How large would a world have to be for a sailing ship to never make it past the horizon?

Question

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;

Answer 1

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.

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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.

Answer 2

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:

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.

Answer 3

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.

Answer 4

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 :)

Answer 5

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)

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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).

Answer 11

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.

Answer 12

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.

Answer 13

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.)