If Sweden was to magically float away, at what altitude would it be visible from the southern hemisphere?
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If a chunk of land mass the size and position of Sweden took off and floated away in a straight line, exiting the atmosphere and orbiting earth, at what distance would it be visible from, let's say, South Africa?
Alternatively, what size would our planet have to be to allow for such a floating land mass to be visible from the proportionate equivalent of South Africa, while still within the bounds of the planet's (unchanged) atmosphere?
Edit: Thanks for all your detailed and interesting answers! Didn't expect my silly question to blow up like that, tbh.
I was quite tired when I first posted and realize that I haven't specified some things properly.
Originally, I had intended for floating Sweden to remain geostationary, orbiting with the planet rather than around it (because magic).
I thought perhaps the atmosphere's refractivity (which only @Chronocidal mentioned as far as I've seen) might allow for a high enough Sweden to be visible on the horizon, if the observer was positioned at the northernmost point of South Africa, with no obstacles in his line of vision, at maybe 1500 meters above sea level.
But in light of the response here, I decided to just have my floating island properly orbit the planet somewhere within the thermosphere instead.
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asked May 28 at 10:06
PangolinPangolin
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If a chunk of land mass the size and position of Sweden took off and floated away in a straight line, exiting the atmosphere and orbiting earth, at what distance would it be visible from, let's say, South Africa?
Alternatively, what size would our planet have to be to allow for such a floating land mass to be visible from the proportionate equivalent of South Africa, while still within the bounds of the planet's (unchanged) atmosphere?
Edit: Thanks for all your detailed and interesting answers! Didn't expect my silly question to blow up like that, tbh.
I was quite tired when I first posted and realize that I haven't specified some things properly.
Originally, I had intended for floating Sweden to remain geostationary, orbiting with the planet rather than around it (because magic).
I thought perhaps the atmosphere's refractivity (which only @Chronocidal mentioned as far as I've seen) might allow for a high enough Sweden to be visible on the horizon, if the observer was positioned at the northernmost point of South Africa, with no obstacles in his line of vision, at maybe 1500 meters above sea level.
But in light of the response here, I decided to just have my floating island properly orbit the planet somewhere within the thermosphere instead.
geography moons astronomy
asked May 28 at 10:06
PangolinPangolin
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– L.Dutch♦
Jun 1 at 3:16
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If a chunk of land mass the size and position of Sweden took off and floated away in a straight line, exiting the atmosphere and orbiting earth, at what distance would it be visible from, let's say, South Africa?
Alternatively, what size would our planet have to be to allow for such a floating land mass to be visible from the proportionate equivalent of South Africa, while still within the bounds of the planet's (unchanged) atmosphere?
Edit: Thanks for all your detailed and interesting answers! Didn't expect my silly question to blow up like that, tbh.
I was quite tired when I first posted and realize that I haven't specified some things properly.
Originally, I had intended for floating Sweden to remain geostationary, orbiting with the planet rather than around it (because magic).
I thought perhaps the atmosphere's refractivity (which only @Chronocidal mentioned as far as I've seen) might allow for a high enough Sweden to be visible on the horizon, if the observer was positioned at the northernmost point of South Africa, with no obstacles in his line of vision, at maybe 1500 meters above sea level.
But in light of the response here, I decided to just have my floating island properly orbit the planet somewhere within the thermosphere instead.
geography moons astronomy
asked May 28 at 10:06
PangolinPangolin
1651 gold badge2 silver badges5 bronze badges
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If a chunk of land mass the size and position of Sweden took off and floated away in a straight line, exiting the atmosphere and orbiting earth, at what distance would it be visible from, let's say, South Africa?
Alternatively, what size would our planet have to be to allow for such a floating land mass to be visible from the proportionate equivalent of South Africa, while still within the bounds of the planet's (unchanged) atmosphere?
Edit: Thanks for all your detailed and interesting answers! Didn't expect my silly question to blow up like that, tbh.
I was quite tired when I first posted and realize that I haven't specified some things properly.
Originally, I had intended for floating Sweden to remain geostationary, orbiting with the planet rather than around it (because magic).
I thought perhaps the atmosphere's refractivity (which only @Chronocidal mentioned as far as I've seen) might allow for a high enough Sweden to be visible on the horizon, if the observer was positioned at the northernmost point of South Africa, with no obstacles in his line of vision, at maybe 1500 meters above sea level.
But in light of the response here, I decided to just have my floating island properly orbit the planet somewhere within the thermosphere instead.
geography moons astronomy
geography moons astronomy
asked May 28 at 10:06
PangolinPangolin
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asked May 28 at 10:06
PangolinPangolin
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edited Jun 18 at 15:02
Pangolin
asked May 28 at 10:06
PangolinPangolin
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asked May 28 at 10:06
PangolinPangolin
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asked May 28 at 10:06
PangolinPangolin
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– L.Dutch♦
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– L.Dutch♦
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– L.Dutch♦
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Given Sweden has a latitude of 60° N and south Africa of a bit over 30° S, you can never see one from the other no matter how high one is and no matter how small the planet is (as long as it still is big enough to allow you to neglect the distance between your eyes and the surface).
The Sweden simply rises further away in the hemisphere invisible from South Africa.
That is if Sweden rises simply above and continues to stay above the used-to-be-Sweden, i.e rotates with Earth. If it starts orbiting Earth forming new moon, it will probably pass over South Africa sooner or later. Obviously, I am magically hand-waving away the atmospheric friction.
Edit: very sophisticated visualization:
Edit 2: for a more general solution, given both points have the same longitude:
cos (difference in latitude) = (Earth radius, 6378 km)/(Sweden distance from Earth center, i.e. height + Earth radius).
For point directly below equator this yield
0.5 = cos(60 - 0) = 6378/(h+6378) -> h = 6378 km
waaaay above atmosphere.
edited May 29 at 13:12
Machavity
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answered May 28 at 12:25
MoriMori
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– L.Dutch♦
May 30 at 13:03
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There is no atmospheric height any object of any size can be, perpendicular to the location of Sweden, that could be seen by anyone standing anywhere in South Africa.
I don't need math to prove this. Just take out a piece of paper, draw a circle, and use a ruler to start drawing lines. That combination cannot be done.
Alternatively, visit maps.google.com, scroll out until you can see the whole planet, and rotate it until Sweden is in the middle. It'll look something like this (credit: Google):
Molot's answer (which is the mathematically pure answer, go vote for it) brings out that tall things slightly beyond the edge of the hemisphere defined by Sweden acting as the pole or center of the circle will see Sweden if it's high enough. But below that it's geometrically impossible — there's a planet in the way.
And that example, above, assumes we send Sweden so far away that the entirety of its defined hemisphere can see it. If we limit that height to keep it within the atmosphere, then the total area that can see it at any height within the atmosphere (~300 mi thick) is something more like this:
Compared to the radius of the Earth, our atmosphere is proverbially as thick as a sheet of paper. Variations in land height don't matter, either, as those variations are less than the dimples on a golf ball compared to the radius of the Earth. When you're talking about the entire planet, it can be reasonably modeled as a smooth surface.
