how determine stars diameter by moon occultation?
957015
The occultation needs time t. The intensity of the light from the star decreases. That measurable time depends on\n1) how fast the moon moves  compared to the background stars. That is known. (v)\n2) the diameter of the star. (d)\nCalculate d with d = vt\n\nTh
740323
Assume a flag is 1 meter across, and the moon is 3.6x10^8 m distance at perigee.  Your telescope needs a resolving power of 1/3.6x10^8 radians.  \n\nThe resolving power of a typical telescope is ~1.22lambda/D radians where lambda is the wavelength of light being observed and D is the diameter of the telescope.  \n\nSolve for D;  D= 3.6x10^8*1.22*lambda  \n\nAssume a visible light telescope sensitive at 600nm = 6x10^-7 m\nand your telescope must be 263m in diameter.  You will want to orbit that puppy or phase errors from turbulence will kill you, with current adaptive optics technology.
655459
You don't really need the Star's speed, only it's spectral class. The spectral classes are O, B, A, F, G, K and M, with O being the brightest and M the dimmest. If you know how bright the star is intrinsically from it's spectral class, then you can calculate how far away it is from it's apparent brightness in the sky. But this method if fraught with difficulties. Red giants and white dwarfs are off the main sequence of stars. That means that a red star may be much brighter that it's spectral type would indicate and a white dwarf is much dimmer than it's spectral class would indicate. Also, there may be interstellar dust and gas between you and the star that makes it look dimmer that it would through clear space. But if you are careful, all these factors can be taken into account and you can get an approximate distance.\n\nFor the distance to a remote galaxy that is too far away for individual stars to be seen, the red shift of the overall glow of the galaxy can be measured and used to determine the galaxies radial velocity, or the speed at which it is moving away from you. Multiplying that by the Hubble constant gives you an approximate distance.\n\nIf you want a really accurate distance to a star, the only sure way is by measuring it's parallax. But this can only be done for comparatively nearby stars. The Hipparcos satellite measured thousands of stellar parallaxes to a precision not obtainable from the ground, and the catalog is freely available online.
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.
148361
Size of Star depends on how it was formed originally. I don't think it could be said precisely. U can take the size of our sun as a measure. Radies=6400kms
1165410
Ugh I cannot believe people actually ask this retarded questions.\nTell you what sparky, go outside at night, when there are plenty of stars. Now, take a picture of a friend with a flash or using an external light (as they did on the moon) Know what you'll find when the photos come back? NO STARS!\nWhen you understand what FOCAL LENGTH means, you'll know why the photos on the moon have no stars in them. I hope the website helps you understand...
708252
If you watch a lunar eclipse, you can see the Moon entering the shadow of the Earth, which is a circle on the sky.  If they could figure out the phases of the Moon and the reason for it, they could figure out that the Earth was round.
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)
370062
Well, for one, you need the sun. After that, I'd get some sidewalk chalk, then make a marking on the ground. Then place the object whose shadow you are measuring in a position where the top of the shadow meets the marking. Then make another marking at the bottom of the shadow. Then, using a ruler, measure the distance.\n\nIf you're looking to do this mathematically, you would need to know the height of the object and the angle at which the sun is (which you can approximate based on the time of day).
259557
have u heard about right Ascension and declination?\nthose are used 4 finding stars.right asection is measured in hours and declination is measured in degrees.\nu must know abt meridian 2 find stars.It is an imaginary curve that passes across head 2 north-south.\nfind a planisphire.it's a star chart which is very easy 2 use.
778505
For any planet with a moon, the planet's mass can be approximated by GmM/r² = mv²/r = mrω² = mr(2π/P)², where G is the gravitational constant, m the mass of the moon, M the mass of the planet, r the distance between planet and moon, and P the period of the moon. The mass of the moon cancels out, leaving\nM = r³(2π/P)²/G\n\nThe masses of moonless planets have to be determined by Perturbation Theory, the effect of one planet's orbit on the orbits of other planets.\n\nThe diameter of another celestial object can be estimated by multiplying the angle subtended by the object by its distance from us.  A closer estimate is obtained by\nd/2 = Dsin(α/2) \nwhere d is the diameter, D is distance to the object, and  α is the subtended angle.
178225
Try this.\nhttp://www.calculatorcat.com/moon_phases/moon_phases.phtml
591541
While refraction by the atmosphere is too small to affect the apparent diameter of the moon except perhaps when it is within one diameter of the horizon (and in this case, the moon will appear stretched a bit in the direction perpindicular to the horizon) it does appear larger to our brains when on the horizon because we see it in reference to distant large objects.  Gives the impression of it looking bigger, however if you could look straight up at the moon, and see a bunch of trees a mile up in sky between you and the moon it would look just as large as when it is near the horizon.  (If you ever do see this by the way, seek shelter or psychiatric treatment immediately!)\n\nThe moons orbit IS elliptical, however, and at times will be closer to the earth than others.  This phenomenon is called "lunation" and there is a great still and a gif animation of this at the Astronomy Picture of the Day.\n\nhttp://antwrp.gsfc.nasa.gov/apod/ap041021.html\nhttp://antwrp.gsfc.nasa.gov/apod/ap991108.html\n\nThe moon's surface brightness (brightness per area) also varies with phase. Because larger particles tend to reflect a bit more light directly back at the light source than in other directions (look up "Mie" scattering) a given bit of the moon will appear brighter when the earth is in the direction of the sun as seen from the moon. (All three are in a line with the Earth more or less between). This occurs when the moon is full or near full. This brightening might make the moon look larger.
354664
go to this website: http://tycho.usno.navy.mil/vphase.html and type in the information, they'll tell you what the shape is for that day.  Then go to this website to look up the name: http://www.calculatorcat.com/moon_phases/moon_phases.phtml
451278
here is a calculator for determining planet and orbit sizes to scale.  i hope this helps.\n\nhttp://www.exploratorium.edu/ronh/solar_system/
169454
there are books out there in astrology, it shows the phases of the moon and how to interpret it. go on any astrology website. Go on google and type astrology chart and it explains how to read it. hope this helps.
903470
r=sin 11θ for the smaller petals and r=Xsin11θ where X>1\nthis will give you your larger petals.