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I have a sailing ship. It leaves port. As it goes away, it grows smaller and smaller. In our world, it eventually is hidden by the curvature of the Earth, such that the top mast stays visible longer. I want it to simply grow smaller and smaller and smaller, but (from the perspective of a human with a human lifespan of 100 years or so) "never" makes it past the horizon.

How big does my world have to be for that to happen?

Ignore the real-world problems such as normal planets that size turning into hydrogen-filled Jupiters, stars or black holes aspect of it. I'm going for a hollow Dyson world built (out of unobtanium of course, with tens of miles of rock, water and stuff piled on top) around a sufficiently massive object (say a star or a black hole) to give 1g gravitational acceleration at the (external facing) surface.

Edit: To preempt the question, yes, advanced optics are available, so the visual acuity of the human observer is not the issue here: as long it it can be reasonably detected and distinguished from the sea background using light, you can assume that it will. So no magic detectors, but imagine a high-performing scope, well built, but subject to the problems real optic instruments have -- air attenuation was pointed out in comments below. I'll assume the sails to be white, if that helps.

Later Edit: Since air attenuation might be a factor as pointed out in the comments, let's assume our ship is much like Santísima Trinidad: Galleon; Length: 51 m; 2,000 tons; Comp.: 400-800; Armament: 54 guns;

Serban Tanasa
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  • Great question! I'm in a hurry so I can't post an answer but I have some ideas. – fi12 Mar 10 '16 at 14:04
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    What kind of advanced optics here. We have optics that can see stars in other galaxies. – bowlturner Mar 10 '16 at 14:13
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    @bowlturner, we do? Maybe the occasional supernova, but resolving actual stars into disks might be a reach even for Andromeda... Heck, even for our own galaxy. – Serban Tanasa Mar 10 '16 at 14:19
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    If the limit is "if it can be detected, it will", then your world must be an infinite, flat plane for an object to never pass beyond the horizon. – Frostfyre Mar 10 '16 at 14:20
  • @Frostfyre, see small edit, lemme know if you think that satisfices. – Serban Tanasa Mar 10 '16 at 14:21
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    here is a page that will calculate horizon using distance above the sphere and the diameter of the sphere. http://www.neoprogrammics.com/spheres/distance_to_horizon.php – bowlturner Mar 10 '16 at 14:22
  • ...So, if I paint the ship varying shades of blue so that it matches the sea and sky and put it in the harbor, does that count as no longer being distinguished from the sea background using light? – Frostfyre Mar 10 '16 at 14:23
  • @Frostfyre, sounds like a Romulan cloaking device. – Serban Tanasa Mar 10 '16 at 14:24
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    @Frostfyre Light interacts with the atmosphere. If the distance is big enough the ship will not be detectable. – Taemyr Mar 10 '16 at 14:26
  • Based on the comments to Mike L.'s answer, it seems that the OP wants this answered in terms of atmospheric attenuation. @SerbanTanasa if that's correct, could you please edit your question to clarify? – CAgrippa Mar 10 '16 at 14:31
  • @Taemyr We routinely correct for atmospheric interference with current satellites and telescopes, though we haven't tried looking through that much atmosphere. I don't know what the limit would be for correction. – Frostfyre Mar 10 '16 at 14:32
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    If it's a hollow dyson world or ringworld, surely the ship should be on the inside and therefore always visible. The sea will curve upwards and there will be no horizon. – pjc50 Mar 10 '16 at 15:54
  • @pjc50, the assumption is I'm using the external surface – Serban Tanasa Mar 10 '16 at 15:54
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    is there any way to introduce curvature of light into this? i.e. so that the sphere is actually much smaller, and the ship is over the horizon to a straight line, but you can still see it because the light from the ship curves around the earth too? – Corley Brigman Mar 10 '16 at 16:58
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    You have a big problem in the form of air transparency: https://en.wikipedia.org/wiki/Visibility. Air is opaque over 300 km. – 2501 Mar 10 '16 at 17:54
  • @2501 the article seems to be about unaided vision – Serban Tanasa Mar 10 '16 at 18:04
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    @SerbanTanasa Contrasts decreases exponentially with the distance of the object. No optics is going to help you. At a certain distance air is going to be like a brick wall to light. – 2501 Mar 10 '16 at 18:11
  • @2501 I'm fine with that as well, care to develop that into an answer? – Serban Tanasa Mar 10 '16 at 18:12
  • My understanding of the question (and the 2 already posted answers) is that this is purely a mathematics question, – Vincent Mar 10 '16 at 18:49
  • @CorleyBrigman That's exactly what I was thinking. Curvature due to the Earth's atmosphere let's you see past the actual horizon a ways on Earth. Maybe there'd be some way to match the curve due to the air to the curve of the planet's surface. – DCShannon Mar 10 '16 at 19:12
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    @Vincent By that logic, every question asking for help in calculating the position of a habitable planet in a solar system would be off-topic. What makes calculating the size of a planet/structure off-topic, as opposed to its position? – Frostfyre Mar 10 '16 at 19:16
  • Couldn't we just make the inhabitants and sailing ships of this hypothetical planet much smaller? or would the sailing ships be able to reach escape velocity and leave the planet? – emory Mar 11 '16 at 00:34
  • Even on disc world, the ship will eventually sail over the edge of the disc. So I guess you would have to live on an infinite plane. – Steven Alexis Gregory Mar 11 '16 at 07:33
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    Re interior or exterior surface... If it's exterior, it's gonna be awfully dark. – user2338816 Mar 11 '16 at 08:57
  • @pjc50 with a ringword the rotation simulates gravity so you can live on the inner surface, but a sphere can only be rotated about a single axis, so you only get that simulated gravity around the equator. The further you are from the equator, the less the simulated gravity, until you reach a point at which the gravity of the central star is greater, and you fall off the surface into the star. So you can live inside a ring or a cylinder, but a sphere only makes sense for living on the outside. – trichoplax is on Codidact now Mar 11 '16 at 11:53
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    Dies your Sea have waves? – PlasmaHH Mar 11 '16 at 14:50
  • @PlasmaHH presumably, it is a sea – Serban Tanasa Mar 11 '16 at 19:04
  • Then you sooner or later run into the problem that waves in the front appear to be bigger than the far away ship – PlasmaHH Mar 11 '16 at 19:59
  • @PlasmaHH If the observer is well above the waves, which isn't that hard (Lifeguard tower) then close waves will never block the horizon. (Unless Tsunami...) – Muuski Mar 11 '16 at 20:40
  • I guess I don't understand, because... physics. If an object is travelling away from a fixed point on a curved world, that object will eventually disappear from sight as it passes the horizon. There's no way around this - it's just the way reality works. – Omegacron Mar 11 '16 at 21:19
  • @Omegacron You could have the world infinitely large thus effectively making any direction appear to go on infinitely in a straight line. – Simply Beautiful Art Mar 11 '16 at 23:57
  • If the ship is on the outside of a Dyson sphere, what is the source of light that makes it visible at long distances? – David K Mar 12 '16 at 23:33
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    @CorleyBrigman Theoretically this is absolutely doable, you just need the planet to have enough gravity so that escape velocity at the eyeline is the speed of light. Such bodies are called black holes. – Todd Wilcox Mar 13 '16 at 04:05
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    @Omegacron he didn't say that the ship never disappears, only that the time it takes is approximately a lifetime. – JDługosz Mar 13 '16 at 22:23

