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2004-Mar-07, 07:30 PM
Right now, I am studying for a competition called Science Olympiad. It is a science tournament held at the regional, state, and national levels. I was reading Astronomy Today 4th Edition by Eric Chaisson and Steve McMillan to do some studying and there were a few things i didn't understand in the first few chapters.

1. The first one was the orbital resonance of Mercury. Why isn't it tidally locked like the Earth &amp; Moon system? Why does it display a 3:2 resonance instead?
2. Could someone explain how the Kirkwood Gaps in the asteroid belt and the ring divisions in the rings of Saturn are formed?
3. Ok, I know that the synodic period is the time between oppositions. For example, I heard that the synodic period of Mars and Earth is 780 days, but I don't understand how that number is derived. I also don't understand how the formula to find it is derived.
4. This really stumped me. Can someone explain in detail the eclipse seasons and the Saros cycle in detail? I understand it somewhat, but I think i need more clarification.
5. I understand this pretty well, but I want to know the difference between an absorption line and an emission line in more detail.

Thanks a lot to anyone who replies. :D

Manchurian Taikonaut
2004-Mar-07, 08:38 PM
Science Olympics..what the hecks, is this America olympic sport?

The saros idea is about the regression of the nodes. It was used by ancient peoples and cultures, and used to show the cycles of the Moon, and Sun. A cycle of 18 years in which the Sun, Moon and nodes of the moons orbit return to the almost same relative positions. It is due to regressions of nodes. These anicent peoples such as the Babylonian astronomers, used this system to predict eclipse, since the same patterns happen some 18 years and 11.3 days again in the futures. They can also predict slight variations from cycle to each cycle, so each eclipse is not exactly identical.

sorry, my English is not good enough, can you understand?

milli360
2004-Mar-08, 01:16 AM
1. The first one was the orbital resonance of Mercury. Why isn't it tidally locked like the Earth &amp; Moon system? Why does it display a 3:2 resonance instead?
Here's a thread (http://www.badastronomy.com/phpBB/viewtopic.php?p=157401#157401) where we discussed that last October. Short answer is that a permanent deformation of Mercury causes it to speed up in part of its orbit, offsetting the tidal slowing in the rest of the orbit.

3. Ok, I know that the synodic period is the time between oppositions. For example, I heard that the synodic period of Mars and Earth is 780 days, but I don't understand how that number is derived. I also don't understand how the formula to find it is derived.
Mars (http://planetscapes.com/solar/eng/mars.htm) orbital period is about 686.98 (earth) days, Earth (http://planetscapes.com/solar/eng/earth.htm) orbital period is 365.256 days. After the planets match up, the Earth goes around the Sun faster but by the time the
Earth returns to that point, Mars has almost gone halfway around the Sun--and the Earth won't catch it before the Earth goes all the way around again. So the Earth goes around the Sun one more time than Mars does, before the Earth catches Mars again. That is, 686.98(x)=365.25(1+x), and x would equal 365.25/(686.98-365.25), and the number of days is x times 686.98, or 779.93, So, the formula is AB/(A-B), and because of the ellipticity of the orbit, it is an approximation--more of an average value.

Good luck.

AGN Fuel
2004-Mar-08, 02:47 AM
5. I understand this pretty well, but I want to know the difference between an absorption line and an emission line in more detail.

Take a star at random, light will be produced from the photosphere in a continuous spectrum - in other words, an uninterrupted range of wavelengths.

As light from the photosphere passes through the gaseous atmosphere of the star toward you as observer, atoms in that atmosphere will absorb light at very specific energy levels (as electrons in the atom are elevated from low energy states to higher energy levels). As the electrons cascade back down to the lower energy level, the re-emit a photon of the same wavelength, BUT (and this is the tricky bit they never really seem to emphasise in the textbooks), the photon is re-emitted in a random direction. There is a very low likelihood that the photon will be re-emitted toward you as observer. The result of this is that the continuous spectrum from the star will be punctuated by dark lines corresponding to the absorbed/re-emitted wavelength. This is an absorption spectrum.

Imagine now that you are observing a nebula offset from the line of sight to that star. You will not see the continuous spectrum from the star (because you aren't looking at it). What you will see though is the same process of absorption &amp; re-emission - now however, because the background is the blackness of space and hence your contrast is vastly increased, the relatively few photons that are re-emitted in your direction will show up as bright lines against a black background.

So, at the simplest level, an absorption spectrum will indicate the atoms in a gas lying between you and a star, while an emission spectrum will tell you the atoms in a gas offset from the line of sight between you and the star.

