how do we determine the temperature at a certain distance from the sun?
604131
Extrapolate the formula, using the surface temoerature, and instead of the earth distance, substitute the distance you want to caluclate the temperature at.  \n\n\nMeasuring the Sun's Temperature\nOne method, called Wien's law, uses the wavelength of the peak emission, wavelengthpeak, in the Sun's continuous spectrum. The temperature in Kelvin = 2.9 × 106 nanometers/wavelengthpeak.\n\nAnother method uses the flux of energy reaching the Earth and the inverse square law. Recall from the Stellar Properties chapter that the flux is the amount of energy passing through a unit area (e.g., 1 meter2) every second. From the Inverse Square Law of Light Brightness, you find that the solar flux at the Earth's distance = the Sun's surface flux × (Sun's radius/Earth's distance)2 = 1380 Watts/meter2. Since the Sun's photosphere is approximately a thermal radiator, the flux of energy at its surface = sigma × (the Sun's surface temperature)4, where sigma is the Stefan-Boltzmann constant. Rearranging the equation, the photosphere's temperature = [(solar flux at Earth)/sigma) × (Earth distance/Sun's radius)2]1/4.\n\nThese two methods give a rough temperature for the Sun of about 5800 K. The upper layers of the photosphere are cooler and less dense than the deeper layters, so you see absorption lines in the solar spectrum. Which element absorption lines are present and their strength depends sensitively on the temperature. You can use the absorption line strengths as an accurate temperature probe to measure a temperature of about 5840 K. \n\nRead the link, you'll have to scroll down to come up to the stuff I have cut and paste above.\nst\nHope you find this useful.
111333
Begin by subtracting 32 from the Fahrenheit number. \nDivide the answer by 9. \nThen multiply that answer by 5.
1154551
By the spectrum of the light that it gives off.\nThink of a burner on a stove. As the temperature increases the light that it gives off changes (red to orange to bright orange). In exactly the same way we can look at the light of the sun (or distant stars) and by examining the peak wavelengths of light it gives off, we can determine what the actual temperature is.\n\ndo a google search for black body radiation or black body spectrum. or check out the wikipedia link\nhttp://en.wikipedia.org/wiki/Black_body
501846
lets say the temperature difference between yestar and today (0degrees) is x. then tomorrow's temperature will be '0-2x' degrees.\n\nmore example:\nyest temp    = 10 degree.\ntoday's temp =  0 degrees.\ntom's temp   = -20 degrees.
769934
It's the actual temperature with humidity, wind chill factor (if applicable), dew point and other conditions taken into consideration. There's actually a mathematical equation to figure it out but it may be hard to find since the calculations are done by computers these days.
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)
533060
You need to know the temperature and use the variants of PV=nRT.
881785
10 degrees\nor \npi/18 radians
326233
They can basically find out the composition of the sun ie.., what material it is made up of, which is because different elements have unique spectral characteristics.And they can estimate its surface temperature.
278956
This has been asked and answered many times before:\n\nhttp://answers.yahoo.com/search/search_result;_ylt=Amylv82zdJRUVkmYew.javjpy6IX?p=temperature+of+the+sun
1136926
A simple solution is to mulitply the celsius temperature by 2 and then add 31 (32 - 1). It's not the exact science, but It will give you the right answer 97% of the time.
812605
°F = (9/5)*°C + 32\n\nSo\n\n°F = (9/5)*(37.8) + 32 = 100.04 °F
65568
18.8 degrees F., (-7.3 degrees C)
441948
98.4degrees farenheit or 37 degrees celsius.it can vary a little.
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).
941285
This website shows relative distances from the sun. You can then calculate their distances from each other by using (Farther planet's distance)-(closer planet's distance).\n\nhttp://www.idahoptv.org/ntti/nttilessons/lessons2000/lau4.html
581211
Astronomers determined the size and shape of our galaxy and measured distances to other stars using the “parallax method” (and eventually other methods using the star’s light spectrum).  Unfortunately, these methods do little to help us find out the exact distance to our very own star, the Sun.  Surprisingly, to get the distance to the sun, astronomers first had to determine the exact distance to Venus!\n\nWith the invention of radar, scientists were able to measure the distance to Venus very precisely.  By timing how long it took the radar beam to travel at the speed of light to Venus and back, the total distance to the object can be determined from the equation:\n\nDistance = (Speed of Light) X (Total Time)/2\n\nThe reason the total time is divided by two is to get just the distance from the Earth to the object.\nEarth-Sun Distance = (Earth-Venus Distance) / (Cos q)\n\nUsing trigonometry, astronomers now know that the “Astronomical Unit” = 92,955,820.5 miles (149,597,892 kilometers).  This incredible degree of accuracy is possible because the speed of light is known very precisely and extremely accurate clocks are used to time the radar reflection.  Radar can’t be used to determine the distance to the Sun directly because the Sun doesn’t have a solid surface to reflect the radar efficiently.