At what magnitude would adjacent large geostationary communications satellites see each other?

Question

Imagine I'm sitting on a geostationary satellite. I can see the Earth in front of me. It's about the size of a soccer ball at arm's length. To the left and right, I can see similar satellites in the immediately adjacent slots. They're 0.1 degrees or 74 km away. You can almost discern their shape since their solar cells span about 25 meters and human visual resolution is about 1 arcsecond. But how bright will they appear? [ Maybe I can get this by looking up their magnitude seen from Earth and using the inverse square law? ]

Answer 1

Here is a profoundly simplistic way to get a feeling for how bright a sunlit object 74 km away might appear in space. Your milage may vary by one or two orders of magnitude depending on details of satellite shape and materials, and geometry of the sun-satellite-observer angle.

In this 0th order model the illumination by Earthshine is ignored, but can be added later. There will be simpler expressions for this, certainly from radar texts for example.

Assumptions:

• The observed satellite is a Spherical Cow with a radius $R_{cow}$ of 2 meters. Hereafter known as CowSat.
• CowSat is a Holstein with 70% of it's area acting as a simplified diffuse reflector - 50% of the light striking the white area is reflected hemi-isotropically into $2\pi$ sr.
• Geometry is optimal for cow brilliance. You are between CowSat and the Sun so that CowSat is fully illuminated.
• The Sun's visual magnitude is -27
• You are smart and so never look directly at the sun to keep your night vision. This leaves your pupils with a radius ($R_{pupil}$) of 0.003 meters. This actually factors out in the end anyway.
• You are 150 million kilometers or 1.5E+11 meters ($R_{SunEarth}$) from the sun.
• At that distance, your pupil takes in $(\pi R_{pupil}^2)/(4\pi R_{SunEarth}^2)$ or 1.0E-28 of the Sun's output.
• At the same distance, CowSat receives $(\pi R_{CowSat}^2)/(4\pi R_{SunEarth}^2)$ or 4.4E-23 of the Sun's output, and reflects 0.7*0.5 of that (or 1.6E-23) into $2\pi$ sr.
• At a distance $R_{satsep}$ of 74,000 meters, your pupil takes in $(\pi R_{pupil}^2)/(2\pi R_{satsep}^2)$ or 8.2E-16 of CowSat's reflected light.

Direct to Pupil: 1.0E-28 suns Reflected to Pupil: 1.6E-23 * 8.2E-16 = 1.3E-38 suns

CowSat/Sun = 1.3E-10 or 24.7 magnitudes dimmer.

-27 + 24.7 = -2.3 magnitude. CowSat will certianly be visible and could be very bright - as in Venus seen from Earth bright. In reality it will vary quite a bit, but in general next-door-neighbor geostationary satellites one or two tenths of a degree apart will be easily visible to each other - even using a cellphone camera, a significant fraction of the time when the basic geometry is favorable (viewed satellite is front or at least side-lit from the sun as seen by viewer satellite).

To put things in perspective, even a simple 100 mA lensed green LED 18 kilometers away will still appear to be as bright as a 0 magnitude star!

above: CowSat, without question, outstanding in its field. From here.

Answer 2

thanks @uhoh. I used to keep Jersey dairy cows (brown rather than black and white, but the albedo is probably similar) so I just love the spherical-cow back-of-the-envelope approach and, with my British roots certainly relate to the Monte Python link. I found this information about geostationary satellites on http://www.satobs.org/geosats.html: "Typically the satellite will be in the mag. +11 to +14 range". The correction of square-law brightness with distance is

$$2.5log_{10}((R_{GEO}/r_{nn})^2) \approx 14$$

where the geosynchronous radius and nearest-neighbor satellite separation $R_{GEO}$ and $r_{nn}$ are 42164 and 74 kilometers respectively. This would bring the brightness range between 0 and -3 magnitude, in very good agreement with CowSat.

The magnitude would fall off as

$$2.5log_{10}\left(\left(\frac{2 \ R_{GEO} \ sin(N \ \theta / 2)} {2 \ R_{GEO} \ sin( \theta / 2)}\right)^2\right) \approx 5 log_{10}\left(\frac{sin(N \ \theta / 2)}{sin(\theta / 2)}\right) \approx 5 log_{10}(N)$$

for the nearer satellites. So the brightness (magnitude) drops off quite slowly, e.g. 0, 1.5, 2.3, 3, 3.5, 3.9, 4.2, 4.5, 4.8, ... so, in the densely populated part of the orbit, quite a few of these will be visible.

All these satellites will appear to the observer as being in a straight line, but, most interestingly, they will appear all equispaced, which I wasn't expecting. This arises because the chord of a circle subtends the same angle everwhere on the circumference. Or, conversely, an observer fixed on the circumference will see a chord of a given length (74 km) subtending the same visual angle (0.05 degrees) wherever the chord is placed on the circle. So someone sitting on one of these satellites and looking around would instantly recognise their neighbors not only because they didn't rotate every 24 hours like the rest of the stars, but also because of strange linear equispaced pattern with monotonically diminishing brightness.