I'm asking this question as a reference for use in worldbuilding when developing world size, atmospheric content, or alien optometric abilities (aka, vision).

While I'm specifically asking two questions, their relevance is so close that it's more practical to ask them once, together.


  • A sphere 10 Km in diameter
  • It's homogeneous (the material isn't relevant)
  • Having an albedo of 70% (similar to fresh snow)

How far away can that sphere be...

A) In space, with Sol at the observer's back, the observer 1 AU from Sol?

B) In space as in (A), but assuming for the purposes of this question, that space is filled with an aether in all ways equivalent to Earth's atmosphere at sea level?

And still be seen (even if its shape cannot be clearly distinguished, i.e., a "point") by the average human eye?


1. I understand that eye sensitivity to light and focus vary between people. I don't know how to specify the "average." If someone can provide metrics that would clarify this issue, please leave a comment and I'll add them into the question.

2. For the purpose of this question, please ignore the lack of ground. I understand that light reflecting from surfaces between the oberserver and the observed will contribute (substantially, I suspect) to whether or not an object can be seen at a distance. I could be wrong, but I believe the basic setup I've proposed represents a best-case scenario.

3. For the purpose of this question, please ignore the fact that the sun isn't above the observer. In other words, it's shining from behind the observer rather than above or on top of the observer. This effects the amount of Rayleigh Scattering that would also contribute to visibility. Once again, I suspect this makes the setup a best-case scenario.

4. Ignore the fact that our hapless observer is sans-spacesuit. In other words, there's nothing about his/her environment as the observer that's affecting his/her vision (no faceplate). Lucky dude, as otherwise the eyes would have a bit of trouble in the vacuum of space.

5. The atmosphere between the oberver and the observed is uniform. This is different from the conditions of a planet (convex) or the inner surface of a Dyson sphere or ringworld (concave) where the atmosphere density is not uniform along the sight path.

6. Please ignore the field of stars (assume they aren't there). This isn't a test of how well a human can identify one tiny mote from a field of tiny, shining motes.

A curious thought...

Please note that there may be a considerable difference between the sun behind the observer and the sun above and midway between the observer and the observed as the reflection off the sphere may (may...) be greater in the later case. However, it would only be greater on the top half of the sphere (closest to the sun) while it would be lower on the bottom half (and all the Rayleigh scattering issues come into play... and the sun's in your eyes...). At this time, I'm thinking that placing the light source behind the observer produces the highest reflectivity and lowest optical distortion for the greatest distance. If the math says I'm wrong, please let me know.

  • $\begingroup$ B breaks my brain a little. The diffusion/diffraction of light if the entirety of space in that region was identical to Earth's atmosphere would be incredible. $\endgroup$ Commented Nov 11, 2018 at 17:37
  • $\begingroup$ From a sufficient distance, the myriad of stars is going to make it pretty hard to tell the difference between that sphere and stars. $\endgroup$
    – RonJohn
    Commented Nov 11, 2018 at 17:46
  • $\begingroup$ I don't know the answer, but this will point you in the right direction: livescience.com/33895-human-eye.html $\endgroup$
    – RonJohn
    Commented Nov 11, 2018 at 17:48
  • $\begingroup$ @JesseWilliams, but that's the point, isn't it? There's a limit to how far any human can see anything due to reflected light. Whether it's the resolution of the eye or the diffusion of light. Standing on a planet means the curvature of the planet is getting in the way of line-of-sight. I'm specifically removing that. $\endgroup$
    – JBH
    Commented Nov 11, 2018 at 18:37
  • 1
    $\begingroup$ @JesseWilliams, good point for clarifying, though. (B) is not a vaccuum. It's not even gravity-less, from that perspective. It's a contrived condition for a point of reference. Thanks for your interest. $\endgroup$
    – JBH
    Commented Nov 12, 2018 at 15:37

2 Answers 2


Calculating apparent magnitude

Your alignment of the sun, the observer (in space), and the object is as so

           1 AU                    r AU
(Sun) ------------- (obsv) ---------//------------(obj)

Given that, there are three separate calculations. First is the magnitude of reflected incident energy from the sun. The second is the magnitude of reflected light, given the difference in visible disk of the object, compared to the sun. The third is the brightness of this light to the observer.

An issue here is that luminosity is a full spectrum measure of energy output, but we are only interested in energy from the visible spectrum. Since I can't find a visual spectrum only luminosity for the sun to do an energy calculation, we can use the absolute magnitude of the moon (+0.25) as our starting point.

We must adjust the magnitude of the moon by four factors; as each of these factors increases, the object will be relatively dimmer:

  • Squared ratio of distance from sun to object to distance from sun to moon
  • Ratio of surface area of the moon to surface area of the object
  • Albedo of moon to albedo of object
  • Squared ratio of distance of object to Earth to distance of moon to Earth

A difference of $n$ visual magnitudes is equal to a difference of a factor of $100^{n/5}$ in luminosity. Thus, the logarithm base $100^{1/5}=2.512$ of these ratios is summed. We will use $\log$ to represent $\log_{2.512} = 2.5\log_{10}$ for simplicity.

