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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 the 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 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.

In light of the response here, I decided to just have my floating island properly orbit the planet somewhere within the thermosphere and think of a magical reason that would allow for life to exist there.

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.

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.

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 R_E

(R_E >> 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.

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.