Conclusion
There is no height within the atmosphere that would allow Sweden to be seen from any point within what we would normally call the Southern Hemisphere. There is a height where it could be seen by a couple of points in the Southern Hemisphere (see that area below the dashed line in Africa? Yeah, the dashed line is the equator), but Sweden would be so far outside Earth's atmosphere that it likely doesn't matter for your story.
This assumes you do not subscribe to the Flat Earth theory. I don't.
answered May 28 at 16:12
JBHJBH
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I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
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– Innovine
May 28 at 19:41
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Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
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– Cœur
May 29 at 1:43
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@Cœur 😀 Touché.
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– JBH
May 29 at 1:45
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By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
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– reirab
May 29 at 18:29
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This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
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– Fattie
May 30 at 12:32
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This answer is fundamentally flawed in that it assumes that the 'horizon' formula is valid for any value of h.
Unfortunately, this is simply not true, and as the image shows, it is only a valid approximation when h is much less than RE
(RE >> h)
If you can see something depends on how high you are, and how high is what you look at:
Africa is reasonably flat*, so assume a person 2 meters tall (or he may be standing on something ;) )
You can look and fiddle with the numbers here with nice calculator that does the math for you. With Observer at 2 meters and object at the edge of space (100km) visual horizon is at the distance of 1134km. Distance from Sweden to South Africa is 10058 km, so it is about ten times too far to have a line of sight.
To get the line of sight, you will need to put Sweden at height of about 7922km. That's outside atmosphere, but inside Van Allen radiation belt, so long time life support would be hard for at least these two reasons.
* Of course for more precise result you should include the mean or max elevation of South Africa, and decide where, exactly, your observer is placed to account for mountains, valleys, trees and so on. The above calculation is meant as an estimate, and to show a way how to calculate it yourself.
The question says only "from South Africa" so we can assume optimal point of observation, Thabana Ntlenyana 3482 meters high (+2m person) for the flying height of "mere" 3452km. No trees or other mountains to significantly change this altitude.
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@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
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– Mołot
May 28 at 14:41
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In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
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– Brythan
May 28 at 14:47
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I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
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– brhans
May 28 at 18:42
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To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
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– Michael Seifert
May 28 at 18:48
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Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
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– Mark
May 29 at 20:13
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If you imagine an idealised planet with a star at an infinite distance, how much of the planet can see it? The answer is half: the star projects a cone that rests on an equatorial line perpendicular to the direction it lies in. Those standing on that equatorial line are looking along the tangent to the curve of the surface of the planet.
So the basic question is, what is the minimum distance at which this is still possible, or rather, given two opposite points on a circle, where do their tangents intersect? This is effectively squaring the circle that forms said equator, which simplifies down to the square root of 2r^2 (minus the radius of the Earth if you want the height above it).
In the case of Earth, that comes to 2,639km, however there are other factors regarding the shape of the orbit that might have an effect. Obviously, given an orbit it will not be visible at all times to the whole planet. However, if that orbit is geostationary (e.g. proportional to the rotational speed of the planet) it will only be visible to half the planet when above said height.
Given orbits are elliptical, it's also possible that it is not above that height the entire orbit. If the orbit resonates with the rotation of the planet and has a low point below said height, this could create dead spots that will never see it in the sky.
Finally, there's always the possibility of mountain ranges and atmosphere practically limiting visibility along the line of the tangent.
answered May 31 at 11:44
DanikovDanikov
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You have three questions here.
1. How high can a geostationary Sweden be visible from South Africa?
Answer: 68,100 km
This is a question that humanity has pondered since we first climbed down from the trees.
The other answers are correct that if two points are more than 90 degrees apart on the globe, they cannot see each other at any height (unless the earth has a big groove in it or atmospheric refraction does something funny, but we don't have to consider that in this answer).
The correct way to calculate angular distances between two points on the globe is to use the haversine formula. This takes both latitude and longitude into account. I'm feeling a little lazy today, so I will just take the distance measured on Google Maps, divide by 40,075 kilometers and multiply by 360 degrees. This is equivalent, and I've checked that the math lines up.
If Sweden goes straight up from the point of view of its center point, the important location of Sweden is Flataklocken at 62°23′15″N 16°19′32″E.
I found a point near South Africa's Zimbabwe that is 9,472 km away from Flataklocken, which comes out to 85.09 degrees.
You're in luck! You can see Sweden from South Africa. It will be low in the sky, within 5 degrees of the horizon. It will be brighter than the full moon since it's a lot closer.
If you intend to actually do this, I'd recommend buying a hot air balloon to take South Africans up to view.
Some math:
a = 85.09 degrees (this is the angular distance between South Africa and central Sweden)
r = 6371 km (this is the radius of the earth)
find h, the height of Sweden
cos(a) = r / (r + h)
Therefore
h = r (-1 + 1 / cos(a)) = 68,100 kilometers.
This is over 10 times earth's radius. It's far outside earth's atmosphere, but low enough it's not going to crash into the moon.
2. If Sweden were floating in a geostationary position in the upper atmosphere, how far would it be visible?
Answer: 2400 km
Earth's atmosphere only goes up 480 km, so if you want Sweden to stay in a geostationary position in the atmosphere, it won't be visible from South Africa.
If you want the people of Sweden to have enough air to breathe, it will have to be even lower. I'm not going to run the numbers for that, because that's not the question you asked.
The equation is the same, except in this case, we know h and r, and we're trying to find a.
r = 6371 km (this is still the radius of the earth)
h = 480 km (Sweden floating in the edge of the atmosphere)
cos(a) = r / (r + h)
Therefore
a = arccos(r / (r + h)) = 21.57 degrees.
This comes out to a distance of 2400 kilometers.
This isn't far enough to see from the southern hemisphere, but you would still be able to see southern Sweden from parts of Algeria and Tunisia. As far as I'm concerned, that's good news!
3. Can a Sweden orbiting in the atmosphere be visible from South Africa?
Answer: No
If you can make Sweden orbit at this low elevation, it can be visible from every point in the world. But you can't make Sweden orbit in the atmosphere.
Orbit means something is spinning around a world without rockets or thrusters because it is going so fast that gravity can't pull it down. You can calculate the orbital speed for any height in a vacuum. Unfortunately, orbit is not possible within the atmosphere because the air resistance will eventually slow Sweden down so that it will crash into the earth.
Don't despair. You can still make Sweden fly. You'll need to buy big jets or something else that uses energy to move it around. I don't know what your budget is, but I think it's still worthwhile. Think of the spectators in South Africa.
answered Jun 10 at 17:05
JetpackJetpack
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Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
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– Jetpack
Jun 11 at 0:11
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That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
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– Pangolin
Jun 19 at 13:54
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5 Answers
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Given Sweden has a latitude of 60° N and south Africa of a bit over 30° S, you can never see one from the other no matter how high one is and no matter how small the planet is (as long as it still is big enough to allow you to neglect the distance between your eyes and the surface).
The Sweden simply rises further away in the hemisphere invisible from South Africa.
That is if Sweden rises simply above and continues to stay above the used-to-be-Sweden, i.e rotates with Earth. If it starts orbiting Earth forming new moon, it will probably pass over South Africa sooner or later. Obviously, I am magically hand-waving away the atmospheric friction.