13 Answers13

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Sailing ship are now much faster than they used to be. So it's hard to give a definite answer. Nevertheless, if we consider 18th Century ships, according to that blog post, ships travelled at roughly $8$ mph (about $13$ km/h).

We consider the 18th Century, because before people were not so much travelling to the other side of the world, and after engines started to be more dominant.

Now, let's imagine the worst case scenario. A ship sailing in a single direction, and a human, who does nothing else than checking that ship every day for 100 years.

Talk about an obsession.

According to Wikipedia, the distance $d$ from a height $h$ to the horizon line of a planet of radius R would be

$$d = \sqrt{h(2R+h)}$$

In the extreme case when the ship reaches just the horizon line after those 100 years, the ship would have travelled

$$s = R\tan^{-1}\frac{d}{R} $$

with $s$ the distance on the planet (whereas $d$ is the distance from the height $h$). But taking $h=2m$, we can approximate that $s\approx d$. Thus,

$$R=\frac{1}{2}(\frac{d^2}{h}-h)$$

100 years at $13$ km/h leads to $d\approx11,395,566$ km. This leads to a radius

$$R\approx 32.5 \times 10^{15}\mbox{ km}$$

That's... a lot.

As comparison, the Sun has a radius of around $696,342$ km. So that's almost $5\times10^{10}$ times the size of the sun. And the largest known star would still be at a factor $3\times10^{7}$. So 30 million times more than the largest known star.

