Astronomical Math for Dummies?
413063
This is usually called tidal friction.\n\nThe gravitational flux between two bodies is simply the well-know gravitational equation.  Calculating the heat absorbed/lost by a planet is another much more complicated matter.  There are several mathematical models used, but since some constants aren't well-known, the models usually have a large amount of error.  These are problems that people get Ph.D's just for doing a small amount of work on them, so they're not simple or easily solved systems.\n\nThe earth's oceans are currently assumed to absorb a lot of the energy, so it doesn't go directly into the Earth's crust.  \n\nSee the (vague) reference below.
76505
The astronomy.com and NASA websites are very good starting points.  The NASA website also has a kids section which is very useful for students.
896136
try http://www.webmath.com/\nit gives you help on math on almost any time of math.
169843
You would need to perform the calculations yourself.  But, here is a useful website that will help you understand the inputs you need for making those calculations:\nhttp://pages.stern.nyu.edu/~adamodar/
874618
I came up with a magnification of about 16\n\nThis interactive website will calculate what you need ==>http://www.csgnetwork.com/telescopemagcalc.html
151685
not sure of web site but think it's http://www.hubble.com,or you can try NASA's web site,either should might be of use/help.
983210
Hey! i am totally in the same situation, but i'm a senior (hey smartypants ...)\n\nwell, try these sites:\nwww.clarku.edu/~djoyce/trig/\nstars.troy.edu/stars/CRSLISTS/UAB/MA-106.pdf\nwww.mvla.k12.ca.us/apps/pubs.asp?Q=2&T=Trig+Course+Information+Sheet&P=208 \n\nhope this helps you!\n\n~miza~
994577
I think what you’re wanting is a way to calculate the Sun’s altitude and azimuth for a given time and location.  The calculations are bit involved—you’d need to determine the Julian (Ephemeris) Date, solve Kepler’s equation to determine eccentric anomaly, then true anomaly, the Sun’s radius vector to obtain the Sun’s longitude, the obliquity of the ecliptic to determine latitude, then transform the RA and dec coordinates to altitude and azimuth.  I’m probably leaving something out.  Anyway, I do have the formulae, but thought I’d see if that’s what you’re really wanting before I do a lot of typing.\n\nPerhaps a more sane approach would be to modify a telescope (a Meade or other) and let its mounting do the tracking for you?  Just an idea.\n\n[edit]\n\nMy apologies for the delay in updating my answer.  Here’s how to calculate the Sun’s RA and dec:\n\nFirst, calculate the Julian Day (JD).  (note: Julian Day begins at Greenwich mean noon, or 12h Universal Time).\n\nLet\nYYYY = year\nMM = month\nDD.dd = day (fraction of day derived from hours, minutes, seconds past 00:00:00 UT; convert local time to Universal Time before doing this)\n\nIf MM > 2, then\ny = YYYY\nm = MM\n\nIf MM = 2 or 1, then\ny = YYYY – 1\nm = MM + 12\n\nIf date >= October 15, 1582 (in your case it will be), calculate\nA = INT(y / 100)\nB = 2 – A + INT(A / 4)\nelse (not in your case)\nB = 0\n\nJulian Day is then:\n\nJD = INT(365.25 * y) + INT(30.6001 * (m + 1)) + DD.dd + 1720994.5 + B\n\nNext, calculate the time T (measured in Julian centuries of 36525 ephemeris days from the epoch 1900 January 0.5 ET):\n\nT = (JD – 2415020.0) / 36525\n\nNote: since T is expressed in centuries, it should be calculated with a sufficient number of decimals; an error of 0.0001 in T corresponds to an error of 0.37 day in the time.\n\nThen, geometric mean longitude of the Sun (in degrees):\n\nL = 279.69668 + 36000.76892 * T + 0.0003025 * T^2\n\nSun’s mean anomaly (in degrees):\n\nM = 358.47583 + 35999.04975 * T – 0.000150 * T^2 – 0.0000033 * T^3\n\nEccentricity of the Earth’s orbit:\n\ne = 0.01675104 – 0.0000418 * T – 0.000000126 * T^2\n\nHere, I opted for an easier method that doesn’t require the solution of Kepler’s equation; the Sun’s radius vector isn’t needed:\n\nSun’s equation of center C (in degrees):\n\nC = + (1.919460 – 0.004789 * T – 0.000014 * T^2) * sin(M) + (0.020094 – 0.000100 * T) * sin(2 * M) + 0.000293 * sin(3 * M)\n\nSun’s true longitude (stl) is:\n\nstl = L + C\n\n(The true anomaly (M + C) is not needed.)