How to know the right ascension and delineation of a heavenly body [Star]?
403081
Right ascension if like longitude and declination is like latitude, only on the celestial sphere instead of the Earth. Right ascension is traditionally measured in hours and not degrees because it is determined by the time that a star passes overhead.
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!
198142
Perhaps an illustration will help.\nhttp://home.cvc.org/science/kepler.htm
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)
380895
Ask Roy Kerr:\n\nhttp://www.astro.ku.dk/~cramer/RelViz/text/geom_web/node4.html
559281
http://skyandtelescope.com/observing/ataglance/article_110_1.asp will give you a summary of what is up each week. \n\nhttp://skyandtelescope.com/observing/skychart/article_1220_1.asp will give you a star chart customized for your location.\n\nThat should get you started. \n\nIf you like space and the moon, try\nhttp://groups.yahoo.com/group/lunar-observing/\n\nChuck
161934
Optical telescope based information about these bodies largely comes from the frequency spectrum of the light we receive.  Peaks and dips at particular frequencies can identify relative abundance of some of the elements in the star which emit those frequencies, and substances in the light path that absorb specific frequencies.  Frequency shifts of these features indicate velocities.\nOther information comes from distortion of the image due to gravitational "lensing" (bending of the light by objects in the light path).
70857
everything is related to each other if u know the mass, temperature or type of star it is then by using the russell-hertzsprung diagram u can figure out the luminosity
889433
According to western astrology:\n\nYour Ascendant is 5°22' AQUARIUS ....\nSun = 18°16'49" PISCES\nMoon = 19° 40' CAPRICORN\n\nPS: According to Hindu astrology your Ascendant is:\n11°45' Capricorn\n\nYou live in the western culture so stick with the Aquarius Ascendant
309954
Go to http://skyandtelescope.com and click on "Observing", then open their Interactive Sky Chart.  It will give you a chart of the stars in your exact location at any time of the night that you choose.  It's the best sky chart that I have found.  I use it every night I go out.
171995
3K background radiation measurement instruction:\n\n\n(1) take a full spectrum snapshot of the sky sphere.\n\n(2) substract all known active radiation sources (stars)\n\n\n(3) take the resulting frequency spectrum I(n) and\n\n(3.1) assume that you know allmost each existing active radiation source\n\n(3.2) assume that the universe doesn't radiate outwards and nothing radiates into it\n\n(3.3) then you can speak of a radiation field equilibrium and are allowed to apply 'blackbody radiation' theory\n\nhttp://en.wikipedia.org/wiki/Black_body\n\n\n(4) apply 'blackbody radiation' theory to the frequency spectrum: I(n) of step (3)\n\n(4.1) fit Planck's function of blackbody radiation: I(T,n) by variation of the remaining variable T\n\n\n(5) if you get a proper fit, call the temperature T, the temperature of the universe.\n\n\nuntil now, all such measurements led to T = 3 Kelvin.\nBut this doesn't mean that empty space would have a temperature itself.
884800
Go to www.astro.com and make a birth chart. From there, look in your 6th and 10th houses, and your Midheaven. The 6th house will show how you work. The 10th house is the house of career and public images, which should be around where your Midheaven is. The Midheaven will tell you what career will work best for you. \n\nGood luck!
1145113
This should help you:  http://www.le.ac.uk/physics/faulkes/web/stars/o_st_overview.html
605585
It sounds like you need three things\n\nA starting point for each star.\n\nA vector for each star (that takes care of direction and speed)\n\nGravity (The more dense a star is in relation too its mass the greater its gravitational pull)\n\nOnce you have worked out the specific gravity for each object (star) the only things that vary are the vector of each star and thus its new position.\n\nGood luck!! I know any gravitational calculation with three or more object gets very complicated very quickly.
217356
Either get a starguide book or google for on the internet and it will tell you the position of the stars and planets in the sky.  It should let you figure out there exact position for the sky where you live any for any day and time you want to look.  There are also many programs that you can get for your computer that do that and more!  Google has great maps of the moon and there are many astronomy web sites or forums where you can learn lots about the planets, stars, how to use your telescope and lots of other cool stuff!  Hope you have fun with your new toy!!
408710
Hold your arm out in front of you with your thumb pointed up and look at a distant building (or whatever). Now look at your thumb with one eye,and then the other, going back and forth between the two eyes. The thumb will appear to jump back and forth on the building. This is parallax. If you hold your thumb closer, the jump looks larger and if it is further away, the jump is smaller.\n\nThe same thing happens when you look at nearby stars (thumb) compared to much more distant stars (building) from different ends of the earth's orbit (eyes). The main difference is that the amount of the jump is very small since the star is so far away. It turns out that the angle of the jump is haved if the distance to the star is doubled. By knowing the size of the earth's orbit and the amount of jump, we can determine how far away the star is.
860017
You first need to apply a mass - luminosity relationship. I believe Kepler suggests M = k L^3.5. My results show Gamma:Delta luminosity ratio is 1:3184.\n\nAs for absolute magnitudes you need to log things I believe.\n\nYou will need m - M = -5 * 5Log(Distance)\n\nThis sounds like first year astrophysics coursework. You should attend workshops if you are having trouble.