I need to find where mars and saturn will be on a certain date.
129142
Some of them can be identified by the naked eye . . . They are brighter, seem to twinkle less (in my opinion), and have a tinge of color. After someone points out Mars, Jupiter, Venus and Saturn, you can find them just by looking for them.\n\nWhen you look for planets on a certain date, you need to know your latitude and the time that you will be looking becuase both of these affect the position of the planet in the sky.\n\nCheck out the sources for exact locations given time and latitude in your area.
232953
Try an almanac.  Or maybe there's an almanac.com or something like that.
353841
You might also look around at http://www.spaceweather.com/
3904
Here is a link that should be perfect! Scroll down a bit and there is a table with all the data you could possibly want!  http://en.wikipedia.org/wiki/2003J6%28moon%29--_statistics_of_Jupiter%27s_Moon
431505
March 4th.\n\nMarch forth.
38882
http://www.astrology.com
560096
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)
252649
Very good question.  I went searching and came up with this:\n\nBTW, the person that answered 687 is correct for EARTH days, not MARS days.  It appears to be closer to 668.6 Martian days.
452096
astro can definately tell marriage dates
118284
Go to Heavens Above.  They have a lot of information about the planets, as well as satellites that have been placed in orbit or sent out into the solar system.\n\nhttp://www.heavens-above.com
498766
See the Cassini web site.
168236
From the web page below:\n\n"Mars is 20 degrees above the western horizon at sunset and sets near 11:30 p.m. Look for Mars near the new moon on May 2 and again on May 31. Take care to avoid looking at the Sun as both the Moon and Mars are low on the horizon after sunset. On May 30, Mars is at the left end of a long celestial line consisting of Mars, Moon, Pollux, and Castor, the two brightest stars in the constellation Gemini."
619497
12/17/2006\n\n\nIf you use spreadsheets in your computer then you can easily figure out future dates with one simple formula.
190243
see for yourself.\n\nhttp://marsrovers.nasa.gov/home/index.html
441671
go to the sirus sattalite website and see if the have any info
464259
http://www.heavens-above.com\n\nis all you need.  It works with your city/town/village/locality\n\nIt has 2 million places, so you will be there.  My village is tiny and in New Zealand, and it has that.  \n\nThen you do not need to know you lat/long.\n\nIt has a star map for the current time at your location, but you can set any time, future or past.\n\nIt shows times for all major satellites, comets, phases of moon, etc etc.\n\nTry it.
9377
http://science.howstuffworks.com/mars-rover.htm\n\nhttp://marsrovers.jpl.nasa.gov/spotlight/dayinthelife01.html