I How many degrees / radians is one full orbit of the Earth?

AI Thread Summary
One full orbit of the Earth around the Sun is slightly more than 360°, specifically 360.9856°, due to the Earth's axial rotation and its orbital motion. The Earth's orbit is inclined relative to the Sun's orbit around the Milky Way, which complicates the relationship between these movements. The Sun and Solar System orbit the Milky Way in a circular path with a period of approximately 220 million years, oscillating above and below the Milky Way's spiral arms every 30 million years. The differences in orbital measurements, such as between a sidereal year and a galactic year, are negligible, allowing for practical simplification to 360°. Understanding these dynamics provides insight into the complex motion of our solar system.
Patrick Aberdeen
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One day consists of slightly more than 360° of rotation (360.9856°) on it's axis (due to Earth's orbit around the Sun).

I imagine that one orbit is also either > or < 360° around the sun, relative to the motion of the sun around some object. Is this true, or is the orbit of the Earth perpendicular to the motion of our solar system relative to other celestial objects (solar systems, galaxies etc.)?
 
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The Sun, along with the rest of the Solar System, orbits the Milky Way in a somewhat complicated orbit: more or less circular around the center, with a period ~ 220 MYr. But it also oscillates above and below the MW's spiral arms and disk with a period ~ 30 MYr. (FWIW, we are headed in the direction of the constellation Hercules at this point in the Sun's orbit.)

The plane of the Earth's orbit is somewhat inclined to the plane of the Sun's orbit around the MW, but much less than 90 degrees.

It does not make sense to talk about periodic orbits for other objects that include the Sun. It is true that the MW and Andromeda Galaxy are interacting gravitationally, but that is not at all a periodic orbit. They will have a more-or-less head-on collision in several GYr. (So there are no radians involved in that.)
 
JMz said:
The Sun, along with the rest of the Solar System, orbits the Milky Way in a somewhat complicated orbit: more or less circular around the center, with a period ~ 220 MYr. But it also oscillates above and below the MW's spiral arms and disk with a period ~ 30 MYr. (FWIW, we are headed in the direction of the constellation Hercules at this point in the Sun's orbit.)

The plane of the Earth's orbit is somewhat inclined to the plane of the Sun's orbit around the MW, but much less than 90 degrees.

It does not make sense to talk about periodic orbits for other objects that include the Sun. It is true that the MW and Andromeda Galaxy are interacting gravitationally, but that is not at all a periodic orbit. They will have a more-or-less head-on collision in several GYr. (So there are no radians involved in that.)

Thank you! That gives me a big head-start on learning more about our solar system's motion. I'm guessing from your answer that the Earth's orbit around the sun is therefore very very slightly more or less than 360°, due to its incline to the plane of the Sun's orbit around the MW. Compared to the difference between a sidereal day and a solar day (56 minutes), I'm getting from your answer that the difference would be between a sidereal year and a MW year. Is that right? And that this difference would be so trivially tiny (given the 220 MYr period) that practically speaking we can leave it at 360°? Thanks again for for the explanation!
 
Patrick Aberdeen said:
Thank you! That gives me a big head-start on learning more about our solar system's motion. I'm guessing from your answer that the Earth's orbit around the sun is therefore very very slightly more or less than 360°, due to its incline to the plane of the Sun's orbit around the MW. Compared to the difference between a sidereal day and a solar day (56 minutes), I'm getting from your answer that the difference would be between a sidereal year and a MW year. Is that right? And that this difference would be so trivially tiny (given the 220 MYr period) that practically speaking we can leave it at 360°? Thanks again for for the explanation!
That's exactly right. (Sometimes called a "galactic year", BTW -- though of course it's the Solar System that has the orbit, not the Galaxy.)
 
Thank you so much. It was driving me nuts not being able to find the answer for that one :)
 
OK. I imagine it's not an FAQ for very many sources. :-)
 
Patrick Aberdeen said:
Thank you! That gives me a big head-start on learning more about our solar system's motion. I'm guessing from your answer that the Earth's orbit around the sun is therefore very very slightly more or less than 360°, due to its incline to the plane of the Sun's orbit around the MW. Compared to the difference between a sidereal day and a solar day (56 minutes), I'm getting from your answer that the difference would be between a sidereal year and a MW year. Is that right? And that this difference would be so trivially tiny (given the 220 MYr period) that practically speaking we can leave it at 360°? Thanks again for for the explanation!

For what it's worth, the difference between a solar day and a sidereal day is only 4 minutes, not 56 minutes.
 
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phyzguy said:
For what it's worth, the difference between a solar day and a sidereal day is only 4 minutes, not 56 minutes.
Right: Length is 23 hours and 56 minutes.
 
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