If a spacecraft is in low earth orbit ~400 Kilometres above earth.

How big does it need to be visible from earth with the human eye without telescopes, with enough of its details to clearly distinguish the craft from stars.

// EDIT:

The ISS is visible to the human eye as a fast moving object, but its features cannot be distinguished clearly.

I would like some information on what size the craft would have to be at 400km above earth so that it's big enough for humans to recognise as a craft.

Just to give an example, if the craft was something akin to the USS Enterprise. What size would it have to be for a human on earth to see the clear outline of its shape.

  • What is the spacecraft's albedo? Naked eye or telescope? – Scott Downey Nov 7 '14 at 9:56
  • @ScottDowney With the naked eye, I have clarified. – Kami Nov 7 '14 at 10:07
  • I definitely know you can see ISS passing above you using naked eye. And you can tell it apart from stars, because it moves fast and preictable – Pavel Janicek Nov 7 '14 at 10:10
  • I guess there's an issue though, as he said 'other stellar objects' are we to believe that this space craft must be distinguishable from the ISS? If so it will need to move at a different speed, or with radically different orbit, or be large enough for someone to actually see that it is not the ISS. – sydan Nov 7 '14 at 10:20
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    A 2cm object at 1m distance has a viewing angle of about 1.146 degrees - this is equivalent to the average thumb being held out in front of you - if pointed at space, at a distance of 400km, it would cover a region of space around 8km in width - is 'covered by a thumb' big enough to be identified as a spacecraft? – Scott Downey Nov 7 '14 at 10:42
up vote 29 down vote accepted

The angular resolution of the naked human eye is approximately one arc minute, or 1/60th of a degree. (Also a nice illustration of what this means in practice, which just so it happens also shows the ISS for scale.) This is going to be different for different people, but should serve as a baseline (and it is also a convenient number to work with).

Given an angular size $a$ (in seconds of arc), object diameter $d$ and distance $D$, we know that $$ \frac{a}{206265} = \frac{d}{D} $$ where 206,265 is the number of arc seconds in one radian ($ 206265 = 60 \times 60 \times \frac{360°}{2\pi} $) and $d$ and $D$ are in the same units. We know that $ a = 60 $ (arc seconds) and $ D = 400 \text{ km} $, and if we plug these in and use the here more useful unit of meter then we get $$ \frac{60}{206265} = \frac{d}{400000 \text{ m}} $$

Solving for $d$ gives us $$ d = 400000 \text{ m} \times \frac{60}{206265} \approx 116 \text{ m} $$ which tells us that the smallest resolvable size for the human eye at distance 400 km is approximately 116 m. This is, in a manner of speaking, the "pixel size" of the human eye at that particular distance.

Hence, any spacecraft features must be at least 100 meters or so large for them to be discernable by the naked eye at a distance of 400 km. For comparison, the International Space Station measures 109 meters truss length by 73 meters solar array length and nominally orbits at 418 to 423 km altitude. While the ISS is visible, its shape is not discernable.

Note that the size figures you get for this assumes that $D$ is the distance to the object. If the spacecraft is directly above the ground-based observer, the distance will be equal to the altitude, but if the spacecraft is not directly overhead then the distance to it from the observer will be larger. (Trigonometry is your friend here.)

Also note that the brightness of the object makes a potentially very large difference. Many objects smaller than this theoretical value are visible to the naked eye because they put out, or reflect, enough light to be registered by the rods or cones of the eye; however, those appear as point light sources without any discernable features, just like remote stars. A spacecraft is likely to be made of reflective material, which will be reflecting sunlight making it much easier to spot.

Atmospheric conditions, light pollution and other factors will also contribute to whether an object is visible. Clear wilderness skies are obviously better for this purpose than light- and smog-polluted city landscapes, for instance.

