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By that I mean, if you are travelling about on a continent on a planet with a circumference similar to the Sun, how much further would you be able to see, and would it be noticeable, without more complex scientific instruments? Would standing on a mountaintop, or the shoreline of a Atlantic ocean sized body make it noticeable enough?

Most other challenges (Gravity for one) of a planet with that size and mass are fixed or handwaved by a number of other things, but for the sake of argument you just wake up in a coniferous filled woodland, and as you wander, you find hills, plains and mountains, lakes and oceans, with pretty much the same array of features, and the same range in terms of height and depth, as on earth.

--Crazy background follows-- Feel free to ignore if this complicates the answer too much. I'm mostly interested in the geometry, but the TL;DR is, math is still the same, space doesn't bend in weird ways, and neither does light, but throw out pretty much most of what you know about astronomy.

This is for a very much magical world where people from our normal earth were brought. Continent sized 'floating' islands are in a orbit of sorts around a star-like central object after fracturing eons ago, with their surface pointed away. Even before the fracture and the dispersal of the 'islands', the planet was the size of the sun (Or just very very big). So would someone from Earth be able to easily notice this difference? Most knowledge regarding astronomy, save the math, is useless in this world (No stars; the "sun" at the center of the islands provides 'sunlight' by reflecting it unto a moon like object that disperses it onto some of the islands some of the time. The glittering lights in the night sky are the crystallized (frozen over?) remains of the planet that did not settle into continents

