Last Friday I had an argument with two co-workers about planetary motion. Most of the fight, however, was about me trying to get myself understood. Once we had gotten past that, we basically agreed that we just didn’t know. I hope I’ll be able to get my point across more easily now. Here goes:
Assume a sun and a planet. Let the planet move around the sun in such a fashion that the same side keeps facing the sun (like the Moon vs. the Earth). After a year, how many times has the planet turned around its own axis?
The answer seems to depend on your frame of reference. One co-worker said the planet has not turned at all since the same side keeps facing the sun. I said it has turned exactly once in relation to “absolute space”. (Let’s keep this Newtonian; you can replace “absolute space” with “the surrounding stars and galaxies assumed as fixed.”)
Also: How many days have there been for the planet, one or none? Is a day a revolution of the planet around its own axis or is it a full light/dark cycle?
We say that a year on Earth has 365 days (neglecting the .24). Is that true in astronomical terms? Or are there 364 or 366 days/revolutions? (I forget the direction.)
I suppose it all boils down to what definitions astronomers use. Can anyone tell me what they are? I think if you’re observing other objects than just the Earth and the Sun, the “absolute” frame of reference is more useful. But is it used?
Well, I hope my question is clear now. I’d hate to have to shout at you like I did on Friday…
I think it’s just a question of word meaning. I looked up day in my trusty dictionary, and found a dozen or so definitions. It was described as the period of light between two nights, the period of time between two noons, etc. It also metioned analogous periods on other worlds: martian day. So a day is more of a relative thing then some absolute motion. For a planet where one side always faces its sun there’s no such thing as a day. For planets that don’t rotate (relative to some external reference point), one day per year.
E1, I take issue with your answer with respect to the parent star. The key is the condition:
The moon does rotate with respect to the earth; its orbital period simply matches its rotational period. (Otherwise the men on the lunar missions could not have experienced Earthrise.) Thus, the planet has rotated exactly once with respect to its parent star, unless we really are positing a planet with a rotational period of zero; but that wasn’t the condition specified.
Phil (may I call you Phil?): read my post again. I said exactly the same as you. The planet has rotated ONCE around it’s parent star, but as regard with “how many days have passed?” my answer was NONE.
I’ll come back with more later. There’s a power outage right now, and my “no-break” only gives me 15 minutes. See ya’.
A sidereal day is the time it take the Earth to rotate exactly 360 degrees. It lasts 23 hours, 56 minutes.
A solar day lasts from when the Sun is on the meridian at a point on Earth until it is next on the meridian. A solar day is exactly 24 hours (by definition). Because of the Earth’s revolution, a solar day is slightly longer than a sidereal day. In every day life, we use solar time.
Because the Earth moves in its orbit around the Sun, the Earth must rotate more than 360 degrees in one solar day .
The Earth must rotate an extra 0.986 degrees between solar crossings of the meridian. Therefore in 24 hours of solar time, the Earth rotates 360.986 degrees.
Because the stars are so distant from us, the motion of the Earth in its orbit makes an negligible difference in the direction to the stars. Hence, the Earth rotates 360 degrees in one sidereal day .
A sidereal day lasts from when a distant star is on the meridian at a point on Earth until it is next on the meridian. A sidereal day lasts 23 hours and 56 minutes (of solar time), about 4 minutes less than a solar day.
So, the Earth does 366 360-degree spins in a year. But since a 24-hour day is when the Earth spins 360.986 degrees, there are 365 of these in a year. (Fractional days/leap days excluded from this discussion.)
sidereal day: “the interval between two successive transits of a point on the celestial sphere (as the vernal equinox) over the upper meridian of a place” (Mirriam-Webster)
Thus, a planet that orbits the sun once, with the same face turned to the sun always, has experienced one sidereal day, without experiencing any ‘solar’ days.
For really funky discussions of ‘day’ look at the motions of Venus (retrograde spin) and Uranus (axis tilted over 90 degrees from perpendicular to the plane of orbit).
Come again??? The earth’s position in the sky relative to any place on the moon stays at almost exatly the same apparent celestial longitude, varying only small amounts. This is why we always see the same side of the moon. Men on the moon itself did not experience ‘earthrise;’ men in ORBIT around the moon did.
Nevertheless, the moon does complete one ‘rotation’ during the period it revolves around the Earth, for the same reason the planet hypothesized in the OP completes a rotation - the best way to visualize this is to look down at it from above and then remove the nearly circular motion of the orbit and view the relative position of the moon to the fixed background of the sky.
I knew there had to be a word for what I was talking about. Now there’s just one question left open: Which concept do astronomers use, solar or sidereal time?
I don’t know what the hell I was thinking–E1 is correct and I am wrong. One year has been completed, but from the POV of someone on the surface the sun has never passed the meridian.
The answer, as usual, is it depends, but mostly they use the sidereal time.
Stars rise and set on a sidereal schedule, so to locate stars and other non-solar system objects you use sideral time. First look up their declination and right ascension (kind of a celestial latitude and longitude) then you convert the right ascension to the hour angle by subracting the current local sideral time. A little trig converts the hour angle and declination to the altitude (how far up) and azimuth (how far sideways) for the telescope.
The process for solar system objects is very similar except that the R.A. and declination change with time due to their orbital motion. (Stars move a little bit, too, but not enough to worry about here.) So for near objects you have to determine their R.A. and declination for the time you want to view them and then proceed as above.
If it’s the motion of the earth around the sun that you are concerned with then you would (obviously) use solar time.