Any individually discernable features will need to be at least this minimum size however to be clearly visible. Note that sunlight reflecting off the spacecraft may make individual features more difficult to make out even if they nominally might be large enough, so you may need to make them even larger to be readily visible.

A spacecraft in low Earth orbit will move very differently compared to a star, so assuming that it is visible in the first place, it will be easy to tell that it is not a star. Primarily, it will be moving much faster across the observer's field of vision than any star would.

For the grand finale, borrowing from TildalWave's answer on the ISS over on Space Exploration also linked to earlier:

According to Human Photoreceptor Topography, Curcio et al., 1990 (PDF) that lists several sources as well as own measurements of the spatial density of cones and rods in whole-mounted human retinas, the greatest cone density recorded was 324,100 cones/mm2. That gives us acuity (or row-to-row spacing we need to discern at least two individual features) of 86.3 cycles/°. So for our best case, with a great eye, neglecting any atmospheric effects, the ISS right above us when it's closest, and optimal contrast with the background sky, we get minimum separation of objects of 74.83 m. If there was no air between the observer and the station!

Do note that TildalWave uses 370 km as the ISS' orbital altitude, which is less (though not significantly less) than the 400 km used in this question. Said differently, the two are in close enough agreement for our purposes: at 400 km distance, ignoring such pesky details as air, you need features to be at least about 100 meters in size in order to be visible to the naked eye (and that is an absolute minimum). Allowing for the obscuring effects of air, and quite possibly for the problem of making things out near a bright point light source such as the surface of a spacecraft in sunlight, you need to make them larger than that.

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    Albedo makes a big difference -- how bright is the object? It sounds like the above "limit of resolution" is for a non-reflective object. Artificial satellites are shiny metal and much easier to see. There are historical cases of isolated people who had never heard of satellites noticing when they appeared in the 1950s. Atmospheric conditions matter too, clear wilderness skies are much better than a city with smog and light pollution. – Royal Canadian Bandit Nov 7 '14 at 10:39
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    @RoyalCanadianBandit Obviously this isn't the only thing that matters. It does, however, establish a lower bound on the size of an object of unknown construction for that object to be detectable. If the purpose of the object is to shine a visibly detectable laser at the observer, clearly it wouldn't need to be anywhere near as big as this to be detectable. – a CVn Nov 7 '14 at 10:42
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    @RoyalCanadianBandit Actually I don't think so. There are two separate problems here. Brightness determines whether you can actually see that the object exists. Something can be the size of the moon but completely black and you only see it by occlusion of objects behind it. On the other hand something tiny shining a bright light can be seen. – Tim B Nov 7 '14 at 12:31
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    The separate problem though is whether you can actually see the shape of the object. The smallest shape you can see at that distance is 100 meters so the object would need to be absolutely huge for any real shape to be seen with the naked eye. – Tim B Nov 7 '14 at 12:32
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    What the angular size does do is define the minimum size below which any object will appear a point source. – Dan Neely Nov 7 '14 at 18:39

Taking Michael Kjorling's great answer, I'll expand it. As was noted the space station is about as small as is likely visible, a pixel as Michael said, so to have something to see you will likely a few pixels worth. A ship about about 10 x 10 pixels would be a square kilometer. This would be large enough to begin to identify what it is, not just guessing by its location and movement.

It also means the ships coming in around the Earth in ID4 would have been big enough to identify before they even entered the Earth's atmosphere.

Just looked it up. The City Destroyers in ID4 were 25km/15miles across.

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    So a Super Star Destroyer at 17 km long would be easy to identify. – Zan Lynx Nov 28 '14 at 2:33
  • @ZanLynx Depends. If it were directly approaching earth then you would be looking at it nose on, the cross section is much smaller than 17km (still easily visible at 400 km away though). Of course if the ship had to obey physics it would be facing AWAY from us and using it's engines to slow down, which would be very visible from much farther away, but Star Wars ships don't seem to need to do this :P – Jason K Jun 9 '16 at 19:08

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