To get to larger scales, that's about 3,400 light-years (more than $800$ times the distance to Proxima Centauri). $7.2\times10^6$ times the maximum distance of Neptune to the sun. And that's 4 to 7 % of the Milky Way's radius.

Update

As the OP added some information about it, and there were some discussions in the comments. I think I should add some details about the optics. Can our human still see the ship after all that time? Well not unless they have magic eyes.

Now, Mike's answer presents some details about optics and I don't want to be repeating it. However, in a simple case without atmospheric effects (diffusion, attenuation, etc.), we can see that the angular resolution of any instrument is given by $\alpha_{min}$. Since it is rather small, we can approximate that $tan(\alpha)\approx\alpha$. For a ship of length $L$ at a distance $d$, we have

$$d=\frac{L}{\alpha}$$

This means that for a given instrument, $d_{max}=\frac{L}{\alpha_{min}}$ is the maximum distance where that instrument can still distinguish the ship. And thus $t_\mbox{inst}=\frac{L}{v\alpha_{min}}$ is the time for the ship to reach that distance.

So they could follow the ship for (see telescopes angular resolution)

  • 13 hours with the naked eye,
  • 16 days, switching to a 4 inches telescope,
  • 163 days with a 36 inches telescope,
  • 1 year and 3 months with Hubble,
  • almost 45 years with PIONIER (100m).

How that observer came around such telescopes is beyond me.

Community
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clem steredenn
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    Comments are not for extended discussion; this conversation has been moved to chat. – Monica Cellio Mar 11 '16 at 05:11
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    @MonicaCellio I'd love to see the discussion inline -- often they are as interesting as the question to me. – Peter - Reinstate Monica Mar 11 '16 at 09:26
  • @PeterA.Schneider you can see, and continue, the discussion by clicking the link in my comment. Enjoy! – Monica Cellio Mar 11 '16 at 15:22
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    @Serban Tanasa what would be the point of embarking on a sailing journey that will take more than 100 years? A generational starship might be half plausible, but how could a 18 century sailing ship, and its crew, (and the next few generations that would have to arise in that ship) survive during the 100 year journey? – X-27 is done with the network Mar 11 '16 at 17:07
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    You are only taking the height of the observer into account, not the height of the ship. Even if part of the ship is behind the horizon, the deck is above sea level and may still be visible. On Earth this effect causes a lot of ships to appear to be exactly at the horizon even if they are several km in front of or behind the horizon. – kasperd Mar 12 '16 at 22:42
  • @kasperd I indeed only considered the height of the observer. When the ship is past it, it will still be visible, but partially. So you'd lose the illusion of flat planet. – clem steredenn Mar 12 '16 at 23:45
  • This particular answer relies on the speed staying roughly the same as on the Earth. I would expect such a place would have either drastically higher winds or drastically lower (i.e. basically standstill) winds. – corsiKa Mar 13 '16 at 16:53
  • While there might be many climatic differences there would also probably be technical differences. And it would be unlikely to sail towards such a distant destination. We have to start somewhere. And speed differences wouldn't change so much the answer. – clem steredenn Mar 13 '16 at 18:03
42

Far more plausibly than being trillions of solar masses, you can make your planet really small!

If it's small enough relative to the ship, the observer can see the ends poking out even if it's on the far side of the planet, like so:

Boat sitting on a very small planet

Of course you need an extremely dense material to make your tiny planet from, else the atmosphere would escape and the oceans boil off. That would make sailing impossible.

FLHerne
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    Why make the planet small, when you can make the boat really big? – SE - stop firing the good guys Mar 11 '16 at 12:55
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    @Hohmannfan At what point does the definition of the system become a port on a sphere that is sitting atop an irregularly shaped planet? ;) – Casey Kuball Mar 11 '16 at 19:44
  • @Hohmannfan The larger the planet, the taller the observer has to be. – Muuski Mar 11 '16 at 20:46
  • @Muuski I don't think the edit defining the ship's size existed when Hohmannfan wrote that. – FLHerne Mar 11 '16 at 21:02
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    If my doodling is correct, the ship length needs to be (4r + 2h)/tan(acos(r/(r+h))), where r is the planet's radius and h is the eye height of the person. Assuming a really tall (2m to eyes) person on Earth, the ship only needs to be 32 million kilometers long. That means it extends about 5x beyond the moon's orbit on each side. – FLHerne Mar 11 '16 at 21:15
  • Is that all? 5x beyond the moon's orbit on each side seems totally reasonable..... – Erik Mar 11 '16 at 23:19
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    Thanks for the ship. Can I also have a sheep, to eat the baobabs? – Serban Tanasa Mar 12 '16 at 11:53
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    @Darthfett To dock at a different port, put some wheels on the bottom of the ship to spin the planet around... – corsiKa Mar 13 '16 at 16:54
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The limiting factor to how far you can see an object with the naked eye is your angular resolution.

