Astrophys: Concerning the comparison of the brightness, apparent magnitude, and parallax angle of 2 stars?
356232
Magnitude is a logarithmic scale and negative numbers are brighter than positive numbers.  Hence a magnitude 1 star is ten times brighter than a magnitude 2 star and 100 times brighter than a magnitude 3 star.  A magnitude 0 star is 10 times brighter than a magnitude 1 star and a magnitude -1 star is ten times brighter than a mag 0 star.\nAldebaran is simply a lot brighter (bigger and hotter) than formalhut, despite being much further away.  If the two stars were at the same distance, i.e. you're talking about relative magnitude, aldebaran would have a much lower rel. magnitude than formalhut and therefore appear much brighter.  Are you confused enough yet?\nThink of relative magnitude (also known as absolute magnitude i think) as the magnitude the star would have if it was 10 times further from us than the sun is- that enables the astronomers to compare like with like.\nParallax is the angle through which the star appears to move against the background of 'fixed' stars.  As the Earth moves around the sun, our perspective of the universe changes.  Nearer stars appear to move relative to the ones that are furthest away.  Hence, Formalhut, a nearby star, appears to move through a greater parallax angle than distant Aldebaran, which is fairly fixed against the stellar background.
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.
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
596808
type of star\nspectral type\ndistance from star to Earth
425588
There are only 2 that are bright enough to see with the naked eye, that is, they have a magnitude less than 3. Alphard is the brightest star of the constellation, with an apparent magnitude of 1.98. R Hydrae is intermittently visible because it has a varying magnitude between 3.5 and 10.9.
49829
check this site...\nhttp://www.bbc.co.uk/science/space/stars/index.shtml
380895
Ask Roy Kerr:\n\nhttp://www.astro.ku.dk/~cramer/RelViz/text/geom_web/node4.html
864425
One source that breaks down these classifications by temperature, or maybe it was only color photometry (I don't recall) is Kenneth Lang's Astrophysical formula:\n\nhttp://www.amazon.com/Astrophysical-Formulae-Radiation-Processes-Astrophysics/dp/3540296921/sr=1-1/qid=1163374728/ref=sr_1_1/102-7735916-8685767?ie=UTF8&s=books\n\nVolume I has some tables I think you will find helpful.  You might need to do a little interpolation...\n\nanother place to look would be the iau website http://www.iau.org . The IAU is the kind of group that would define a standard, though this might be too detail oriented for them to mess with.\n\nYou will be hard pressed to find an equation, however.  You are going to be stuck with a lookup table.
905593
zee_prime's got a good answer there.  Very precise and correct.  I'm gonna take another approach to it, though, and try to make it intelligible for those who don't have a lot of math or physics behind them.  My answer'll be correct, too, it'll just use smaller words--and more of them!  (-;  \n\n~~\nPARSEC:\nA parsec is a standard unit of astronomical distance equal to 3.26 light years. It is also the distance a star has to be from earth so that its parallax is one arcsecond.\n\nOuch!  What the heck does all that mean? \n\nFirst off, we need to know about "parallax".  Parallax is easy.  It's just the change in the apparent position of an object when your position changes. \n\nA great example is easy to show: hold a thumb up in front of your face, about 20 centimeters away. Close your left eye, and move your thumb so that while looking through your right eye it is lined up with some distant object like a tree. Now close your right eye and look at your thumb through your left eye. See how it no longer lines up with the tree? That's because the angle formed by your eye, your thumb and the tree has changed. That angle is called the parallax angle.\n\nParallax angle depends on two things: (1) how far apart your eyes are (the baseline), and (2) how far away your thumb is. If your eyes are really far apart, the angle gets bigger. If the thumb is held farther away, the angle gets smaller. \n\nWhen you are in a car and you look at trees nearby, they appear to race past you. Far away trees appear to move slowly. This is because for the nearby trees the parallax angle is large and so it changes rapidly if the car moves a little bit. Trees farther away appear to change less as the car moves along. Their parallax angle is small. \n\nThis parallax method has been used for over a hundred years to find the distances to nearby stars. The Earth has an orbit about 300 million kilometers across. That's a long baseline! Even nearby stars will appear to change position due to parallax when you have such a long baseline. \n\nAstronomers needed some sort of unit to measure this change in position. What they decided was to make a standard unit for the distance. A star at that distance would change position by 1 arcsecond due to parallax. The distance is 3.26 light years, and is abbreviated parsec for PARallax SECond. (Ummm, one arcsecond is 1/3600 of a degree--which comes from 360 degrees of a full circle.) \n\nStars farther away than one parsec would shift by less than 1 arcsecond, and closer stars by more. The nearest star to the Sun is Proxima Centauri, which is about 4.3 light years away, or 1.3 parsecs away. That means that even the closest star has a parallax less than an arcsecond. Stars out to a distance of about 100 parsecs have had their distance measured this way. \n\nTo imagine how small a shift that really is, imagine someone standing 125 miles (about 200 kilometers) away from you and then moving to the left by one centimeter. That is the same shift! \n\nGot it?
