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Earth moves in a few different ways through the universe. Earth orbits the Sun (plus the orbit precesses), the Sun moves through the Milky Way, and the Milky Way moves through the universe.

Compared to my position exactly one Earth revolution around the sun, approximately how far have I actually moved relative to my starting point? Or more specifically, what is my displacement relative to my position when Earth was last in the same position within its orbital path around the Sun?

The movement of the solar system and/or the galaxy might be negligible over the course of a year, but negligible on those scales would presumably still translate to big distances relative to the scale of a person. Though I imagine the margin of error might make such a calculation impractical.

Another way to think about it is if I was able to jump forward in time one year, would I wind up floating in space or would I still be on the surface of Earth?

If the answer is insignificant or not determinable, how many years would you need to consider for the displacement to be significant / determinable?

Can assume that a year ago and right now I am at the exact same geographic coordinates on the surface of Earth.

shim
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I all depends on the frame of reference you choose, and the larger it is, the higher are the relative velocities observed.

Earth surface (at the equator) moves at .5 km/s relative to the center.

The earth moves around the sun at 30 km/s. After 12 h, you are not circa one, but approximately 100 earth diameters from the place you were.

The solar system moves with ~230 km/s around the milky way.

The milky way has about 550 km/s relative to the cosmic microwave background.

(Current speed of Voyager I: 17 km/s.)

If you chose the CMB as a reference, your displacement after a year would in fact be 78 astronomical units, far outside of Pluto.

Karl
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If we consider position relative to the fixed stars (more specifically, the Sun) and disregard the slow galactic rotation, one year's duration of orbit roughly moves you back to your initial position.

The most important remaining motion, is the combination of the Earth's annual extra quarter-day (roughly) rotation, and the offset of the Earth-Moon barycenter (about 3000 miles) times the progression of the Earth-moon direction vector.

The quarter-day of Earth rotation amounts to a displacement of circa 6000 ground miles (if you start at the equator) or less, and would leave you still at a comparable altitude (probably you could breathe).

It is only the Earth-moon barycenter that really follows a simple orbit, not either the Earth or Luna individually, though, so at a synodic month period of 29.53 days, one year mismatches a repeat by 0.37 of a revolution; straight-line distance, that's a couple of thousand miles. So, probably NOT within atmosphere, though it would be roughly in the ecliptic; direction depends on the phase of the moon.

Whit3rd
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Displacement is a vector quantity, i.e. the difference between two points. In the rest frame of the sun an exact solar year would put you right back where you started, ignoring effects like Earth's orbital precession, wobble about axis of rotation, etc. In the rest frame of the center of mass of the Milky way, however, there would be a net non-zero displacement.

Albertus Magnus
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Spatial distances without reference system of rest are undefined in special, the more so in general relativity.

The spatial distance between two events $(t1,x1),(t2,x2)$ , one year apart in time, perhaps an astronomical unit or 6 light minutes in space, eg measured in the milky way system of reference, is microscopic small with respect to time distance of 1 year. So it depends mainly on the system of reference. Its zero for an observer moving on a straight line between the two events with constant velocity.

Roland F
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