Edit: very sophisticated visualization:
Edit 2: for a more general solution, given both points have the same longitude:
cos (difference in latitude) = (Earth radius, 6378 km)/(Sweden distance from Earth center, i.e. height + Earth radius).
For point directly below equator this yield
0.5 = cos(60 - 0) = 6378/(h+6378) -> h = 6378 km
waaaay above atmosphere.
edited May 29 at 13:12
Machavity
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answered May 28 at 12:25
MoriMori
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– L.Dutch♦
May 30 at 13:03
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Given Sweden has a latitude of 60° N and south Africa of a bit over 30° S, you can never see one from the other no matter how high one is and no matter how small the planet is (as long as it still is big enough to allow you to neglect the distance between your eyes and the surface).
The Sweden simply rises further away in the hemisphere invisible from South Africa.
That is if Sweden rises simply above and continues to stay above the used-to-be-Sweden, i.e rotates with Earth. If it starts orbiting Earth forming new moon, it will probably pass over South Africa sooner or later. Obviously, I am magically hand-waving away the atmospheric friction.
Edit: very sophisticated visualization:
Edit 2: for a more general solution, given both points have the same longitude:
cos (difference in latitude) = (Earth radius, 6378 km)/(Sweden distance from Earth center, i.e. height + Earth radius).
For point directly below equator this yield
0.5 = cos(60 - 0) = 6378/(h+6378) -> h = 6378 km
waaaay above atmosphere.
edited May 29 at 13:12
Machavity
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answered May 28 at 12:25
MoriMori
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– L.Dutch♦
May 30 at 13:03
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Given Sweden has a latitude of 60° N and south Africa of a bit over 30° S, you can never see one from the other no matter how high one is and no matter how small the planet is (as long as it still is big enough to allow you to neglect the distance between your eyes and the surface).
The Sweden simply rises further away in the hemisphere invisible from South Africa.
That is if Sweden rises simply above and continues to stay above the used-to-be-Sweden, i.e rotates with Earth. If it starts orbiting Earth forming new moon, it will probably pass over South Africa sooner or later. Obviously, I am magically hand-waving away the atmospheric friction.
Edit: very sophisticated visualization:
Edit 2: for a more general solution, given both points have the same longitude:
cos (difference in latitude) = (Earth radius, 6378 km)/(Sweden distance from Earth center, i.e. height + Earth radius).
For point directly below equator this yield
0.5 = cos(60 - 0) = 6378/(h+6378) -> h = 6378 km
waaaay above atmosphere.
edited May 29 at 13:12
Machavity
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answered May 28 at 12:25
MoriMori
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Given Sweden has a latitude of 60° N and south Africa of a bit over 30° S, you can never see one from the other no matter how high one is and no matter how small the planet is (as long as it still is big enough to allow you to neglect the distance between your eyes and the surface).
The Sweden simply rises further away in the hemisphere invisible from South Africa.
That is if Sweden rises simply above and continues to stay above the used-to-be-Sweden, i.e rotates with Earth. If it starts orbiting Earth forming new moon, it will probably pass over South Africa sooner or later. Obviously, I am magically hand-waving away the atmospheric friction.
Edit: very sophisticated visualization:
Edit 2: for a more general solution, given both points have the same longitude:
cos (difference in latitude) = (Earth radius, 6378 km)/(Sweden distance from Earth center, i.e. height + Earth radius).
For point directly below equator this yield
0.5 = cos(60 - 0) = 6378/(h+6378) -> h = 6378 km
waaaay above atmosphere.
edited May 29 at 13:12
Machavity
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answered May 28 at 12:25
MoriMori
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edited May 29 at 13:12
Machavity
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edited May 29 at 13:12
Machavity
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edited May 29 at 13:12
Machavity
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1,0314 silver badges14 bronze badges
answered May 28 at 12:25
MoriMori
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answered May 28 at 12:25
MoriMori
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answered May 28 at 12:25
MoriMori
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– L.Dutch♦
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– L.Dutch♦
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– L.Dutch♦
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– L.Dutch♦
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There is no atmospheric height any object of any size can be, perpendicular to the location of Sweden, that could be seen by anyone standing anywhere in South Africa.
I don't need math to prove this. Just take out a piece of paper, draw a circle, and use a ruler to start drawing lines. That combination cannot be done.
Alternatively, visit maps.google.com, scroll out until you can see the whole planet, and rotate it until Sweden is in the middle. It'll look something like this (credit: Google):
Molot's answer (which is the mathematically pure answer, go vote for it) brings out that tall things slightly beyond the edge of the hemisphere defined by Sweden acting as the pole or center of the circle will see Sweden if it's high enough. But below that it's geometrically impossible — there's a planet in the way.
And that example, above, assumes we send Sweden so far away that the entirety of its defined hemisphere can see it. If we limit that height to keep it within the atmosphere, then the total area that can see it at any height within the atmosphere (~300 mi thick) is something more like this:
Compared to the radius of the Earth, our atmosphere is proverbially as thick as a sheet of paper. Variations in land height don't matter, either, as those variations are less than the dimples on a golf ball compared to the radius of the Earth. When you're talking about the entire planet, it can be reasonably modeled as a smooth surface.
Conclusion
There is no height within the atmosphere that would allow Sweden to be seen from any point within what we would normally call the Southern Hemisphere. There is a height where it could be seen by a couple of points in the Southern Hemisphere (see that area below the dashed line in Africa? Yeah, the dashed line is the equator), but Sweden would be so far outside Earth's atmosphere that it likely doesn't matter for your story.
This assumes you do not subscribe to the Flat Earth theory. I don't.
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
$endgroup$
3
$begingroup$
I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
$endgroup$
– Innovine
May 28 at 19:41
6
$begingroup$
Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
$endgroup$
– Cœur
May 29 at 1:43
1
$begingroup$
@Cœur 😀 Touché.
$endgroup$
– JBH
May 29 at 1:45
2
$begingroup$
By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
$endgroup$
– reirab
May 29 at 18:29
2
$begingroup$
This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
$endgroup$
– Fattie
May 30 at 12:32
|
show 2 more comments
$begingroup$
There is no atmospheric height any object of any size can be, perpendicular to the location of Sweden, that could be seen by anyone standing anywhere in South Africa.
I don't need math to prove this. Just take out a piece of paper, draw a circle, and use a ruler to start drawing lines. That combination cannot be done.
Alternatively, visit maps.google.com, scroll out until you can see the whole planet, and rotate it until Sweden is in the middle. It'll look something like this (credit: Google):
Molot's answer (which is the mathematically pure answer, go vote for it) brings out that tall things slightly beyond the edge of the hemisphere defined by Sweden acting as the pole or center of the circle will see Sweden if it's high enough. But below that it's geometrically impossible — there's a planet in the way.
And that example, above, assumes we send Sweden so far away that the entirety of its defined hemisphere can see it. If we limit that height to keep it within the atmosphere, then the total area that can see it at any height within the atmosphere (~300 mi thick) is something more like this:
Compared to the radius of the Earth, our atmosphere is proverbially as thick as a sheet of paper. Variations in land height don't matter, either, as those variations are less than the dimples on a golf ball compared to the radius of the Earth. When you're talking about the entire planet, it can be reasonably modeled as a smooth surface.
