Imagine someone putting a battleship over a country, always in sight, in order to threaten them to not try anything funny. Imagine tech that allows to suspend the ship is not a problem.

How high is it possible to put a starship in the atmosphere in order for it to be visible clearly on sunny day? As a dark figure cast against clear sky.

What's the radius in which it will be visible, on the ground, assuming Earth is the planet?

What size should the ship be in order to be the size of moon from the surface?

UPD: To clarify. It's not the ship's size that is a problem. It's the atmosphere. It scatters light from the sun, otherwise the skies will be dark. But, this means if we place ship outside atmosphere, it will probably not be visible unless directly between the sun and observer.

Let's agree that the ship does not shine (is dark) and is visible because it blocks or absorbs light.

  • $\begingroup$ I think you don't need the new "containment"-tag that you added. I think the "control"-tag is good enough. $\endgroup$
    – Secespitus
    Mar 13, 2017 at 10:25
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    $\begingroup$ The answers to all of these questions are entirely dependent on how big the ship is. The biggest ship I know of in science fiction is 90 million miles long ("Golden the Ship Was—Oh! Oh! Oh!" by Cordwainer Smith) and would be clearly visible while it was still well outside the solar system. How big is your ship? $\endgroup$
    – Mike Scott
    Mar 13, 2017 at 10:25
  • $\begingroup$ Your question can be improved by telling what the spaceship's shape is and what country it is hovering above. A sphere hovering above China will have a different set of parameters from a cylinder above Lichtenstein. It's reasonable to assume a spaceship hovering in the atmosphere will be below the Karman Limit of 100 kilometres. Actually solutions might be easier to find if it hovers outside the atmosphere. But we shall see. $\endgroup$
    – a4android
    Mar 13, 2017 at 10:36
  • $\begingroup$ @MikeScott we can't see Jupiter it daylight, can we? $\endgroup$
    – alamar
    Mar 13, 2017 at 17:58
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    $\begingroup$ Perhaps interesting, looming is a psychological term describing a particular kind of movement. Obviously it has other non-scientific meanings as well, but I found the psychological term to be interesting. For instance, dragonflys approaching prey will do so along a vector which causes them to appear to loom. $\endgroup$
    – Cort Ammon
    Mar 13, 2017 at 18:09

4 Answers 4


The answer to this question could vary depending on the size of the country in question. Due to the curvature of the earth I would expect the following to be in reverse.

Height (in Feet) / distance (in miles) the horizon is from you. (I have approximated)

  • 10 / 3.9
  • 100 / 12.3
  • 1,000 / 39
  • 10,000 / 123
  • 100,000 / 390
  • 1,000,000 / 1,230

For the UK, from top to bottom, approx. 450miles. For an object to be seen from the southern coast to the north coast, the object could be under 100,000 ft, however somewhere like China, approx. 2000-3000 miles North to South, the object would need to be higher than 1,000,000 ft in the air to be seen across the whole country.

If you disregard the China example and state that for larger countries you would either place them over more populated areas or cover with multiple objects, we can continue with the UK example.

You could possibly see an aeroplane from around 32 miles away as a tiny dot, (more likely to see the contrails from them,) with the plane length to be 150 ft, to have something that looked the same size from the extremes of the country, the object would be required to be 10 times larger. 1,500 ft.

1,500 ft object would not dominate the skies across the whole country. Now to make a large enough black smudge across the UK skies, 150,000 ft sized object would be noticeable.

[Edit] Didn't notice that you would require moon sized, but with my above example, the object would be greater than moon size when directly under, but a black smudge to the extremes.


The Moon has an angular size of about 0.5 degrees (29'56''), which using the relationship $delta = 2arctan(d/2D)$

enter image description here

gives you a way to estimate the distance and dimension of the object, once you fix its angular size.

For an object distant 100 meters it gives you that it has to be 0.87 meters wide to be seen as big as the full Moon. If the distance is 100 km, that the size has to be about 870 meters.

Long story short: choose one among distance or size, figure out the other one.

Regarding visibility, it's not the color in itself to hamper it, but the contrast between the object color and the background. Think of sharks: they are dark in the upper part of their body, and more light colored in the bottom part, because they aim to be unseen when observed from above (against a dark background) and from below (against a light background).

So if your ship is dark against the bright blue sky it can be visible. It may be more difficult to be seen at night with a dark sky, but still it will be made visible by the amount of stars it eclipse, looking literally like a pitch black patch in the starry sky, unless it has lights to mimic perfectly the background stars.

  • $\begingroup$ See update to the story. Unfortunately this addresses just half of the question. $\endgroup$
    – alamar
    Mar 13, 2017 at 15:01
  • $\begingroup$ I don't get it. The Moon is out of the atmosphere and it is still visible. $\endgroup$
    – L.Dutch
    Mar 13, 2017 at 17:57
  • $\begingroup$ @alamar, check my edited answer $\endgroup$
    – L.Dutch
    Mar 13, 2017 at 18:04
  • $\begingroup$ @LDutch you can't make ship visible as dark against the bright sky if it's behind that sky (i.e. atmosphere). So the question is, how high you can go in the atmosphere while still being "in front" of it. I imagine it's a couple km, which will naturally limit visibility radius. $\endgroup$
    – alamar
    Mar 13, 2017 at 18:12
  • $\begingroup$ You might want to consider adding a slight mention on reflectivity also. I.e. a shiny underside would be more visible at night — saves the need to produce its own light, — but would make it glare during the day, and thus the viewers would tend not to look towards it. Unless, of course, it had shutters to cover the reflective surfaces, or rotated them away, or something. $\endgroup$ Mar 17, 2017 at 4:42

There is a very long mathematic/physics answer to this. But the easiest would be to first choose the country you want to scare. If you want to say, dominate Belgium you would need something in the size of a Pyramid. So it can be visible from every part of the country. But if you would like to make impression on Russia... You would need to take under consideration not only the latitude but also longitude.
So the spaceship should be seen from Pskov and Petropavlovsk, that's almost 130 degrees in longitude. We know that moon is visible from both of these places so you could cut in half both it's dimensions and distance from earth.

But now you have created a thing that would be better as a tidal control than as a watchful omnipresence.

Also if the object would be black and absorb the light it would only by visible against the sun. Which would require to calculate gravitational lensing into he object size to not appear smaller.


Or you can have a smaller object with a magnified projection. This projection will work on the free ions of the atmosphere. If intimidating a country is actually the designated purpose, this will do so.

  • $\begingroup$ The idea is to avoid cheating, because cheating is cheap and might inspire false hopes in the people down below. Frankly I just want to know the precise stats of the vessel and the effective intimidation range. $\endgroup$
    – alamar
    Mar 17, 2017 at 18:37

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