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.
741917
This one is tough.  I've looked at it three different ways and still don't get it.\n\nFirst try.  If P(x1, y1) and Q(x2, y2) are given, and y^2 = 4ax, then the derivative is 2y dy = 4a dx  ==>  m = dy/dx = 2a/y.\n\nSo (y - y1) / (x - x1) = m1 = 2a/y1.  Solve for y to get\n\ny = (2a/y1) (x - x1) + y1  (equation of a tangent line)\n\nRepeat for (x2, y2), then set the two y-values equal to each other, and solve the resulting equation for x at the point where the two lines intersect.  That result is\n\nx = [y2 (y1y2 + 2ax1) - y1 (y1y2 + 2ax2)] / [2a(y2 - y1)]\n\nso it's possible to find the point of intersection T, but the algebra is very messy.  That's where I stopped.\n\nSecond try.  Parallel rays of sunlight striking a parabolic mirror will get reflected off the tangent and redirected to the focus.  Snell's law says the angle of incidence equals the angle of reflection.  That gave me a lot of angle relationships, but I didn't make much progress.\n\nThird try.  I got it to work for one special case.  Let P(a,2a) (the top end of the latus rectum) and Q(4a, -4a) be given.  Then the slope of the tangent is given by m = 2a/y, and\n\nm1 = 2a/y1 = 1 and m2 = 2a/y2 = -1/2\n\nFrom this we find that the two tangent lines intersect at T(-2a,-a).\n\nTo get the angles POT and QOT (where O is the focus at (a,0)), I got three vectors OP = <2aj>; OQ = <3ai - 4aj>; and OT = <-3ai - aj>.  The magnitudes are OP = 2a, OQ = 5a, OT = a sqrt(10).  Then using the inner (dot) product:\n\nOP dot OT = <2aj> dot <-3ai - aj> = -2a^2 = 2a^2 sqrt(10) cos POT, so cos POT = -1/sqrt(10) and angle POT = -108.435 degrees.\n\nOQ dot OT = <3ai - 4aj> dot <-3ai - aj> = -5a^2 = 5a^2 sqrt(10) cos (QOT), so cos QOT = -1/sqrt(10) and angle QOT = -108.435 degrees.\n\nIn this special case, the angles subtended by PT and QT are equal.\n\nAnd that's better than nothing.
989734
Two I have seen are AU where 1 AU or Astronomical Unit is the distance between the Earth and Sun, and LD which is the Lunar Distance, approx 250000 miles.\n\nLargest? Parsec: A unit of astronomical length based on the distance from Earth at which stellar parallax is one second of arc and equal to 3.258 light-years, 3.086 × 10^13 kilometers, or 1.918 × 10^13 miles.
180255
The only way that I know to do this with a normal telescope is to compare the star with others nearby of known magnitude and try to decide if it's brighter or dimmer until you find the same.\n\nOtherwise you'd need to measure the amount of light that comes from it, and I think you need computers to do that.\n\nThe magnitude scale is exponential, so deciding a magnitude ain't that simple. (ie a star of magnitude +2 is ten times brighter than a star of magnitude +3)
250137
It's not, unless you count pattern matching.  You look at the dots (connected by lines) on the star chart, then you look at the stars and try to see the same pattern.  It took me many nights over the whole summer that Halley's Comet was in the sky (20 years ago) to get the hang of it.
782266
Parallax\n\nIf you choose to go to the following web site for the Wikipedia free encyclopedia; I like the part on "stellar parallax" best.\n\nThe following is from that page. But the actual web site is better than this copy of it.\n\nStellar parallax\n \nStellar parallax motionOn an interstellar scale, parallax created by the different orbital positions of the Earth causes nearby stars to appear to move relative to the more distant stars. However, this effect is so small it is undetectable without extremely precise measurements.\n\nThe annual parallax is defined as the difference in position of a star as seen from the Earth and Sun, i.e. the angle subtended at a star by the mean radius of the Earth's orbit around the Sun. Given two points on opposite ends of the orbit, the parallax is half the maximum parallactic shift evident from the star viewed from the two points. The parsec is the distance for which the annual parallax is 1 arcsecond. A parsec equals 3.26 light years.\n\nThe distance of an object (in parsecs) can be computed as the reciprocal of the parallax. For instance, the Hipparcos satellite measured the parallax of the nearest star, Proxima Centauri, as .77233 seconds of arc (±.00242"). Therefore, the distance is 1/0.772=1.29 parsecs or about 4.22 light years (±.01 ly).\n\nThe angles involved in these calculations are very small. For example, .772 arcseconds is roughly the angle subtended by an object about 2 centimeters in diameter (roughly the size of a U.S. Quarter) located about 5.3 kilometers away.
139607
Venus is brighter ,but the space station when you saw it might have been turn in a way that it was'nt getting a big reflection.Maybe it could be the time of year because of the axis of the earth .Just a guess.
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.