Conclusion
There is no height within the atmosphere that would allow Sweden to be seen from any point within what we would normally call the Southern Hemisphere. There is a height where it could be seen by a couple of points in the Southern Hemisphere (see that area below the dashed line in Africa? Yeah, the dashed line is the equator), but Sweden would be so far outside Earth's atmosphere that it likely doesn't matter for your story.
This assumes you do not subscribe to the Flat Earth theory. I don't.
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
$endgroup$
3
$begingroup$
I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
$endgroup$
– Innovine
May 28 at 19:41
6
$begingroup$
Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
$endgroup$
– Cœur
May 29 at 1:43
1
$begingroup$
@Cœur 😀 Touché.
$endgroup$
– JBH
May 29 at 1:45
2
$begingroup$
By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
$endgroup$
– reirab
May 29 at 18:29
2
$begingroup$
This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
$endgroup$
– Fattie
May 30 at 12:32
|
show 2 more comments
$begingroup$
There is no atmospheric height any object of any size can be, perpendicular to the location of Sweden, that could be seen by anyone standing anywhere in South Africa.
I don't need math to prove this. Just take out a piece of paper, draw a circle, and use a ruler to start drawing lines. That combination cannot be done.
Alternatively, visit maps.google.com, scroll out until you can see the whole planet, and rotate it until Sweden is in the middle. It'll look something like this (credit: Google):
Molot's answer (which is the mathematically pure answer, go vote for it) brings out that tall things slightly beyond the edge of the hemisphere defined by Sweden acting as the pole or center of the circle will see Sweden if it's high enough. But below that it's geometrically impossible — there's a planet in the way.
And that example, above, assumes we send Sweden so far away that the entirety of its defined hemisphere can see it. If we limit that height to keep it within the atmosphere, then the total area that can see it at any height within the atmosphere (~300 mi thick) is something more like this:
Compared to the radius of the Earth, our atmosphere is proverbially as thick as a sheet of paper. Variations in land height don't matter, either, as those variations are less than the dimples on a golf ball compared to the radius of the Earth. When you're talking about the entire planet, it can be reasonably modeled as a smooth surface.
Conclusion
There is no height within the atmosphere that would allow Sweden to be seen from any point within what we would normally call the Southern Hemisphere. There is a height where it could be seen by a couple of points in the Southern Hemisphere (see that area below the dashed line in Africa? Yeah, the dashed line is the equator), but Sweden would be so far outside Earth's atmosphere that it likely doesn't matter for your story.
This assumes you do not subscribe to the Flat Earth theory. I don't.
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
$endgroup$
There is no atmospheric height any object of any size can be, perpendicular to the location of Sweden, that could be seen by anyone standing anywhere in South Africa.
I don't need math to prove this. Just take out a piece of paper, draw a circle, and use a ruler to start drawing lines. That combination cannot be done.
Alternatively, visit maps.google.com, scroll out until you can see the whole planet, and rotate it until Sweden is in the middle. It'll look something like this (credit: Google):
Molot's answer (which is the mathematically pure answer, go vote for it) brings out that tall things slightly beyond the edge of the hemisphere defined by Sweden acting as the pole or center of the circle will see Sweden if it's high enough. But below that it's geometrically impossible — there's a planet in the way.
And that example, above, assumes we send Sweden so far away that the entirety of its defined hemisphere can see it. If we limit that height to keep it within the atmosphere, then the total area that can see it at any height within the atmosphere (~300 mi thick) is something more like this:
Compared to the radius of the Earth, our atmosphere is proverbially as thick as a sheet of paper. Variations in land height don't matter, either, as those variations are less than the dimples on a golf ball compared to the radius of the Earth. When you're talking about the entire planet, it can be reasonably modeled as a smooth surface.
Conclusion
There is no height within the atmosphere that would allow Sweden to be seen from any point within what we would normally call the Southern Hemisphere. There is a height where it could be seen by a couple of points in the Southern Hemisphere (see that area below the dashed line in Africa? Yeah, the dashed line is the equator), but Sweden would be so far outside Earth's atmosphere that it likely doesn't matter for your story.
This assumes you do not subscribe to the Flat Earth theory. I don't.
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
edited May 28 at 16:45
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
answered May 28 at 16:12
JBHJBH
57.5k9 gold badges131 silver badges279 bronze badges
57.5k9 gold badges131 silver badges279 bronze badges
3
$begingroup$
I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
$endgroup$
– Innovine
May 28 at 19:41
6
$begingroup$
Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
$endgroup$
– Cœur
May 29 at 1:43
1
$begingroup$
@Cœur 😀 Touché.
$endgroup$
– JBH
May 29 at 1:45
2
$begingroup$
By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
$endgroup$
– reirab
May 29 at 18:29
2
$begingroup$
This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
$endgroup$
– Fattie
May 30 at 12:32
|
show 2 more comments
3
$begingroup$
I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
$endgroup$
– Innovine
May 28 at 19:41
6
$begingroup$
Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
$endgroup$
– Cœur
May 29 at 1:43
1
$begingroup$
@Cœur 😀 Touché.
$endgroup$
– JBH
May 29 at 1:45
2
$begingroup$
By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
$endgroup$
– reirab
May 29 at 18:29
2
$begingroup$
This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
$endgroup$
– Fattie
May 30 at 12:32
3
3
$begingroup$
I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
$endgroup$
– Innovine
May 28 at 19:41
$begingroup$
I don't think it'll be visible at any height. Just like the north star isn't visible, and that's so far away it might as well be at infinity. Molots maths is wrong.
$endgroup$
– Innovine
May 28 at 19:41
6
6
$begingroup$
Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
$endgroup$
– Cœur
May 29 at 1:43
$begingroup$
Geometry is part of mathematics. So when you're drawing lines, you're doing maths.
$endgroup$
– Cœur
May 29 at 1:43
1
1
$begingroup$
@Cœur 😀 Touché.
$endgroup$
– JBH
May 29 at 1:45
$begingroup$
@Cœur 😀 Touché.
$endgroup$
– JBH
May 29 at 1:45
2
2
$begingroup$
By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
$endgroup$
– reirab
May 29 at 18:29
$begingroup$
By the way, while 300 miles is still technically inside the atmosphere in the sense of being inside the thermosphere, it's very much not "inside the atmosphere" in the sense of having useful amounts of air pressure for purposes such as flying or human survival without a space suit. For comparison, the International Space Station orbits at around 250 miles. 300 miles is, for most practical purposes, outer space. A common definition of outer space is above the Kármán line, which is around 60 miles. Ability for a human to survive ends before 10 miles.
$endgroup$
– reirab
May 29 at 18:29
2
2
$begingroup$
This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
$endgroup$
– Fattie
May 30 at 12:32
$begingroup$
This is the correct answer. Simply look at a map centered on Sweden. High-Sweden will, obviously, only ever be visible from the half of the world you see on the map. It's just that simple.
