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TL;DR: assuming circular, coplanar orbits, for any planet A, the sun is on average closer to that planet than any other planet B, because planet B spends a lot

TL;DR: assuming circular, coplanar orbits, for any planet A, the sun is on average closer to that planet than any other planet B, because planet B spends a lot of time on the other side of the sun from planet A, and given the way orbital rates work out the closer a planet is to the sun, the lower its average distance is to any planet farther out than it. In other words, Mercury is on average closer to Neptune than, say, Saturn is, and indeed Mercury is on average closer to every planet than any other planet is. This is both somewhat surprising (I'd have thought that there would be a symmetry here - that the average distance between Neptune and Saturn would be the same as the average distance between Neptune and the Sun), and also (I think) uninformative. The way orbital dynamicists think about "close" orbits is in terms of delta-V, which is to say, how big a change in velocity does it take to change one orbit to a different one. Or, in more concrete terms, how much fuel would it take for a spacecraft to travel from a body in one orbit to a body in a different orbit. That's a completely different way to think about closeness than the OP 's metric. Spaceflight is non-Euclidean like that. The OP's metric would be helpful for estimating average time-of-flight communications delay, but over distances like this I don't think that averages are that informative. The total variation is quite large; basically it varies as |r1|-|r2| to |r1|+|r2|. For Earth-Mars that's from 4.5 to 20 minutes. Not sure knowing the "average" distance is all that helpful there. For Mars-Neptune it's 4 hours to 4.4 hours, and for Earth-Neptune it's 4 to 4.3 hours. Even though the distance is much greater, the orbital radius of the outer planets is so large that the variation as the inner planet goes around the sun is almost in the noise. Takeaway: for comms delay to the the outer planets, just calculate the time-of-flight distance to the sun.

sandworm101 8y ago
>>Spaceflight is non-Euclidean like that. Spaceflight maybe, the traveling between objects part, but other space operations are very much euclidean. The folks trying to talk to mars rovers are very aware of exactly how far mars is away from earth, as is anyone trying to image planets using telescopes.
- GlenTheMachine 8y ago
  Yes, but I suspect they don't care that much about average distance(1). They care about current distance, and about minimum and maximum distance. Maybe that's being too pedantic though. (1) I've never driven a Mars rover.
Majromax 8y ago
> TL;DR: assuming circular, coplanar orbits, for any planet A, the sun is on average closer to that planet than any other planet B, because planet B spends a lot of time on the other side of the sun from planet A, and given the way orbital rates work out the closer a planet is to the sun, the lower its average distance is to any planet farther out than it. It doesn't come from orbital rates; it comes from basic geometry. Imagine that the orbits of Earth and Mercury are perfect circles. Draw them on a piece of paper, and mark the position of Earth. Now, take a compass, place one leg at Earth, and draw a circle (with the Earth at the center) that passes through the Sun. This is a circle of radius 1AU. Now note that slightly less than half of Mercury's orbit is inside this circle: Mercury spends more of its time (in this idealized model) more than 1AU away from Earth. That plus an equivalent result for Venus leads to the finding of this link. This arises because of our old friend, the Pythagorean theorem. When Mercury->the Sun->Earth make a right triangle (that is, when Mercury is neither ahead of nor behind the Sun), the distance between Earth and Mercury is the hypotenuse of that triangle, which has legs of 1 AU and one Mercury-orbit-radius. That's obviously greater than 1AU.

sandworm101 8y ago

>>Spaceflight is non-Euclidean like that. Spaceflight maybe, the traveling between objects part, but other space operations are very much euclidean. The folks trying to talk to mars rovers are very aware of exactly how far mars is away from earth, as is anyone trying to image planets using telescopes.

GlenTheMachine 8y ago
Yes, but I suspect they don't care that much about average distance(1). They care about current distance, and about minimum and maximum distance. Maybe that's being too pedantic though. (1) I've never driven a Mars rover.

Majromax 8y ago

> TL;DR: assuming circular, coplanar orbits, for any planet A, the sun is on average closer to that planet than any other planet B, because planet B spends a lot of time on the other side of the sun from planet A, and given the way orbital rates work out the closer a planet is to the sun, the lower its average distance is to any planet farther out than it. It doesn't come from orbital rates; it comes from basic geometry. Imagine that the orbits of Earth and Mercury are perfect circles. Draw them on a piece of paper, and mark the position of Earth. Now, take a compass, place one leg at Earth, and draw a circle (with the Earth at the center) that passes through the Sun. This is a circle of radius 1AU. Now note that slightly less than half of Mercury's orbit is inside this circle: Mercury spends more of its time (in this idealized model) more than 1AU away from Earth. That plus an equivalent result for Venus leads to the finding of this link. This arises because of our old friend, the Pythagorean theorem. When Mercury->the Sun->Earth make a right triangle (that is, when Mercury is neither ahead of nor behind the Sun), the distance between Earth and Mercury is the hypotenuse of that triangle, which has legs of 1 AU and one Mercury-orbit-radius. That's obviously greater than 1AU.