According to wikipedia, the angular resolution of a naked eye is 1 MOA, or 0.0003 radians (which corresponds well to what we were taught about the HVS, so I'll trust the wiki on this one).

Let's say a typical sailing ship is 50m long. Through some basic trigonometry, we find that the distance at which it has the apparent size of 0.0003 radians is some 83.3km. Beyond this distance, you will not be able to make the ship out even as a dot even on a perfectly clear day.

Using the geometrical model, we can find the minimum diameter of the planet so that the distance to horizon is at least this.

Taking the height of the observer $h=2m$ and requisite distance to horizon $d=83.3km$ and plugging them into the following formula:

$d^2 = h(D+h)$

we express the diameter D of the planet:

$D= \frac{d^2}{h} - h$

which finally gives us a value of $D=3,472,222.4km$, or something over 11 light seconds. That's quite the planet you have there.

Frostfyre
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Mike L.
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    Thanks for the answer, but I made a brief edit about 15 minutes before you answered - perhaps you can briefly revise? – Serban Tanasa Mar 10 '16 at 14:20
  • Note that the given 3,4 million kilometers is already three times the diameter of the sun. – Erik Mar 10 '16 at 14:22
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    @fi12 Bilbo's answer measures how far a sailing ship will sail in 100 years. Mine is about how far you could still plausibly see it. – Mike L. Mar 10 '16 at 14:26
  • @SerbanTanasa Well, that's a little bit unfair. The next biggest limiting factor is atmospheric attenuation, but if you posit that a ship is always detectable, then it will forever be detectable and the question can not be answered. – Mike L. Mar 10 '16 at 14:27
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    @SerbanTanasa - Mike's answer still stands, really. If optics are sufficiently advanced that any curvature can be detected, then (as Frostfyre noted) the "planet" has to be a flat plane. If on the other hand there is a limit to observational ability based on something other than fog or other obstacles, it can be calculated by using the formulae Mike proposed. Just change the numbers to fit, do the math, and you'll have your answer. If on the other hand there is some other kind of interference, some other sort of reason why observation would not detect the curvature, you need to explain that. – CAgrippa Mar 10 '16 at 14:29
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    Nope, I'm perfectly willing to have air-caused attenuation as a factor – Serban Tanasa Mar 10 '16 at 14:29
  • To be more exact, there has to be a limit to when the ship is "detectable by light". This can be measured by spatial resolution or incident radiance, but the latter will always be non-zero, so your perfect detector will always see something that is not obscured. – Mike L. Mar 10 '16 at 14:29
  • This is it. Even with advanced optics .. if you're pointing them "parallel" over a lot of water ... with winds, different temperatures, increased humidity ... they will be almost useless. Perhaps go with 0.000001 radians angular resolution to account for these super optics, but that should do it. – Daniel Jour Mar 10 '16 at 18:22
  • Might be worth noting that ships are generally long and narrow, so if it's sailing away from you you'd see the narrower beam width - say 5-10m. This brings the visible distance limit down quite a bit. – Andrew is gone Mar 10 '16 at 21:18
  • @PeterA.Schneider deleted mine too (I'll delete this before long so we don't start an infinite loop...). – trichoplax is on Codidact now Mar 11 '16 at 13:17
  • Good news: you don't need to recalculate for when the observer has an observatory. According to WIkipedia, 100km is quite good visibility - the sort of thing you get on a clear Arctic day. So the numbers above estimating 80km of visibility is about right. ("visibility" is "how far you can see a 1000 candela light against an unlit background" - a fairly generous definition.) https://en.wikipedia.org/wiki/Visibility – PotatoEngineer Mar 12 '16 at 00:41
  • @PaulMarshall That's probably a good ballpark. Of course, the question is whether I want to subscribe to the same definition of visibility and what the air really is like. I have tried and failed to get some solid numbers of atmospheric optical depth, but that would have been the way to go. – Mike L. Mar 13 '16 at 09:12
  • @Mike L. Slight technical point, if the planet was a quarter the size, then the bottom 2m of ship would be hidden by horizon at the distance the 50m ship was a barely visible speck. If the 50m ship is all just below the horizon at this point, the planet is 96 thousand km in diameter, ie 7.6x earth size (so 434x more volume) Your calculations are for nothing below the horizon. Nothing visibly below the horizon allows a much smaller planet. – Donald Hobson Mar 01 '22 at 00:18
12

This might count as cheating, but...