$endgroup$
– Fattie
May 30 at 12:32
|
show 2 more comments
$begingroup$
This answer is fundamentally flawed in that it assumes that the 'horizon' formula is valid for any value of h.
Unfortunately, this is simply not true, and as the image shows, it is only a valid approximation when h is much less than RE
(RE >> h)
If you can see something depends on how high you are, and how high is what you look at:
Africa is reasonably flat*, so assume a person 2 meters tall (or he may be standing on something ;) )
You can look and fiddle with the numbers here with nice calculator that does the math for you. With Observer at 2 meters and object at the edge of space (100km) visual horizon is at the distance of 1134km. Distance from Sweden to South Africa is 10058 km, so it is about ten times too far to have a line of sight.
To get the line of sight, you will need to put Sweden at height of about 7922km. That's outside atmosphere, but inside Van Allen radiation belt, so long time life support would be hard for at least these two reasons.
* Of course for more precise result you should include the mean or max elevation of South Africa, and decide where, exactly, your observer is placed to account for mountains, valleys, trees and so on. The above calculation is meant as an estimate, and to show a way how to calculate it yourself.
The question says only "from South Africa" so we can assume optimal point of observation, Thabana Ntlenyana 3482 meters high (+2m person) for the flying height of "mere" 3452km. No trees or other mountains to significantly change this altitude.
$endgroup$
3
$begingroup$
@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
$endgroup$
– Mołot
May 28 at 14:41
3
$begingroup$
In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
$endgroup$
– Brythan
May 28 at 14:47
39
$begingroup$
I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
$endgroup$
– brhans
May 28 at 18:42
2
$begingroup$
To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
$endgroup$
– Michael Seifert
May 28 at 18:48
3
$begingroup$
Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
$endgroup$
– Mark
May 29 at 20:13
|
show 9 more comments
$begingroup$
This answer is fundamentally flawed in that it assumes that the 'horizon' formula is valid for any value of h.
Unfortunately, this is simply not true, and as the image shows, it is only a valid approximation when h is much less than RE
(RE >> h)
If you can see something depends on how high you are, and how high is what you look at:
Africa is reasonably flat*, so assume a person 2 meters tall (or he may be standing on something ;) )
You can look and fiddle with the numbers here with nice calculator that does the math for you. With Observer at 2 meters and object at the edge of space (100km) visual horizon is at the distance of 1134km. Distance from Sweden to South Africa is 10058 km, so it is about ten times too far to have a line of sight.
To get the line of sight, you will need to put Sweden at height of about 7922km. That's outside atmosphere, but inside Van Allen radiation belt, so long time life support would be hard for at least these two reasons.
* Of course for more precise result you should include the mean or max elevation of South Africa, and decide where, exactly, your observer is placed to account for mountains, valleys, trees and so on. The above calculation is meant as an estimate, and to show a way how to calculate it yourself.
The question says only "from South Africa" so we can assume optimal point of observation, Thabana Ntlenyana 3482 meters high (+2m person) for the flying height of "mere" 3452km. No trees or other mountains to significantly change this altitude.
$endgroup$
3
$begingroup$
@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
$endgroup$
– Mołot
May 28 at 14:41
3
$begingroup$
In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
$endgroup$
– Brythan
May 28 at 14:47
39
$begingroup$
I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
$endgroup$
– brhans
May 28 at 18:42
2
$begingroup$
To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
$endgroup$
– Michael Seifert
May 28 at 18:48
3
$begingroup$
Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
$endgroup$
– Mark
May 29 at 20:13
|
show 9 more comments
$begingroup$
This answer is fundamentally flawed in that it assumes that the 'horizon' formula is valid for any value of h.
Unfortunately, this is simply not true, and as the image shows, it is only a valid approximation when h is much less than RE
(RE >> h)
If you can see something depends on how high you are, and how high is what you look at:
Africa is reasonably flat*, so assume a person 2 meters tall (or he may be standing on something ;) )
You can look and fiddle with the numbers here with nice calculator that does the math for you. With Observer at 2 meters and object at the edge of space (100km) visual horizon is at the distance of 1134km. Distance from Sweden to South Africa is 10058 km, so it is about ten times too far to have a line of sight.
To get the line of sight, you will need to put Sweden at height of about 7922km. That's outside atmosphere, but inside Van Allen radiation belt, so long time life support would be hard for at least these two reasons.
* Of course for more precise result you should include the mean or max elevation of South Africa, and decide where, exactly, your observer is placed to account for mountains, valleys, trees and so on. The above calculation is meant as an estimate, and to show a way how to calculate it yourself.
The question says only "from South Africa" so we can assume optimal point of observation, Thabana Ntlenyana 3482 meters high (+2m person) for the flying height of "mere" 3452km. No trees or other mountains to significantly change this altitude.
$endgroup$
This answer is fundamentally flawed in that it assumes that the 'horizon' formula is valid for any value of h.
Unfortunately, this is simply not true, and as the image shows, it is only a valid approximation when h is much less than RE
(RE >> h)
If you can see something depends on how high you are, and how high is what you look at:
Africa is reasonably flat*, so assume a person 2 meters tall (or he may be standing on something ;) )
You can look and fiddle with the numbers here with nice calculator that does the math for you. With Observer at 2 meters and object at the edge of space (100km) visual horizon is at the distance of 1134km. Distance from Sweden to South Africa is 10058 km, so it is about ten times too far to have a line of sight.
To get the line of sight, you will need to put Sweden at height of about 7922km. That's outside atmosphere, but inside Van Allen radiation belt, so long time life support would be hard for at least these two reasons.
* Of course for more precise result you should include the mean or max elevation of South Africa, and decide where, exactly, your observer is placed to account for mountains, valleys, trees and so on. The above calculation is meant as an estimate, and to show a way how to calculate it yourself.
The question says only "from South Africa" so we can assume optimal point of observation, Thabana Ntlenyana 3482 meters high (+2m person) for the flying height of "mere" 3452km. No trees or other mountains to significantly change this altitude.
edited Jun 1 at 15:32
community wiki
4 revs, 2 users 83%
Mołot
3
$begingroup$
@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
$endgroup$
– Mołot
May 28 at 14:41
3
$begingroup$
In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
$endgroup$
– Brythan
May 28 at 14:47
39
$begingroup$
I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
$endgroup$
– brhans
May 28 at 18:42
2
$begingroup$
To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
$endgroup$
– Michael Seifert
May 28 at 18:48
3
$begingroup$
Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
$endgroup$
– Mark
May 29 at 20:13
|
show 9 more comments
3
$begingroup$
@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
$endgroup$
– Mołot
May 28 at 14:41
3
$begingroup$
In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
$endgroup$
– Brythan
May 28 at 14:47
39
$begingroup$
I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
$endgroup$
– brhans
May 28 at 18:42
2
$begingroup$
To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
$endgroup$
– Michael Seifert
May 28 at 18:48
3
$begingroup$
Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
$endgroup$
– Mark
May 29 at 20:13
3
3
$begingroup$
@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
$endgroup$
– Mołot
May 28 at 14:41
$begingroup$
@Brythan do you really believe that trees will have a significant impact on the 7922km needed and the fact that it is higher than the edge of the atmosphere? Also, South Africa has mean elevation of 1,200 m (3,900 ft), and at least 40% of the surface is at a higher elevation, so inhabitants could probably see above any mountains between them and Flying Sweden if they wanted - that's the sense behind my "reasonably flat" assertion. In this context, it is. Sure, a simplification, but one that works.