Alexandra Eagle
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MartinArrJay
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    How does sunlight work on this world? Does it rotate? – Starfish Prime Jan 23 '20 at 13:25
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    At sea, or on the seashore, on a clear day, you will notice immediately (as in, at a glance) that you cannot see the horizon (or, at least, that the horizon is very much farther away). (On Earth the horizon is close enough that it appears as a clear line. On a planet vastly larger than Earth the horizon will be so far that is will be lost in atmospheric haze aka "aerial perspective".) – AlexP Jan 23 '20 at 13:28
  • @StarfishPrime As mentioned, a lot of the normal astronomy just doesn't apply here. It's essentially a different dimension, and magic bends more then a few laws of physics, but that doesn't change math, nor does it do much to geometry (It's still in euclidean space). Everything else, is out the window. The day light might be coming from a star like object, being reflected by a gigantic moon sized mirror, that rotates roughly once a day. – MartinArrJay Jan 23 '20 at 13:38
  • @MartinArrJay this is why details are needed. Does the world rotate? Or does the source of light orbit it? – Starfish Prime Jan 23 '20 at 13:41
  • @StarfishPrime Undecided at this point. I was looking mostly for the possibility that it would be noticeable, by the average joe, and whatever passes for daylight itself, is immediately noticeable as being so weird, that it cannot be used for determining what the curvature is. Average joe doesn't get what exactly is providing daylight, only that it isn't a sun, in classical sense. – MartinArrJay Jan 23 '20 at 13:45
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    @MartinArrJay it can be used for judging atmospheric thickness as it sets as the quality of sunsets will be different. If the world does not rotate then weather patterns will be different and Foucault's pendulums will not work. There's lots going on here. – Starfish Prime Jan 23 '20 at 13:46
  • @StarfishPrime I honestly feel like I've been pretty explicit that a lot of weird stuff is going on, and some stuff is still undecided, and that we're talking about whether it would be noticeable from wandering around. Specifically that I was looking for an answer that related more directly to what Zeiss or Alexp mention. – MartinArrJay Jan 23 '20 at 13:53
  • @MartinArrJay dude, are you telling me that sunsets aren't things you'd notice? Your rules, I guess. – Starfish Prime Jan 23 '20 at 13:55
  • @StarfishPrime I'm trying to tell you that someone who would be able to use atmospheric thickness to gauge the size of the "planet" they are on, would notice that what passes for stars, moons and the sun are behaving so erratically, they wouldn't trust using them for calculating this. And yes, maybe sunsets aren't a thing. I don't know yet – MartinArrJay Jan 23 '20 at 13:58
  • Won't it be more noticeable the gravity pull a bigger planet would have? Besides, the atmosphere would be thicker – maxisalamone Jan 23 '20 at 14:35
  • @maxisalamone It seems I'm not explicit enough in writing my question, so maybe I need some help there, my apologies for that, but gravity was specifically stated as something not to be concerned with. Gravity, atmosphere and such are just "the same" as earth, for whatever reason. I didn't include magic as a tag for this, because the answer shouldn't rely on magic in any way. But this is very much a extra dimensional world, where a lot of things aren't the same. What I wanted was the chance of noticing that the horizon looks weird, and understanding what it means, at least partially. – MartinArrJay Jan 23 '20 at 14:50
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    So a much larger planet would be closer to a flat Earth. I once saw a video debunking flat Earth, and at one point, the presenter used software to simulate a view of a mountain (in California I think?) as it would appear on a flat vs reality. I think he even had a slider and could scrub back and forth between the two curvatures. Having trouble finding it now though. – Harabeck Jan 23 '20 at 21:47
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    Is a hand-held 1800's style nautical telescope a "complex scientific instrument"? During the age of sail, a quick way to describe the distance to a recently sighted vessel was "hull up". That is, with a telescope from the highest point on our ship, we can see the hull of the other ship; since the hull disappears first when a ship goes over the horizon, this gives a bound on distance. If the planet has lower curvature then the hull-up distance is farther away, and this would be very noticeable to an observer familiar with sighting by telescope. – Eric Lippert Jan 23 '20 at 23:12
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    While traveling about is the only time I'd notice. I'd wonder why I can still see downtown Chicago returning from the suburbs 100 miles out. I live street level in the flatlands but the horizon (and sunsets on it, btw) is something I rarely see, and I seldom make it down to the lake. - There's plenty of ways to make this happen, but it not being immediately obvious is definitely on the table for as long as you want to milk it. – Mazura Jan 24 '20 at 03:30
  • If it is the Sun, then the gravity would simply crush things with its 28g. Anything would weight 28 times more. (imagine your poor skeleton trying to support 2+ tons of weight). But it gets worse. The Sun consists of light elements . If you would make the thing out of rock - well.. the g-force will be even more - assuming we ignore the fact that chunk of rock this size will collapse into something like a neutron star by its own gravity. I know it's hand waved, but it's a bit too much to hand wave :p – Alma Do Jan 24 '20 at 13:28
  • beyond the scope of your question, but if your planet is much bigger than Earth (~ > 50% more diameter) you wouldn't be able to reach orbit with a rocket (skip to near the end for that bit) so if you tried to launch a chemical rocket and couldn't reach orbit that would be one way to tell. – Michael Jan 24 '20 at 15:16
  • The drastically different sky and the "sun" that winks on an off as continents move in front if it seem like they would be more noticeable. I also believe your moon need to to me moving at near relativistic speeds to have a 24 hour day which should also be noticable, since light will lag behind the sun. – John Jan 24 '20 at 20:24
  • @Michael - "In the staged version of the rocket equation, the possible achievable delta-V is proportional to the number of stages when mass ratio and engine efficiency is held constant. In short, you can get any delta-V you want by adding additional stages." – On a Super-Earth 1.5x the volume and mass of Earth, would our rocket technology allow us to reach orbit? - step one: be a rocket scientist. 2: spend a billion dollars just to find out that there wasn't enough stages. – Mazura Jan 25 '20 at 22:20
  • An interesting question would be, if the sun sized planet were a hollow sphere made of your choice of steel alloy, how thick would it have to be for gravity to be 1.0 Gee? And, would a sphere of that thickness have the structural strength to not collapse? – Harper - Reinstate Monica Jan 26 '20 at 10:14
  • @Harabeck - "So a much larger planet would be closer to a flat Earth." Absolutely, in fact as the diameter approaches infinity, it would become indistinguishable from a flat earth. – Glen Yates Jan 27 '20 at 14:19
  • An ancient Greek scientist measured the earth's diameter within a reasonable amount. This should be possible here. – NomadMaker Jul 07 '21 at 22:35

8 Answers8

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The biggest visual clue that you're on a much larger planet than Earth is a mismatch between atmospheric haze (related to distance), perspective, and horizon distance. A ship out at sea, for instance, would have much more perspective "shrinking" and haze coverage while still above the horizon than you're used to.

