Based on the fact that the curvature would be vastly different than Earth? On Earth, we can see about 3 miles at sea-level, so I’m assuming we’d be able to see a good bit farther on a Super Earth. Would we even notice a difference?
Yes- the horizon distance is determined by the height of the observer and the radius of the planet.
The gravity might be too high to stand upright.
Well take a super-Earth of 1.4M[sub]e[/sub] with a smilar density. The resulting R[sub]se[/sub]=1.4[sup]1/3[/sup]R[sub]3[/sub] or basically 1.1R[sub]e[/sub].
Even a 2.5M[sub]e[/sub] with the same density would give 1.35R[sub]e[/sub].
Now the horizon distance (d), assuming a smooth sphere, is d = sqrt[a*(2R[sub]3[/sub]+a)] where a is the height of a person and we’ll say R[sub]3[/sub]=6,300,000m and a=2m that gives a d of 5019 m
With d[sub]se[/sub]=d[sub]e[/sub]sqrt(2R[sub]se[/sub]+a)/sqrt(2*R[sub]e[/sub]+a)
For the 1.4M[sub]e[/sub] planet that gives 1.04d[sub]e[/sub] or 5216m and the 2.5 version 1.16d[sub]e[/sub] or 5822.
Note that we’ve ignored the influence of a denser atmosphere on light paths.
No because the Earth is flat.
Just kidding.
I think the equation bump linked to is wrong. It should be sqrt(2Rh + h[sup]2[/sup]) where R is the radius of the planet and h is the hight of the observer. The equation bump link to excludes the h[sup]2[/sup] because compared to the 2Rh it is insignificant.
The Apollo astronauts didn’t mention anything about the horizon on the moon being noticeable nearer. And the moon is a hell of a lot smaller than Earth than a SuperEarth is to Ordinary Earth.
Moon horizon would be about .52R[sub]e[/sub] or 2609m. With their helmets, stark light, landscape and lack of atmosphere I’m not sure you’d notice. Interesting question though.
The distance to the horizon would definitely change, but I’m not sure that that would be something you’d be able to distinguish by eye. You’d need to have some features whose distance you could tell in some other way (for instance, something of known size).
No, but they had a lot of trouble gauging distances because of the lack of both atmosphere and familiar landmarks for scale. There’s Interesting footage from Apollo 16 of the astronauts going to sample what they thought was a medium-to-large rock. As they walked toward it they realized it was much farther away than they thought, and it turned out to be gigantic.
So on a super earth I wonder what the effect would be with normal houses and such for scale. Even here on our regular Earth we can get fooled by objects close to the horizon.
To get a bit insight into questions like the OP, I like to think about the extremes and see what happens.
E.g., instead of a large Earth, think a small one. E.g., a basketball sized one. Standing on top of it, notice how close the horizon is! Then expand the Earth in your mind and notice the horizon moving away.
If you get really big and start approaching an infinite Earth (which may have a crisis), then the horizon asymptotically approaches a plane level with your eye height. So for a good sized planet, you will have a hard time telling the difference between a large planet and an infinite one without other visual clues (e.g., the classic ship masts).
With a basketball-sized Earth, or the “planet” the Little Prince came from, the horizon would be a noticeable angle below “horizontal”. But for the Earth or the Moon either one, that angular dip will be so small as to be undetectable to the naked eye. In other words, by that standard, the Moon is already big enough to be “almost infinite”.