$endgroup$
– Mołot
May 28 at 14:41
3
3
$begingroup$
In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
$endgroup$
– Brythan
May 28 at 14:47
$begingroup$
In the context where two meters of height matters, a ten meter tree matters. And the problem isn't mountains far away. It's the nearby hills. If you are one kilometer above sea level plus two meters of height and there is a hill a half kilometer away that is also one kilometer high with thirty meters of trees at the top, your 1002 meters is actually below the 1030 meters. You are only "simplifying" the parts that hurt your argument. You are bringing in the complexity that helps it. I'm fine with ignoring tree and hill height if you ignore perspective height. But not half including
$endgroup$
– Brythan
May 28 at 14:47
39
39
$begingroup$
I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
$endgroup$
– brhans
May 28 at 18:42
$begingroup$
I think you've failed to account for the fact that the horizon formula you're using (and the online calculator) is only an approximation, and specifically only works for RE >> h (as noted in the image you've used). You're plugging in an h value (7922km) which is not only not much less then RE, it is in fact greater than RE.
$endgroup$
– brhans
May 28 at 18:42
2
2
$begingroup$
To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
$endgroup$
– Michael Seifert
May 28 at 18:48
$begingroup$
To see that @brhans is correct, and that the calculator is using an invalid approximation, try plugging in h_1 = some ridiculously large number (like 10,000,000 km, or any number much greater than the radius of the Earth) and h_2 = 0. Depending on how you define "distance to the horizon" in this case, you should either get a result that is the line-of-sight distance (very close to h_1) or something that indicates that you can see half of the Earth (a number very close to 1/4 the circumference of the Earth, or about 10,000 km.) But you don't.
$endgroup$
– Michael Seifert
May 28 at 18:48
3
3
$begingroup$
Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
$endgroup$
– Mark
May 29 at 20:13
$begingroup$
Unfortunately for the correctness of this answer, you've run into one of the edge cases where the approximation you use in step 1 doesn't work.
$endgroup$
– Mark
May 29 at 20:13
|
show 9 more comments
$begingroup$
If you imagine an idealised planet with a star at an infinite distance, how much of the planet can see it? The answer is half: the star projects a cone that rests on an equatorial line perpendicular to the direction it lies in. Those standing on that equatorial line are looking along the tangent to the curve of the surface of the planet.
So the basic question is, what is the minimum distance at which this is still possible, or rather, given two opposite points on a circle, where do their tangents intersect? This is effectively squaring the circle that forms said equator, which simplifies down to the square root of 2r^2 (minus the radius of the Earth if you want the height above it).
In the case of Earth, that comes to 2,639km, however there are other factors regarding the shape of the orbit that might have an effect. Obviously, given an orbit it will not be visible at all times to the whole planet. However, if that orbit is geostationary (e.g. proportional to the rotational speed of the planet) it will only be visible to half the planet when above said height.
Given orbits are elliptical, it's also possible that it is not above that height the entire orbit. If the orbit resonates with the rotation of the planet and has a low point below said height, this could create dead spots that will never see it in the sky.
Finally, there's always the possibility of mountain ranges and atmosphere practically limiting visibility along the line of the tangent.
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
$endgroup$
add a comment
|
$begingroup$
If you imagine an idealised planet with a star at an infinite distance, how much of the planet can see it? The answer is half: the star projects a cone that rests on an equatorial line perpendicular to the direction it lies in. Those standing on that equatorial line are looking along the tangent to the curve of the surface of the planet.
So the basic question is, what is the minimum distance at which this is still possible, or rather, given two opposite points on a circle, where do their tangents intersect? This is effectively squaring the circle that forms said equator, which simplifies down to the square root of 2r^2 (minus the radius of the Earth if you want the height above it).
In the case of Earth, that comes to 2,639km, however there are other factors regarding the shape of the orbit that might have an effect. Obviously, given an orbit it will not be visible at all times to the whole planet. However, if that orbit is geostationary (e.g. proportional to the rotational speed of the planet) it will only be visible to half the planet when above said height.
Given orbits are elliptical, it's also possible that it is not above that height the entire orbit. If the orbit resonates with the rotation of the planet and has a low point below said height, this could create dead spots that will never see it in the sky.
Finally, there's always the possibility of mountain ranges and atmosphere practically limiting visibility along the line of the tangent.
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
$endgroup$
add a comment
|
$begingroup$
If you imagine an idealised planet with a star at an infinite distance, how much of the planet can see it? The answer is half: the star projects a cone that rests on an equatorial line perpendicular to the direction it lies in. Those standing on that equatorial line are looking along the tangent to the curve of the surface of the planet.
So the basic question is, what is the minimum distance at which this is still possible, or rather, given two opposite points on a circle, where do their tangents intersect? This is effectively squaring the circle that forms said equator, which simplifies down to the square root of 2r^2 (minus the radius of the Earth if you want the height above it).
In the case of Earth, that comes to 2,639km, however there are other factors regarding the shape of the orbit that might have an effect. Obviously, given an orbit it will not be visible at all times to the whole planet. However, if that orbit is geostationary (e.g. proportional to the rotational speed of the planet) it will only be visible to half the planet when above said height.
Given orbits are elliptical, it's also possible that it is not above that height the entire orbit. If the orbit resonates with the rotation of the planet and has a low point below said height, this could create dead spots that will never see it in the sky.
Finally, there's always the possibility of mountain ranges and atmosphere practically limiting visibility along the line of the tangent.
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
$endgroup$
If you imagine an idealised planet with a star at an infinite distance, how much of the planet can see it? The answer is half: the star projects a cone that rests on an equatorial line perpendicular to the direction it lies in. Those standing on that equatorial line are looking along the tangent to the curve of the surface of the planet.
So the basic question is, what is the minimum distance at which this is still possible, or rather, given two opposite points on a circle, where do their tangents intersect? This is effectively squaring the circle that forms said equator, which simplifies down to the square root of 2r^2 (minus the radius of the Earth if you want the height above it).
In the case of Earth, that comes to 2,639km, however there are other factors regarding the shape of the orbit that might have an effect. Obviously, given an orbit it will not be visible at all times to the whole planet. However, if that orbit is geostationary (e.g. proportional to the rotational speed of the planet) it will only be visible to half the planet when above said height.
Given orbits are elliptical, it's also possible that it is not above that height the entire orbit. If the orbit resonates with the rotation of the planet and has a low point below said height, this could create dead spots that will never see it in the sky.
Finally, there's always the possibility of mountain ranges and atmosphere practically limiting visibility along the line of the tangent.
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
answered May 31 at 11:44
DanikovDanikov
4512 silver badges2 bronze badges
4512 silver badges2 bronze badges
add a comment
|
add a comment
|
$begingroup$
You have three questions here.
1. How high can a geostationary Sweden be visible from South Africa?
Answer: 68,100 km
This is a question that humanity has pondered since we first climbed down from the trees.
