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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? ]

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  • $\begingroup$ You forgot to consider your angle towards the panels. They might not appear so big unless they are perpendicular to you (If the sun is at your zenith for exemple). I'm not sure where did you got the 74km figure, but you might want to read this question space.stackexchange.com/questions/2515/… $\endgroup$ – Antzi Dec 12 '16 at 6:53
  • $\begingroup$ @Antzi "were did you got"? A glance at a List of satellites in geosynchronous orbit will show you plenty of examples of satellites which are nominally spaced at... wait for it... here it comes... 0.1° sometimes in groups of 3 and 4 in a row. The OP is also likely to have knowledge of multiplication - so for example $0.1° \ (\pi/180) \ 42164 \text{ km}$ equals, yes you guessed it 74 km. Your comment simply states that you don't know as much as the OP. $\endgroup$ – uhoh Dec 12 '16 at 7:08
  • $\begingroup$ @RogerWood I like your question but the answer would have to be a very wide range. An object sitting in a room is lit by diffuse light from all directions, but in GEO almost all the light will come from two well defined, narrow cones - the Sun, and as you point out - the Earth (as reflected sunlight). Shiny polished metal surfaces will appear dark until the geometries are right and then suddenly a 100X flare may happen. If the satellite is painted white with a matte finish it may have a constant brightness, but if it's mostly shiny polished metal the brightness will vary substantially. $\endgroup$ – uhoh Dec 12 '16 at 7:21
  • $\begingroup$ @RogerWood if the satellite is covered in "crinkly" yet shiny material, then the larger cone angle of the earth (about 17° as you indirectly point out) will allow more of the surface facets to reflect light in your direction than the much smaller cone angle of the Sun. As you two orbit, there will be times when you are between the satellite and the sun, and the solar panels - which are often rotated to keep their normal pointing at the sun - will suddenly appear much brighter - a flare. So you should be prepared for an answer that will come with a large range of values. $\endgroup$ – uhoh Dec 12 '16 at 7:34
  • $\begingroup$ @uhoh Satellites (Even GEO), are not on a 1D plane. $\endgroup$ – Antzi Dec 12 '16 at 8:42
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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!

enter image description here

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

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    $\begingroup$ An early CowSat prototype: youtu.be/YV0LGMGuLN0 $\endgroup$ – uhoh Dec 12 '16 at 11:03
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    $\begingroup$ I am disappointed that you did not work μ into your equations. $\endgroup$ – Organic Marble Jan 31 at 2:23
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    $\begingroup$ @OrganicMarble I had to stop chewing my cud for more than a few seconds to understand your comment; it's udderly perfect and quite pasture typical cleverness. $\endgroup$ – uhoh Jan 31 at 2:28
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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.

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  • $\begingroup$ Wow, I'm a little surprised the agreement is so close, but then again I'd be vexed if it weren't at least in the right ballpark. For the narrow cone of light from the sun, the details of the surface texture (crinkly, shiny, diffuse) are really important and I wasn't very certain what these satellites actually look like in their deployed states. I have a suspicion that there would be bright regions within the "constellation" where sides of the box-like satellites would be near specular geometry wrt the Sun, and like wise for the solar panels. $\endgroup$ – uhoh Dec 14 '16 at 4:17
  • $\begingroup$ The side normals of all satellites would likely remain parallel and perpendicular to the satellite-Earth vectors, and the panel normals remain parallel to the satellite-Sun vectors. This could be really beautiful! Even a small Raspberry Pi camera module with a fisheye lens would be enough to record it, though I'm sure adding one to a GEO satellite would not be as simple as it sounds... radiation, temperature, filtering of UV/IR to reduce image sensor damage, and of course protest by everyone involved with spacecraft reliability. $\endgroup$ – uhoh Dec 14 '16 at 4:26
  • $\begingroup$ This answer by @BrianOttum (of my favorites here) is slightly related. $\endgroup$ – uhoh Dec 14 '16 at 4:35
  • $\begingroup$ I've edited your answer in order to take advantage of stackexchange's implementation of MathJax for your equations. Here is a really helpful tutorial I always use when I can't remember how to do something: meta.math.stackexchange.com/q/5020/284619 $\endgroup$ – uhoh Dec 14 '16 at 5:06
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    $\begingroup$ Assuming the solar panels are kept exactly perpendicular to the Sun, then at the Spring and Fall equinoxes, the reflections from the solar panels will be in the same equatorial plane as the satellites themselves. So, for ~12 hours of the orbit, you should be able to look in a direction directly away from the Sun and see strong specular reflections from other satellites on the far side of the orbit. Maybe there's some geostationary spy satellites with cameras/telescopes that can swivel around and appreciate the view. $\endgroup$ – Roger Wood Dec 15 '16 at 20:01

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