, or a circular orbit with a 7378 km semi major axis.

Plugging all of this into the equation above, I get:

     orbit                   closest distance      visual magnitude
Geosynchronous altitude         36,000 km                +20.6
Low Earth Orbit                  1,000 km                +16.7 

In low Earth orbit the apparent magnitude is almost as bright as Pluto, but it's going to be moving pretty fast. $\sqrt(GM/a)$ gives 7350 m/s, at a distance of 1000 km that's about 0.4 degrees per second. Any large telescope that's being used in astronomy will be tracking the motion of stars, or close to that, so it will be fast +17 magnitude track rather than a dot, and last only a fraction of a second. That probably wouldn't expose a photographic plate or if it did it would be dismissed as an artifact, meteor, or scratch. Visually it wouldn't be noticed.

At GEO type distances and +20.6 magnitude, the object would move about 0.25 degree in a minute, so it might also be captured on a photograph, but by the time the plate was developed it would be impossible to know when it appeared in a long exposure. However if the exposure (say at the Hale 200 inch telescope) were short, it really is possible that at a short-term trajectory on the celestial sphere could be considered. The problem is that nobody would suspect that it was in Earth orbit, and they would extrapolate to a heliocentric orbit and never find it again.

If the plate happened to be a series and there was another exposure of the same patch of sky the next night, then they would see it again and get pretty suspicious that it was in Earth orbit.

However, in a post-Sputnik cold war era radar and optical searches of the sky for objects in Earth orbit became particularly interesting.

So I would say that satellite surveys (both optical and radar) in the 1960's and 1970's would be the first likely candidates to find this 5 meter, 0.1 albedo satellite.

For some insight on optical tracking, see the two videos linked in Are commercial communications satellites in GEO being constantly monitored by telescopes?. Currently these links will take you to a new tab with the YouTube video:



If you would like to see them here, then leave a comment or answer or vote at Interest in looking into adding the YouTube viewer?.


To first order: the ratio of the Moon's radius to distance from Earth is

$ \frac{1740e3}{380e6} = 0.004578947 $

and the ratio of a 5-m radius satellite at geosync orbit is roughly

$\frac{5}{36e6 } = 1.388889e-07 $

This means, for similar albedo, the light reaching your telescope (or eye) would be $(\frac{1.388889e-07}{0.004578947})^2 = 9.200339e-10 $ as much light as the full moon. You're not going to see it even with a good telescope.


As the comment points out, I was too glib there. If you knew where to look, a decent 20 cm (aka 8-inch) telescope can easily show an object of that apparent magnitude. The nice thing about geostationary is that you can spend plenty of nights sweeping the possible sky regions; the satellite won't move.

  • 1
    $\begingroup$ That's apparent magnitude +10, easily visible in a 20cm telescope. $\endgroup$
    – Mike G
    Commented Mar 12, 2019 at 20:11

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