The other answers are correct that if two points are more than 90 degrees apart on the globe, they cannot see each other at any height (unless the earth has a big groove in it or atmospheric refraction does something funny, but we don't have to consider that in this answer).
The correct way to calculate angular distances between two points on the globe is to use the haversine formula. This takes both latitude and longitude into account. I'm feeling a little lazy today, so I will just take the distance measured on Google Maps, divide by 40,075 kilometers and multiply by 360 degrees. This is equivalent, and I've checked that the math lines up.
If Sweden goes straight up from the point of view of its center point, the important location of Sweden is Flataklocken at 62°23′15″N 16°19′32″E.
I found a point near South Africa's Zimbabwe that is 9,472 km away from Flataklocken, which comes out to 85.09 degrees.
You're in luck! You can see Sweden from South Africa. It will be low in the sky, within 5 degrees of the horizon. It will be brighter than the full moon since it's a lot closer.
If you intend to actually do this, I'd recommend buying a hot air balloon to take South Africans up to view.
Some math:
a = 85.09 degrees (this is the angular distance between South Africa and central Sweden)
r = 6371 km (this is the radius of the earth)
find h, the height of Sweden
cos(a) = r / (r + h)
Therefore
h = r (-1 + 1 / cos(a)) = 68,100 kilometers.
This is over 10 times earth's radius. It's far outside earth's atmosphere, but low enough it's not going to crash into the moon.
2. If Sweden were floating in a geostationary position in the upper atmosphere, how far would it be visible?
Answer: 2400 km
Earth's atmosphere only goes up 480 km, so if you want Sweden to stay in a geostationary position in the atmosphere, it won't be visible from South Africa.
If you want the people of Sweden to have enough air to breathe, it will have to be even lower. I'm not going to run the numbers for that, because that's not the question you asked.
The equation is the same, except in this case, we know h and r, and we're trying to find a.
r = 6371 km (this is still the radius of the earth)
h = 480 km (Sweden floating in the edge of the atmosphere)
cos(a) = r / (r + h)
Therefore
a = arccos(r / (r + h)) = 21.57 degrees.
This comes out to a distance of 2400 kilometers.
This isn't far enough to see from the southern hemisphere, but you would still be able to see southern Sweden from parts of Algeria and Tunisia. As far as I'm concerned, that's good news!
3. Can a Sweden orbiting in the atmosphere be visible from South Africa?
Answer: No
If you can make Sweden orbit at this low elevation, it can be visible from every point in the world. But you can't make Sweden orbit in the atmosphere.
Orbit means something is spinning around a world without rockets or thrusters because it is going so fast that gravity can't pull it down. You can calculate the orbital speed for any height in a vacuum. Unfortunately, orbit is not possible within the atmosphere because the air resistance will eventually slow Sweden down so that it will crash into the earth.
Don't despair. You can still make Sweden fly. You'll need to buy big jets or something else that uses energy to move it around. I don't know what your budget is, but I think it's still worthwhile. Think of the spectators in South Africa.
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
$endgroup$
1
$begingroup$
Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
$endgroup$
– Jetpack
Jun 11 at 0:11
$begingroup$
That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
$endgroup$
– Pangolin
Jun 19 at 13:54
add a comment
|
$begingroup$
You have three questions here.
1. How high can a geostationary Sweden be visible from South Africa?
Answer: 68,100 km
This is a question that humanity has pondered since we first climbed down from the trees.
The other answers are correct that if two points are more than 90 degrees apart on the globe, they cannot see each other at any height (unless the earth has a big groove in it or atmospheric refraction does something funny, but we don't have to consider that in this answer).
The correct way to calculate angular distances between two points on the globe is to use the haversine formula. This takes both latitude and longitude into account. I'm feeling a little lazy today, so I will just take the distance measured on Google Maps, divide by 40,075 kilometers and multiply by 360 degrees. This is equivalent, and I've checked that the math lines up.
If Sweden goes straight up from the point of view of its center point, the important location of Sweden is Flataklocken at 62°23′15″N 16°19′32″E.
I found a point near South Africa's Zimbabwe that is 9,472 km away from Flataklocken, which comes out to 85.09 degrees.
You're in luck! You can see Sweden from South Africa. It will be low in the sky, within 5 degrees of the horizon. It will be brighter than the full moon since it's a lot closer.
If you intend to actually do this, I'd recommend buying a hot air balloon to take South Africans up to view.
Some math:
a = 85.09 degrees (this is the angular distance between South Africa and central Sweden)
r = 6371 km (this is the radius of the earth)
find h, the height of Sweden
cos(a) = r / (r + h)
Therefore
h = r (-1 + 1 / cos(a)) = 68,100 kilometers.
This is over 10 times earth's radius. It's far outside earth's atmosphere, but low enough it's not going to crash into the moon.
2. If Sweden were floating in a geostationary position in the upper atmosphere, how far would it be visible?
Answer: 2400 km
Earth's atmosphere only goes up 480 km, so if you want Sweden to stay in a geostationary position in the atmosphere, it won't be visible from South Africa.
If you want the people of Sweden to have enough air to breathe, it will have to be even lower. I'm not going to run the numbers for that, because that's not the question you asked.
The equation is the same, except in this case, we know h and r, and we're trying to find a.
r = 6371 km (this is still the radius of the earth)
h = 480 km (Sweden floating in the edge of the atmosphere)
cos(a) = r / (r + h)
Therefore
a = arccos(r / (r + h)) = 21.57 degrees.
This comes out to a distance of 2400 kilometers.
This isn't far enough to see from the southern hemisphere, but you would still be able to see southern Sweden from parts of Algeria and Tunisia. As far as I'm concerned, that's good news!
3. Can a Sweden orbiting in the atmosphere be visible from South Africa?
Answer: No
If you can make Sweden orbit at this low elevation, it can be visible from every point in the world. But you can't make Sweden orbit in the atmosphere.
Orbit means something is spinning around a world without rockets or thrusters because it is going so fast that gravity can't pull it down. You can calculate the orbital speed for any height in a vacuum. Unfortunately, orbit is not possible within the atmosphere because the air resistance will eventually slow Sweden down so that it will crash into the earth.
Don't despair. You can still make Sweden fly. You'll need to buy big jets or something else that uses energy to move it around. I don't know what your budget is, but I think it's still worthwhile. Think of the spectators in South Africa.
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
$endgroup$
1
$begingroup$
Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
$endgroup$
– Jetpack
Jun 11 at 0:11
$begingroup$
That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
$endgroup$
– Pangolin
Jun 19 at 13:54
add a comment
|
$begingroup$
You have three questions here.
1. How high can a geostationary Sweden be visible from South Africa?
Answer: 68,100 km
This is a question that humanity has pondered since we first climbed down from the trees.
The other answers are correct that if two points are more than 90 degrees apart on the globe, they cannot see each other at any height (unless the earth has a big groove in it or atmospheric refraction does something funny, but we don't have to consider that in this answer).