If you have any ability to measure distance, horizon distance can be used to calculate the actual size of the body you're on -- by knowing the height of your viewpoint and how far off your horizon is, you can pretty quickly and fairly accurately estimate the radius of the surface under your feet (don't even need trigonometry; similar triangles will do the job).

Zeiss Ikon
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    I agree, assuming that all other factors are the same (gravity, atmosphere, smell, etc.), sailors would probably notice first if nobody's directly looking or using scientific instrumentation – Dragongeek Jan 23 '20 at 13:36
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    Assuming normal person height, the horizon is about 5 km away on Earth. The haze and distance might well make unaided eyes unable to see a ship at the point it goes behind the horizon. The horizon would be something-like 100 times as far. You would need EXTREMELY good eyes to see a ship at the about-500 km away. A smallish telescope would be required. But they would certainly be able to see it for much more than the roughly 5 km to the horizon on the Earth. And probably for a couple of days instead of the hour or less for a ship on Earth. – puppetsock Jan 23 '20 at 14:36
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    Flatter Earth would definitely make sea navigation much easier, if you can keep a visual on your destination. Two spires could denote a ship lane at close range, and then you'd use the horizon shape as you get further away. – John Dvorak Jan 24 '20 at 14:30
  • "Similar triangles" is trigonometry is it not? – John O Jan 24 '20 at 17:16
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    @JohnO No, it's simple geometry. You don't need any tables, and don't need to know the value of any of the angles in degrees, radians, or what have you. Just need to know the lengths of the legs and have a way to be sure the triangles are similar, i.e. provably have the same angles. – Zeiss Ikon Jan 24 '20 at 17:17
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The big clue is that you'll never see something disappear over the horizon.

On an Earth-sized planet, the horizon is about 5 km away. Under ideal atmospheric conditions, you can see things up to 300 km away. Seeing something drop below the horizon is no problem.

On a Sun-sized planet, the horizon is now 550 km away, but atmospheric conditions are no different. As things move away from you, they will invariably vanish into the haze rather than drop below the horizon.

(If you're paying attention, "things that vanish into the haze" include the ground. You'll never see a sharp horizon, just a continuous shading from sky to ground.)

Mark
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    While I understand why there should be a 'continuous shading from sky to ground', I struggle to imagine what would the outcome look like. are there any artistic depiction of this phenomena? – Kepotx Jan 24 '20 at 16:03
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    @Kepotx it appears there are some example paintings in https://en.wikipedia.org/wiki/Aerial_perspective – B.Kenobi Jan 24 '20 at 16:30
  • Mark, great insight that you simply would not be able to see the horizon! Good one – Fattie Jan 25 '20 at 00:44
  • If the horizon is 550km away, then wouldn't the sun seem to get dimmer and hazier as it approached the horizon, since there is much more atmosphere in between the viewer and the sun? That would make a greater difference between high-noon sun and an hour before sunset then there is here on earth. I'm not sure if that can be used to calculate the size though. – Moby Disk Jan 26 '20 at 20:04
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Ignoring gravitation and the movement of the sun which I will assume are fixed to be as on Earth, the world would appear very similar to ours, but the differences would be noticable. In a great many places where people live the horizon is blocked by high grounds, woods, vegetation or other buildings. But where the line of sight was uninterrupted such as on plains, some high hills and at sea there would be a difference some of the time.

On Earth on slightly misty or hazy days there are occasions when the sea and the sky blend almost imperceptibly together. This would always be the case on this super Earth even in the clearest weather. There would also probably be different quality of light at sunsets and sunrises as the light would have to pass through a much thicker layer of air at dawn and dusk, so redder longer than Earth after dawn and earlier than Earth before dusk.