\n\nSun’s apparent longitude (sal) is then:\n\nsal = stl - 0.00569 – 0.00479 * sin(259.18 – 1934.142 * T)\n\nThen calculate the obliquity of the ecliptic (oe):\n\noe = 23.452294 – 0.0130125 * T – 0.00000164 * T^2 + 0.000000503 * T^3\n\nTo correct for apparent position:\n\noe = oe + 0.00256 * cos(259.18 – 1934.142 * T)\n\nThen, for RA and dec:\n\ntan(RA) = cos(oe) * sin(sal) / cos(sal)\nsin(dec) = sin(oe) * sin(sal)\n\nor\n\nRA = arctan(cos(oe) * sin(sal) / cos(sal))\ndec = arcsin(sin(oe) * sin(sal))\n\nYou can convert RA from degrees to hours by dividing by 15.\n\nAssuming I didn’t overlook something or typo, this should get you the Sun’s RA and dec for a given time.  Do you also need a method to transform RA and dec to altitude and azimuth for a given time and location?  (I’m thinking yes)
777997
I don't know, but I am now very curious as to where this question comes from. I mean, why those particular angles? And what angles are they anyway? Right ascension expressed in degrees instead of hours? Celestial latitude? And why the extreme precision, specified to within 1/10,000 of an arc second?\n\nAnyway, here is a good web site for showing the sky at any time you like. Now all you have to do is try different times until that alignment appears. What a chore!
192330
in addition to the ones mentioned try these as well .. \n\nhttp://www.nasa.gov/\n\nhttp://www.astronomy.com/\n\nhttp://www.universetoday.com/\n\nhttp://hubblesite.org/newscenter/\n\nhttp://skyandtelescope.com/
354230
Greetings,\n\nFollow this link to Sky and Telescope magazine's website almanac.  Plug in your information and it will give you the info you requested.\n\nhttp://skyandtelescope.com/observing/almanac/\n\n-matt
979714
The Cosmos - Alex Filippenko    - ISBN 0-03-005218-1\n\nVery good introductory book, covers everything from the basics to some more in depth cosmology.\n\nTo accompany a telescope, I'd recommed Philip's Guide to the Night Sky
27065
I've been studying this stuff recently, and I'm pretty sure this is how it goes:\n\nK is now (-2,1) and D is now (-3,-2).\n\nFor future reference on determining rotations, simply turn the graph counter-clockwise one time for 90 degrees, upside down for 180 degrees, and for 270 degrees, just turn the graph clockwise once to see what the new cooridnates are.
1134720
Hi, great question.  I got a new telescope a year ago and have found a couple of excellent sites.  First try 'skyandtelescope.com'  this site has several features including a sky chart.  Enter your info, and it gives hour by hour info on whats up there to see, as well as info on upcoming events.  It offers 'how to advice' for telescopes, binocs and  astro photography.\nThe second is 'astromart.com'   this is one of THE BEST and most helpfull sites I have been to.  What a great forum.  Several active members are well known scientists and professors as well as very knowledgeable amature astronomy buffs.  No question is to stupid for them and you won't find a better place to gain advice and knowledge.   Questions posted are answered very quickly, usually in no more then a few hours, sometimes immediately.  Try both of these, I'm sure you will find what you need.    They are both  free of charge, believe it or not!!!\n\nLet me know what you think,  my s/n on astromart is lilkritter!!  Good Luck!
871968
Here is the math for finding our solar system's barycenter. (It's under the heading "Barycenter").\nhttp://en.wikipedia.org/wiki/Center_of_mass
380895
Ask Roy Kerr:\n\nhttp://www.astro.ku.dk/~cramer/RelViz/text/geom_web/node4.html
61016
I just went to astro.com and it was great. It gave me several different reports or charts.  Try it, you'll probably get the information you're looking for.