Don't make the world a planet. The reason we see a horizon is because of the curvature of Earth. If you have negative curvature, there is no horizon.

One example is the classical (and wrong :)) hypothesis of Hollow Earth - that we're actually living on the inside of the planet, so the curvature is inverted. As the ship sails away, it goes up rather than down. Of course, make sure you don't care too much about the problems of having a hollow planet in the first place - a good checklist would be real-life objections to the Hollow Earth hypothesis :)

For artificial objects, this is a bit more realistic. You can have a ringworld, which gives you plenty of sunlight, while having no horizon. Of course, it's just a strip of a world, so boundaries will be rather obvious - still, to the inhabitants, it would simply appear as "this is our world, and on the sides are the boundaries".

There's other topologies you could play with, just make sure the curvature of the space is negative (for example, toroids give interesting topology, but they will still have a horizon).

One example of note is Hal Clement's Heavy Planet. The story is set on a planet that's absurdly massive, with very high gravity (something like 300-600g). However, the planet is also rotating absurdly fast, so at the equator, the effective gravity is only about 3-6g, low enough for the human explorers to survive (with support, of course). The natives of the planet actually think they're living on the inside of a bowl because of the varying effective gravity - air tends to pool around the poles, so looking from the equator, you see the "horizon" going up - in effect, you're looking "behind the corner" due to the massive light "bending". There are places where the planet looks like the ellipsoid it is, but those are off limits even to the natives - they can't take the gravity at the poles. Of course, if you choose a similar setting, you have to get rid of humanoids (good luck building a humanoid that can survive under 300g's) and you better understand the physics involved perfectly - they have many interesting implications :)

Luaan
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  • Very creative answer. I like this idea a lot! – ApproachingDarknessFish Mar 12 '16 at 06:06
  • Your explanation of Hal Clement's planet is not quite correct. The reason for the light bending is the high gradient of atmospheric density as a function of altitude; you'd get the same effect on a perfectly spherical world with the same high gravity. (Note that the gravitational gradient is not significant on the scale a Mesklinite can see.) And the Mesklinites can travel to the pole -- such a voyage is actually the subject of the novel that introduced the planet. – ruakh Mar 12 '16 at 18:00
  • They do reach the low-gravity poles, and that's where the story begins (and ends on the opposite pole). – JDługosz Mar 12 '16 at 21:15
  • @ruakh It's been a while since I read that book the last time :) You're right, they did travel to the pole, but IIRC, it was entirely alien to them - suggesting they never went there before (though I do remember others lived close to the pole... oh well). Are you sure the illusion would occur on a non-rotating planet? I'm not really equipped to even guess at it right now :D Was it explicitly explained in the book? I don't remember. – Luaan Mar 12 '16 at 22:55
  • @JDługosz I think you're a bit confused (it's an old book :)). They started on the equator (where the "gravity" was weak enough to allow the humans to survive and make contact with the natives). But you're right they went to the pole (to retrieve the scientific data) - however, the poles are definitely the highest "gravity" part of the planet, not the lowest. That's why that science machine was landed there in the first place, and why the humans couldn't retrieve it on their own. – Luaan Mar 12 '16 at 22:57
  • @Luaan: They had never traveled to that hemisphere before, let alone its pole; but there's nothing in the book to suggest that the gravity there is too high for them (though they do face other challenges, such as extremely strong winds). As for the light bending: one of the meteorologists explains it thus: "You know how the layer of hot air just above a road on a sunny day bends sky light back upward at a slight angle, since the hot air is less dense and light travels faster in it; you see the sky reflection and tend to interpret it as water. You get more extensive mirages [continued] – ruakh Mar 12 '16 at 23:38
  • [continued] sometimes even on Earth, but they're all based on the same thing—a 'lens' or 'prism' of colder or hotter air refracts the light. It's the same here, except the gravity is responsible; even hydrogen decreases rapidly in density as you go up from Mesklin's surface. The low temperature helps, of course" [emphasis mine]. And Clement qua writer confirms in his paper "Whirligig World" that this is his reasoning (there he attributes the described effect to "the tremendously rapid decrease of air density with height"). – ruakh Mar 12 '16 at 23:42
  • Ok, the poles were quite conducive to the natives. The equator was impassible by boat as it was a highland. – JDługosz Mar 13 '16 at 01:31
  • Unfortunately for Clement's Heavy Planet, at some point tidal forces become a big problem, and spinning the planet to cancel out the gravity actually makes the tidal forces worse (a person's head would be traveling in a larger circle than their feet). – Todd Wilcox Mar 13 '16 at 04:41
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    @ToddWilcox: If I'm understanding the concepts properly, and if I've done my math right, then the tidal forces on a native Mesklinite (=a centipede-like creature only a few inches "tall") at Mesklin's poles are only about three times those on a human on Earth; the gravitational tidal forces on even a human at Mesklin's equator are actually less than those on a human on Earth; and I think rotational tidal forces at its equator should depend only on its angular velocity, which is high for a planet but well below what a human could notice. – ruakh Mar 13 '16 at 19:03
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I haven't checked @bilbo_pingouin's calculations but it sounds plausible to me. As sailing ships today pass over the horizon in a few hours, for the horizon to be a lifetime away would require a huge planet. So let's eliminate the idea of a planet so large that it takes a sailing ship decades to reach the horizon.