Slarty
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    The hazy horizon is a decent indicator, but is there any way to unequivocally determine that the planet is huge, and that it's not just hazy out? As you point out, this can happen on earth, so a hazy horizon does not in itself indicate that the planet is huge... it might just indicate that the planet is hazy. – Nuclear Hoagie Jan 23 '20 at 20:19
  • Yes you could measure the amount of light that get through the atosphere at different elevations of the sun. The amount of light that reaches the surface will depend on the thickness of the atmosphere which will depend on curvature. But there are easier methods as pointed out by Cecilia – Slarty Jan 23 '20 at 22:55
  • @Slarty At that point we are well beyond simply noticing that something is off, and into taking exact measurements to figure out what is going on. – MartinArrJay Jan 24 '20 at 10:35
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This answer assumes that your planet is spherical and relies on basic trigonometry to estimate the circumference of the planet.

Method 1: Eratosthenes' method

I believe, even with your reflected light sources, you could still estimate the size of the planet using Eratosthenes' method, i.e. at noon, measure the length of the shadow of a standard length pole at two different locations.

How do you determine noon on a foreign planet without time-keeping devices? It's the time of day when the light source is highest in the sky, and the shadow is shortest. Simply measure all day at fixed intervals and make a graph. Keep the shortest measurement.

The mathematical assumptions that must hold are

  1. The light source's ray's are approximately parallel. This works great for the sun which is really far away, but it even works for the moon, which is much closer. You could do some trig to calculate whether this works for your reflected light source
  2. You know the distance between your two measurement locations.
  3. The locations are on the same line of longitude. Measuring longitude without a clock is hard (More information about measuring longitude). Eratosthenes simply assumed that Alexandria is due north of Syene (his two measurement points). If your planet has a magnetic field roughly aligned with the rotational axis of the "islands", you could use a homemade compass composed of a needle floating in water to determine north, then travel in that direction for some distance (the longer the better) and make your second measurement. If your planet doesn't have a rotational axis and magnetic north, this becomes a lot more complicated, and I wouldn't recommend it.

Method 2: Al-Biruni's method

But what if you want the light sources to be closer to the planet, you don't have a working compass, or the rotation of the light sources is wonkey? Use Al-Biruni's method. All you need is a mountain.

From the top of the mountain, he sighted the dip angle which, along with the mountain's height (which he calculated beforehand), he applied to the law of sines formula. This was the earliest known use of dip angle and the earliest practical use of the law of sines Wikipedia

How do you know the mountain's elevation? If you are climbing the mountain anyway, use a barometer which you can build out of a sealed vessel of water with a narrow spout.

Great article about other historical methods for measuring mountain height here

OP's requirements

The question states that the differences should be visual, but I would argue that shadow length at noon and angles from mountain tops are visual, and while they might not be the first thing you notice, they would be a good way to confirm your suspicion that something is fishy.

Also the question excludes complex scientific instruments. Only instruments needed here are a way to measure angles or length, and a stick. Oh, and your homemade barometer.

Cecilia
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  • For Method 1, I think you need either a timekeeping device, or a device to measure longitude. Without a timekeeping device, you can only determine local solar noon, but both measurements need to be taken simultaneously. Solar noon occurs at the same time everywhere on a line of longitude, so you can ensure simultaneity by being at the same longitude. The only other way to ensure it is to use a clock, since if you're at different longitudes, your measurements at solar noon will be taken at different times. – Nuclear Hoagie Jan 24 '20 at 14:16
  • @NuclearWang Good point, I'll add that in. – Cecilia Jan 24 '20 at 16:24
  • This strategy doesn't require you to measure longitude. You just need two measuring sites which happen to be at the same longitude, which is much easier to determine. You don't even need a magnetosphere. Just watch the stars all night, and the star which moves the least is your pole star. You can set successive markers in that direction to get another location which is due N/S. – Lawnmower Man Jan 25 '20 at 22:32
  • Furthermore, one should consider that Syrene is about 3 degrees E of Alexandria, which is about a 1% error in longitude, yet Eratosthenes obtained a very good result. I think the bigger difficulty is the separation of sites. Syene to Alexandria is about 900 km, but you would want a proportional distance on the sun-sized planet, which is more than 100x larger. Marking two locations nearly 100,000 km on the same longitude would be...challenging. – Lawnmower Man Jan 25 '20 at 22:44
  • If you have a straight road from A to B (or S), you can establish equality in longitude by observing how closely the shadow at noon (sun's highest elevation / shortest shadow) aligns with the road direction. No timekeeping required - unlike the general longitude problem for ocean crossing, where tehre are no roads, straight or otherwise. – user_1818839 Jan 26 '20 at 19:24
  • @LawnmowerMan The OP specifies no stars. But I appreciate your calculation of the relative distance necessary. I didn't really think about how this would scale up. 100,000 km is 2.5x the circumference of Earth. Yikes. – Cecilia Jan 26 '20 at 20:22
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Sunrise and sunset - assuming the atmosphere is as thick and as dense as on Earth, you will see the daystar much redder (may be red enough not to see it at all) when it is near the horizon.