Kaptain K
2004-Mar-08, 11:19 AM
2. Could someone explain how the Kirkwood Gaps in the asteroid belt and the ring divisions in the rings of Saturn are formed?
Each asteroidal orbit has its own period. As the asteroid orbits the Sun, its orbit is perturbed when it gets near Jupiter. For most orbits, this perturbation occurs at random points in the orbit. If, however, the orbital period of the asteroid is a whole fraction (1/2, 2/3, 3/4, 3/5, etc) of Jupiter's period, the perturbation will occur at the same point each time and will accumulate over time and eventually move the asteroid out of the resonant orbit.
Imagine a child on a swing. If you give the swing a push at random points on it's arc, some will speed it up and some will slow it down, for no net effect, except for an unhappy child screaming "higher, higher". If you time your pushes, so that they are all at the same point in the arc, they will add up and the swing will go higher and higher.

daver
2004-Mar-08, 09:25 PM
Hey, I like the swing analogy. Assume you have a long swingset, with the chains on the swings being all slightly different lengths, so each swing has a different period. Now, imagine that you have a hose that directs a blast of air at the back of each swinger to give them a push. Now, blast the air at, say, once every three seconds. For most of the swingers, these blasts will sometimes hit when they're heading towards the jet, sometimes when they're heading away, so the net effect will be zero. For a few lucky swingers, though, the blasts of air will coincide with their period (those with a natural frequency of three seconds or six seconds or nine seconds). Those swingers will swing higher and higher and higher, until they fall out of the swings and are carted away in ambulances. So, after several minutes, the only swingers left will be those whose swings don't match the frequency of the air jets. The empty swings are the Kirkwood gaps.

JohnOwens
2004-Mar-09, 03:01 AM
3. Ok, I know that the synodic period is the time between oppositions. For example, I heard that the synodic period of Mars and Earth is 780 days, but I don't understand how that number is derived. I also don't understand how the formula to find it is derived.
Mars (http://planetscapes.com/solar/eng/mars.htm) orbital period is about 686.98 (earth) days, Earth (http://planetscapes.com/solar/eng/earth.htm) orbital period is 365.256 days. After the planets match up, the Earth goes around the Sun faster but by the time the
Earth returns to that point, Mars has almost gone halfway around the Sun--and the Earth won't catch it before the Earth goes all the way around again. So the Earth goes around the Sun one more time than Mars does, before the Earth catches Mars again. That is, 686.98(x)=365.25(1+x), and x would equal 365.25/(686.98-365.25), and the number of days is x times 686.98, or 779.93, So, the formula is AB/(A-B), and because of the ellipticity of the orbit, it is an approximation--more of an average value.

Another way to look at it, probably more complicated/less accurate to calculate, but easier to intuitively understand and go "hey yeah, that makes sense!": Calculate the angular velocity of Earth and Mars, and subtract. Figure out how long it will take to get 2*pi radians or 360 deg, according to your preference, for the difference velocity (could maybe think of it roughly as a "closing rate"). Note how, if you like getting rough answers in your head, the angular velocity of Earth comes mighty close to 1 deg/day. :D
Full treatment, using milli360's numbers (and degrees/day): Earth's omega (the symbol for angular velocity) is 360 deg/365.25 days = 0.985626 deg/day, Mars' is 360 deg/686.98 days = 0.524033 deg/day. Difference is 0.461594 deg/day. 360 deg / (0.461594 deg/day) = 779.907 days (which is also what I get for the AB/(A-B) above, 779.93 notwithstanding, probably rounding in x).

I did Science Olympiad myself, among a few other similar things. Good luck!

2004-Mar-09, 04:11 AM
Thanks a lot for everyone who replied. There is one part about the synodic periods that I don't understand though. :(

Full treatment, using milli360's numbers (and degrees/day): Earth's omega (the symbol for angular velocity) is 360 deg/365.25 days = 0.985626 deg/day, Mars' is 360 deg/686.98 days = 0.524033 deg/day. Difference is 0.461594 deg/day. 360 deg / (0.461594 deg/day) = 779.907 days (which is also what I get for the AB/(A-B) above, 779.93 notwithstanding, probably rounding in x).
Ok, I understand that Earth moves 0.985626 deg/day and Mars 0.524033 deg/day. I also understand the difference. But why do you divide 360 deg by the difference of 0.461594? Thanks if anyone has the time to explain. :)

JohnOwens
2004-Mar-09, 04:38 AM
Thanks a lot for everyone who replied. There is one part about the synodic periods that I don't understand though. :(

Full treatment, using milli360's numbers (and degrees/day): Earth's omega (the symbol for angular velocity) is 360 deg/365.25 days = 0.985626 deg/day, Mars' is 360 deg/686.98 days = 0.524033 deg/day. Difference is 0.461594 deg/day. 360 deg / (0.461594 deg/day) = 779.907 days (which is also what I get for the AB/(A-B) above, 779.93 notwithstanding, probably rounding in x).
Ok, I understand that Earth moves 0.985626 deg/day and Mars 0.524033 deg/day. I also understand the difference. But why do you divide 360 deg by the difference of 0.461594? Thanks if anyone has the time to explain. :)

Think of it from the point of view of someone standing on the Sun. Ignoring orbital eccentricity, he'll see that the angle between the two changes 0.461594 deg/day (let's not try using the 25-day solar day!). Now, let's put him on a nice swivel, and ignore the background stars, and have him stare continuously at Earth. He'll see Mars (when it's in his field of view) always going at that same 0.461594 deg/day. So, it has to make a full, 360 degree circle to get back to the same point where it had been. Divide the angle by the angular velocity, and voila!

milli360
2004-Mar-09, 05:14 AM
779.907 days (which is also what I get for the AB/(A-B) above, 779.93 notwithstanding, probably rounding in x).
No, I used 365.256 is all. I meant to change all those.