Other than the curvature of the Earth, what prevents you from seeing a far away object?

  1. It looks so small that you cannot distinguish it. I'm not sure when we'd reach that point for a ship-sized object with the naked eye, but you say that advanced optics are available, so at that point it's going to be very far. NASA used ground-based telescopes to spot defective tiles on the bottom of the Space Shuttle when it was in orbit. We'd need a very big planet for a ship to not be visible with the most powerful telescopes.

  2. Atmospheric haze so distorts the image that it becomes indistinguishable from the background. On a foggy day, this could be as little as dozens of feet. According to this article: https://en.wikipedia.org/wiki/Visibility the maximum visibility given the Earth's actual atmosphere is about 300 km. I'd think it's your best bet. Still, for 300km to be less than the distance to the horizon, by my calculations that still makes for a very big planet. Assuming your eye level is 6 feet above the surface, i.e. you're standing on the beach, for the horizon to be 300km away, I calculate a radius of the planet of over 15 million miles. (r~=s^2/2h, where r=planet radius, s=distance to horizon, and h=height of eyes above surface)

Jay
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It depends also on the gradient of the refractive index of the atmosphere. If light from the ship is refracted sufficiently and attenuation is low enough, light could continue the whole way around a planet. Even on earth, distant hills can look higher in some weather conditions. Mirages are another associated phenomenon.

Steve
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If you are happy for it to go "out of sight" without "going over the horizon", then the answer is suddenly a lot simpler - the horizon needs to be sufficiently far away that the it fades out (atmospheric effects) before the horizon comes into effect.

This gives you a lot of flexibility - Singapore, on a random day, lists visibility as 7km. Using the formula on the horizon page at Wikipedia, $d=\sqrt{1.5*h}$, rearranging it to $h=d*d/1.5$ (d=miles to horizon; h=feet above sea level), if you are 12.7 feet above sea level, you would never see the ship go over the horizon. Someone whose eyes are 5.7 feet from feet to eyes, would only have to be standing 7 feet above sea level.

In this scenario, it would be 99% impossible to see the ship, and yet it would still be 100% unobscured by the horizon. I think you would be able to push it out substantially further - so long as you can see enough of the boat's hull to make it not obvious. Looking at Wikipedia's image of the ship, there seem to be 4 levels below the deck - if they are 6.35 feet each (for easy calculation), then the horizon would be at the bottom of the third level above the waterline, and the top two levels (plus decks) would be visible from land, if the ship was 14km away.

So, with those constraints, and with 14km visibility, your planet would have to be no larger than Earth-size.

Brythan
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AMADANON Inc.
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3

As a slightly different solution, consider that light doesn't necessarily travel in straight lines. Given the right set of circumstances - curvature of the local space due to gravity, atmospheric composition, thermal gradients in the atmosphere, etc - you could potentially have a situation where the light reflected from an object could circumnavigate the globe.

Unfortunately for most purposes the atmospheric effects - mostly density and thermal gradient - will vary at different times of day in different locations, with additional local inconsistencies due to the environment. The air over small lakes reacts to daylight differently to the air over a desert for instance. As a result the curvature of the paths taken by the light will fluctuate wildly over very long distances, making the view at extreme distance appear to shimmer. As long as atmospheric effects are part of the situation they will most likely prevent the effect you are looking for from ever happening.