The correct way to calculate angular distances between two points on the globe is to use the haversine formula. This takes both latitude and longitude into account. I'm feeling a little lazy today, so I will just take the distance measured on Google Maps, divide by 40,075 kilometers and multiply by 360 degrees. This is equivalent, and I've checked that the math lines up.
If Sweden goes straight up from the point of view of its center point, the important location of Sweden is Flataklocken at 62°23′15″N 16°19′32″E.
I found a point near South Africa's Zimbabwe that is 9,472 km away from Flataklocken, which comes out to 85.09 degrees.
You're in luck! You can see Sweden from South Africa. It will be low in the sky, within 5 degrees of the horizon. It will be brighter than the full moon since it's a lot closer.
If you intend to actually do this, I'd recommend buying a hot air balloon to take South Africans up to view.
Some math:
a = 85.09 degrees (this is the angular distance between South Africa and central Sweden)
r = 6371 km (this is the radius of the earth)
find h, the height of Sweden
cos(a) = r / (r + h)
Therefore
h = r (-1 + 1 / cos(a)) = 68,100 kilometers.
This is over 10 times earth's radius. It's far outside earth's atmosphere, but low enough it's not going to crash into the moon.
2. If Sweden were floating in a geostationary position in the upper atmosphere, how far would it be visible?
Answer: 2400 km
Earth's atmosphere only goes up 480 km, so if you want Sweden to stay in a geostationary position in the atmosphere, it won't be visible from South Africa.
If you want the people of Sweden to have enough air to breathe, it will have to be even lower. I'm not going to run the numbers for that, because that's not the question you asked.
The equation is the same, except in this case, we know h and r, and we're trying to find a.
r = 6371 km (this is still the radius of the earth)
h = 480 km (Sweden floating in the edge of the atmosphere)
cos(a) = r / (r + h)
Therefore
a = arccos(r / (r + h)) = 21.57 degrees.
This comes out to a distance of 2400 kilometers.
This isn't far enough to see from the southern hemisphere, but you would still be able to see southern Sweden from parts of Algeria and Tunisia. As far as I'm concerned, that's good news!
3. Can a Sweden orbiting in the atmosphere be visible from South Africa?
Answer: No
If you can make Sweden orbit at this low elevation, it can be visible from every point in the world. But you can't make Sweden orbit in the atmosphere.
Orbit means something is spinning around a world without rockets or thrusters because it is going so fast that gravity can't pull it down. You can calculate the orbital speed for any height in a vacuum. Unfortunately, orbit is not possible within the atmosphere because the air resistance will eventually slow Sweden down so that it will crash into the earth.
Don't despair. You can still make Sweden fly. You'll need to buy big jets or something else that uses energy to move it around. I don't know what your budget is, but I think it's still worthwhile. Think of the spectators in South Africa.
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
$endgroup$
You have three questions here.
1. How high can a geostationary Sweden be visible from South Africa?
Answer: 68,100 km
This is a question that humanity has pondered since we first climbed down from the trees.
The other answers are correct that if two points are more than 90 degrees apart on the globe, they cannot see each other at any height (unless the earth has a big groove in it or atmospheric refraction does something funny, but we don't have to consider that in this answer).
The correct way to calculate angular distances between two points on the globe is to use the haversine formula. This takes both latitude and longitude into account. I'm feeling a little lazy today, so I will just take the distance measured on Google Maps, divide by 40,075 kilometers and multiply by 360 degrees. This is equivalent, and I've checked that the math lines up.
If Sweden goes straight up from the point of view of its center point, the important location of Sweden is Flataklocken at 62°23′15″N 16°19′32″E.
I found a point near South Africa's Zimbabwe that is 9,472 km away from Flataklocken, which comes out to 85.09 degrees.
You're in luck! You can see Sweden from South Africa. It will be low in the sky, within 5 degrees of the horizon. It will be brighter than the full moon since it's a lot closer.
If you intend to actually do this, I'd recommend buying a hot air balloon to take South Africans up to view.
Some math:
a = 85.09 degrees (this is the angular distance between South Africa and central Sweden)
r = 6371 km (this is the radius of the earth)
find h, the height of Sweden
cos(a) = r / (r + h)
Therefore
h = r (-1 + 1 / cos(a)) = 68,100 kilometers.
This is over 10 times earth's radius. It's far outside earth's atmosphere, but low enough it's not going to crash into the moon.
2. If Sweden were floating in a geostationary position in the upper atmosphere, how far would it be visible?
Answer: 2400 km
Earth's atmosphere only goes up 480 km, so if you want Sweden to stay in a geostationary position in the atmosphere, it won't be visible from South Africa.
If you want the people of Sweden to have enough air to breathe, it will have to be even lower. I'm not going to run the numbers for that, because that's not the question you asked.
The equation is the same, except in this case, we know h and r, and we're trying to find a.
r = 6371 km (this is still the radius of the earth)
h = 480 km (Sweden floating in the edge of the atmosphere)
cos(a) = r / (r + h)
Therefore
a = arccos(r / (r + h)) = 21.57 degrees.
This comes out to a distance of 2400 kilometers.
This isn't far enough to see from the southern hemisphere, but you would still be able to see southern Sweden from parts of Algeria and Tunisia. As far as I'm concerned, that's good news!
3. Can a Sweden orbiting in the atmosphere be visible from South Africa?
Answer: No
If you can make Sweden orbit at this low elevation, it can be visible from every point in the world. But you can't make Sweden orbit in the atmosphere.
Orbit means something is spinning around a world without rockets or thrusters because it is going so fast that gravity can't pull it down. You can calculate the orbital speed for any height in a vacuum. Unfortunately, orbit is not possible within the atmosphere because the air resistance will eventually slow Sweden down so that it will crash into the earth.
Don't despair. You can still make Sweden fly. You'll need to buy big jets or something else that uses energy to move it around. I don't know what your budget is, but I think it's still worthwhile. Think of the spectators in South Africa.
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
edited Jun 10 at 20:04
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
answered Jun 10 at 17:05
JetpackJetpack
6093 silver badges7 bronze badges
6093 silver badges7 bronze badges
1
$begingroup$
Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
$endgroup$
– Jetpack
Jun 11 at 0:11
$begingroup$
That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
$endgroup$
– Pangolin
Jun 19 at 13:54
add a comment
|
1
$begingroup$
Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
$endgroup$
– Jetpack
Jun 11 at 0:11
$begingroup$
That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
$endgroup$
– Pangolin
Jun 19 at 13:54
1
1
$begingroup$
Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
$endgroup$
– Jetpack
Jun 11 at 0:11
$begingroup$
Now that I look into this, 480 km is higher than the International Space Station, so orbits are possible there. Should I redo the math for 100 km, another definition used for the edge of the atmosphere?
$endgroup$
– Jetpack
Jun 11 at 0:11
$begingroup$
That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
$endgroup$
– Pangolin
Jun 19 at 13:54
$begingroup$
That was amazing – by far my favourite answer. Thank you for the formulas especially. No need to redo the math for 100 km though, since I already decided to have Sweden orbit at a height similar to that of the ISS, so roughly 400 km :)
$endgroup$
– Pangolin
Jun 19 at 13:54
add a comment
|
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$begingroup$
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– L.Dutch♦
Jun 1 at 3:16