The distances are straightforward, and are calculated in terms of AU; the moon is roughly 0.00257 AU from Earth. The surface areas of the moon and object are proportional to the radii squared. The object has a radius of 5 km, while the moon is 1738 km. The albedo of the moon is a paltry 0.12, while the object is an incandescent 0.70.

We get the following equation for apparent visual magnitude of the object:

$$0.25 + \log\left(\frac{(r+1\text{ AU})^2}{(1\text{ AU})^2}\right)+\log\left(\frac{1738^2\text{ m}^2}{5^2\text{ m}^2}\right)+\log\left(\frac{0.12}{0.70}\right)+\log\left(\frac{(r \text{ AU})^2}{(0.00257\text{ AU})^2}\right) $$

How far away can we see this object?

In a big city at night you might only be able to see a magnitude 3 object, while on a moonless night in the middle of the ocean you could see a magnitude 8 object. Lets say that you can see a magnitude 6 object. Attenuation by the atmospheric mass directly overhead is about 0.145 magnitudes; that means that if we were in space, we could see a magnitude 6.145 object.

I take the above equation and, for values of r in AU, plot the apparent magnitude of the object on log scale.

enter image description here

The code is here:

>>> import numpy as np
>>> func = lambda r : 0.25 + 2.5*np.log10((r+1)**2)+2.5*np.log10(1738**2/5**2)+2.5*np.log10(0.12/0.70)+2.5*np.log10((r/0.00257)**2)
>>> x = [x/2-10 for x in range(22)]
>>> y = [func(np.exp(i)) for i in x]
>>> plt.plot(x, y)
[<matplotlib.lines.Line2D object at 0x7fdb949d7eb8>]
>>> plt.xlabel("Distance from Earth in exp(x) AU")
Text(0.5,0,'Distance from Earth in exp(x) AU')
>>> plt.ylabel("Apparent visual magnitude")
Text(0,0.5,'Apparent visual magnitude')
>>> plt.show()

The solution for apparent magnitude 6.15 is 0.00027 AU. This is 1/100 the distance from the Earth to the moon, 41,000 km; or, roughly, geostationary orbit.

  • $\begingroup$ @JBH This took like 4 hours, so I hope you say thank you! $\endgroup$
    – kingledion
    Commented Nov 14, 2018 at 3:07
  • $\begingroup$ Um, this does NOT seem correct. a 10km white sphere fully illuminated at geostationary should be about 16 times as bright as Venus at its brightest. $\endgroup$
    – PcMan
    Commented Feb 20, 2021 at 14:31
  • $\begingroup$ I get a rough calculation of 10.5 million km (about 27 times moon distance) for the sphere to hit magnitude 6.15 $\endgroup$
    – PcMan
    Commented Feb 20, 2021 at 14:40
  • $\begingroup$ @PcMan Its been some years since I did this, but throw another answer up and we can compare notes. $\endgroup$
    – kingledion
    Commented Feb 22, 2021 at 14:28

In aviation, part of the daily forecast is the estimation of unobstructed visibility. Obviously at a reasonable altitude, the curvature of the earth is no longer a factor, so physical obstructions, including fog and clouds are generally the obstructions. 10 statute miles is the maximum value for a clear day. I have never seen a value greater than this. It would be a good starting point.

  • $\begingroup$ That is a good starting point, but the curvature of the earth is still in play as it causes a non-uniform atmosphere. In other words, the air may be thin near the plane, then thickens as you look across the area where line-of-sight is closest to the ground, then thins again as you approach the target. Note, though, that people can see things on the ocean horizon, so 10 miles is intrinsically short (remember the size of the sphere). Aviation rules have purposes that compromise this test (e.g., not hitting another plane). $\endgroup$
    – JBH
    Commented Nov 11, 2018 at 18:42
  • 1
    $\begingroup$ True. But to really answer this question you need a definition of what seeing actually is. Is it to distinguish a distant but contrasting and virtually infinite line. Or distinguish a small body and possibly distinct features. $\endgroup$
    – mreff555
    Commented Nov 11, 2018 at 19:54
  • $\begingroup$ Yeah, as I said in my question, that's one of the limitations I'm not sure about. What if we bring the FAA's definitions into your answer? How does the FAA define "unobstructed visibility" and where does it derive that 10 mile reference? That would improve the answer. Cheers! $\endgroup$
    – JBH
    Commented Nov 11, 2018 at 23:40
  • 1
    $\begingroup$ Metars and TAF's data are standardized by the ICAO but It would appear that visibility data is empirical. if you are looking for a mathematical solution it would probably be better to compare the eye to a camera. Despite the fact that many professional photographers warn against this, the human eye roughly resembles the behavior of a 50mm lens (this is somewhat contested). This is a link I found that explains some of the theory behind this. petapixel.com/2012/11/17/the-camera-versus-the-human-eye $\endgroup$
    – mreff555
    Commented Nov 12, 2018 at 2:25
  • $\begingroup$ @mreff555, I'm surprised you've never seen visibility over 10 sm. I just looked up an airport where I know the weather is clear and their METAR gives 15 miles. Vancouver BC is currently reporting 20 miles. $\endgroup$ Commented Nov 14, 2018 at 5:11

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