Gravity - or more precisely the warping of space by matter - will be much more consistent. There could be local perturbations due to different mass distributions, but those would be static as well. The downside is that for gravity to curve the path of the light sufficiently for it to circle the world it would have to be so high that your planet's inhabitants had better be made of neutronium unless you're happy with a planet that's several light years in diameter.

Probably the only way to make this work is in flashes. Set it up so that the constantly changing variables in the equations - atmosphere density, temperature, etc - occasionally come together to form a perfect curvature of the path of light to let you catch glimpses of the back of your own head.

Most likely the requirements would be a fairly heavy world to get gravity high enough to retain a dense atmosphere. If I had to guess (which I do, since I honestly don't know where to start the calculations on this) I'd say that a big planet with lowish gravity would have a deep atmosphere that could produce the effect I'm describing. Trying to achieve it on a planet with similar composition and size as the Earth would probably not be feasible.

AMADANON Inc.
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Corey
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The answer is going to be measured in hours to a few days, unless you have a non-water ocean and no atmosphere.

Story time:

I had the slow canoe. We were crossing from Victoria portage to Iron Point on Lake Winnipeg -- a 12 mile crossing that puts about 2 miles off shore. We were paddline into about a 15 knot headwind which was kicking up 18" waves. Two hours into the crossing I was between 1 and 2 miles behind the 2nd last canoe, and could just barely make it out against the clutter of the waves. I could no longer see the front canoe at all. At the time I was young and had normal vision.

Factors against me:

  • Solar position was putting a lot of sun sparkle off the waves into my eyes.
  • 18" swells are on the same order of size as the height of the hull.

These were Voyageur type canoes, with a crew of 6, and a length of 21 feet, and a beam of 4 feet. No sails to enhance visibility.

If a sailing ship as described is used, then you may be able to extend the visible distance by a factor of 30-40. (Figuring a height of 120 feet compared to the 3 foot visible height of a paddler) This gives a figure from 30 to 80 miles.

Another effect that comes into play, is the non-transparancey of humid air.

https://en.wikipedia.org/wiki/Visibility cites just under 300 km for the clearest possible air due to Raleigh scattering. 70 to 100 km is cited for arctic or mountain air under optimum conditions.

So: In practice, without some new tweak, your visibility is under 100 miles, with 200 miles as your upper bound, and likely to be less than 40 miles with reasonable weather.

Sherwood Botsford
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Given the size as calculated by others it's obviously out of the question to have a planet that big.

However, in the spirit of the question I can come up with a solution:

A Ringworld. Now, Niven's design requires impossible super-materials but one could be built by having it ride on a non-rotating track. (The combination of ring + track would average out to orbital velocity, no super-materials needed.)

Now, a Dyson sphere at first glance also appears to be an option but it's not going to work. The problem is that it would have an extremely low surface gravity. You specified a sailing ship--if there's enough wind to sail by what will the waves be like in such a low gravity???? I note that you are saying 1g on the surface of a Dyson sphere--but you'll need a pretty big black hole (far above what a star can make) in the center to provide that 1g.

Loren Pechtel
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To expand on the Dyson Sphere (of which Ringworlds can be considered a special case). Ignoring issues of how gravity is provided then:

  • Ships will appear to sail upwards.
  • Ships will eventually appear overhead.
  • Atmospheric effects will be most significant when clouds obscure vision.
  • At night the ghostly ships lights can be seen moving like a planet, (Greek 'wanderer') overhead but not following Kepler's laws.
  • The maximum clear atmospheric obstruction will occur when the internal arc between the observer and the ship just grazes the edge of the atmosphere.
  • so ignoring subtended angle distance effects the ship will become clearer as it moves further overhead from that point.
  • In the special case of a Ringorld and in some cases of a Dyson Sphere the ship will be either wholly or partially physically occluded by the sun / power source in it's path. Brightness may also prevent vision.
  • This will lead to either partial or total eclipses of the ship by the sun to an observer who is about perpendicular to the ship in the same plane.
  • If nights exist then the ship may also be also eclipsed by the night producing mechanism (if a physical one such as light baffles).
SimonN
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    Hi SimonN. It's great to see people contributing, but as you are probably aware from contributing to other sites in the network, on Stack Exchange, answers should (attempt to) answer the question as asked. I don't see how this answers the question of how large a world would have to be to get the effect the OP is after. Could you perhaps [edit] to clarify how this answers the question? – user Mar 12 '16 at 13:23
  • You missed the point that they are on the outside. There's no gravity toward the shell on the inside, so sailing is not applicable. – JDługosz Mar 12 '16 at 21:11
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    As I can not comment on others posts yet I can only post directly. To clarify that which I thought was relatively clear - Ringworlds had already been included in answers - I generalised on the special case to the extension to a Dyson Sphere as was made clear in the first line. As to the distance issue the vessel would always be visible (given sufficiently advanced optics unless obscured by clouds, power source or night generating mechanisms. Again I thought that was clear. – SimonN Mar 13 '16 at 11:27
  • As to the outside comment - whilst the OP did state that I was adding to the Ringworld option which had already been accepted - I think it would be a very foolish civilisation that attempted to live on the outer side of a ringworld. – SimonN Mar 13 '16 at 11:29
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Here is another solution, which you can use to make your world any size you want.