fraxinus
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    There are multiple comments, and even some answers, referencing that objects outside of the "planet" itself, behave in an otherworldly way, and that anyone taking a cursory glance at them, would notice it as being VERY weird. This includes the sun not being a actual sun, and maybe not even a star to begin with. – MartinArrJay Jan 24 '20 at 10:30
  • Also, arguably, even if the sun was there, it being redder would not be a noticeable indication that the planet you are standing on is larger, not unless you had a degree in physics, or more than a passing interest, it seems. – MartinArrJay Jan 24 '20 at 10:31
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    In such a setup, the object assuming the sun role will behave a lot like the Sun. And no physics degree will be needed - the world wil be red most of the day. – fraxinus Jan 24 '20 at 10:55
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    Not sure why this is being voted down. This seems correct: the sunrise and sunset would look drastically different, whatever your light source is, whatever you want to call "light near the horizon". With the red sky thing, it might be worth considering that at a point you would rarely even see the sunset or sunrise because of the chance of it being blocked by particulates, clouds, diffraction etc.

    Advantage is that you wouldn't even need equipment or a high spot to notice.

     – pjp Jan 24 '20 at 11:12
  • But what exactly is it one would notice, that could lead to the possible conclusion that you are on a huge planet? Again, there might not be a sunset or sunrise, I cannot stress this enough. So with that, how would a redder horizon, or just a redder sky, be an indicator that the planet is big? – MartinArrJay Jan 24 '20 at 13:25
  • Would the green flash be affected ? https://en.wikipedia.org/wiki/Green_flash – Porunga Jan 24 '20 at 14:13
  • @Porunga maybe, but OP said that the daystar does not move. – fraxinus Jan 24 '20 at 14:35
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Absolutely (given clear environmentals). The sight distance is over 10 times as much (https://planetcalc.com/1198/ if you want to formula) on the sun compared to earth. Granted, your eyes won't be able to see the entirety of this distance even given perfect atmospheric conditions, but the difference will be noticeable.

Summed up another way, the difference of sight distances from standing on Earth to standing on a sun-sized planet would be greater than the sight distance difference of standing on Earth and standing on a 100m tower on Earth. Having been on 100m towers before, I can definitively say that you can easily perceive the sight distance, even without perfectly clear days.

Just to note, this calculation does not take in to account refraction which would have a minor impact on the calculations.

ColonelPanic
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Eclipses of your moons.

Such a large planet would be almost certain to have a number of moons. Their eclipses would be more frequent and long lasting (assuming they were close to the planet, i.e., not typical of interplanetary differences).

Without details, can't calculate the Roche limit, but I suppose it would be a few hundred thousand kilometers (e.g., Roche limit of our moon and our sun is 657,161 km), so orbits won't be particularly close - though having rings because a moon was too close would be visually obvious.

From a simple article written for teachers re: Measuring the Earth's Curvature discusses observing a lunar eclipse (nice picture in the article, not sure of copyright):

the Earth's shadow on the surface of the Moon is obviously curved during these eclipses, which gave ancient astronomers the idea that the Earth must be spherical.

On this super-earth, any curve would be so flat as to be unnoticeable.

Gary Walker
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At ground level you might not immediately notice unless you were observant and/or knew what to look for, especially if air quality is similar to Earth.

I think if you took a flight things would be far more apparent.

Mr. Boy
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