Picture yourself in the middle of the North American prairie stading on a small, lone hill - about 15 ft high - looking at a road beyond the hill that stretches all the way to the horizon on virtually flat land. Notice how the sides of the road appear to get closer and closer to each other, taking the shape along the lines of of a very long isosceles triangle (think of the Infiniti car logo) . The tip of that triangle, at the curvature of the Earth, is where the road seems to end. If you were to take a telescope and point it at the tip of the road (on clear day), you would reaquire accurate depth perception. Now, envision in your mind an alternate version of Earth which is the size of Jupiter - in which 1400 regular Earths could fit. From the same sized hill on Jupiter-sized Earth, A) How much higher “up” in relation to the observer would the line of the horizon be; and B) Would the road that would look like an isosceles triangle on normal Earth just eventually appear to be a very thin line, or would it look like something different?
P.S. If anyone knows of some sort of Google Earth-type program that could accurately portray a Jupiter-sized planet from a ground observational view, or even an illustration of some sort, I would be very appreciative.
Receding parallel lines would appear the same, but if you were to look at them at the horizon, they would appear closer on Jupiter. How much so? If I didn’t screw up my perspective calculations, it would appear about 4 times narrower at the horizon on Jupiter.
Are you asking what the angle would be between a truly “horizontal” line (i.e. perpendicular to a line to the center of the Earth) and a line to the horizon due to the planet’s curvature? If so, the general formula for this angle theta is
theta = sec[sup]-1[/sup](1 + h/R)
where h is the height of your observer above the surface and R is the planet’s radius. This works out to about 2.5 minutes of arc on the Earth (1 degree = 60 arcminutes) and 0.75 arcminutes on the surface of a Jupiter-sized planet. Again, though, this assumes a perfectly spherical planet and no atmosphere.
Your Weight on Other Worlds
A 70kg (155lb) person will weigh 686 Newtons on Earth and 1621.7 Newtons on Jupiter, which is the same as a 165kg (363lb) on Earth, which is slightly less than Aaron Gibson, the heaviest NFL player ever.
This is a start… but if we’re talking about a rocky terrestrial planet the size of Jupiter (which is what the OP was asking about, I think), then you’re going to have much higher numbers. Earth is several times denser than Jupiter.
The Earth has a dense core, primarily Iron and Nickel. If the hypothetical planet were made of the less dense Mantle material, its gravity would be correspondingly lower. About half the Earth’s average density is plausible, which would be closer to only 5 times as heavy.
You could just postulate an artificial planet; this thread made me think of the old novel The World Is Round, set on a hollow artificial world much larger than an Earth type planet. The title refers to the fact that the locals thought their world was flat because it lacked clues like ships sinking over the horizon; they just faded into invisibility before going far enough to do that.