Here on earth, if there are layers of air with different densities, you get a mirage - the light is bent, causing a reflection of an actual object (sometimes distorted).

In your world, if the air densities were just right, it could bend the light about the same amount as the curvature of the planet. The planet could be any size, and still seem to be flat - you could see, in theory, indefinitely.

AMADANON Inc.
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  • Hal Clement's Heavy Planet exploits this in an awesome way. The inhabitants of the titular planet think that they're living in a "bowl" rather than a sphere, because of the complex interaction of the planet's massive gravity (something like 600g or so) and extremely fast rotation (so that on the equator, the effective gravity is only something like 3-6g, survivable for the humans that visit the planet). – Luaan Mar 11 '16 at 09:11
  • Also the case (planet looks like a bowl) on Saraksh in Strugatsky's Noon Universe. I think that one has gravity fairly close to Earth standard; it's the dense atmosphere that does the job. – January First-of-May Mar 13 '16 at 09:25
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I've previously considered a similar question a few years ago in a discussion about Dyson spheres (which is probably worthy of its own question, actually), and what actually happens to far-away objects (even on Earth scales, to an extent) is atmospheric extinction (also known as optical depth).

Atmospheric extinction on Earth sea level is said to be 0.16 magnitudes (for zenith), and 0.01 magnitudes, or 0.15 less than sea level, for Mauna Kea, 4.2 km above sea level (the numbers, for the record, are from here). We can use that figure of 0.15 magnitudes/4.2 km as our best approximation for atmospheric extinction in a constant sea-level atmosphere (as would be the case in the OP's scenario); of course the real figure should be a bit higher, but this works as a first approximation.

The apparent visual magnitude of the Sun is about -26; a far-away ship cannot possibly be brighter (even ignoring the atmospheric effects, which we will now apply). The apparent visual magnitude of the faintest stars visible in our best telescopes is +36. The difference between these two values is 62 magnitudes; this extinction will thus appear at a distance of 4.2*62/0.15 kilometers, or about 1800 km (I'm rounding up a bit again).

So what you need is a planet where the ship is still above the horizon when it's 1800 km away.

I'll shamelessly steal the approximate formula from the current accepted answer: R ~= 1/2 (d^2/h). For d=1800 and h=0.002 (kilometers) we get R=8.1*10^8 (that is to say, 810 million kilometers). This is a bit larger than your typical Dyson sphere (a radius of 810 million km is about 5.5 AU, which is to say, slightly smaller than the orbit of Jupiter), but I've rounded up so much during the previous calculations that the limit might actually be closer to typical Dyson sphere size, or even a bit below it.
For what it's worth, if the 300 km visibility figure given in the other answers is correct (it seems to underestimate both telescope power and brightness of a ship), the result is R'=2.25*10^6, or 22.5 million kilometers (0.15 AU).

A typical sailing ship, given the speed in the current accepted answer, would take about a week to pass a distance of 1800 km; the 300 km distance would be doable in a day (as in 24 hours). It's reasonable enough for someone sufficiently persistent to point their telescope to look at the same ship for several days in a row (at least, with occasional short breaks); and beyond that it shouldn't be visible anyway.
(But it would take centuries - many millenia on the bigger version - for said ship to travel all the way around. It's a very big planet.)

January First-of-May
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  • Great first post and I'd hope I'm not the first to welcome you to Worldbuilding.SE! – fi12 Mar 12 '16 at 23:03