Neptune requires about 165 years to orbit  the sun. How is Keplers law used to find its distance from the sun?
396604
dr_islam_safwat is exactly correct, as far as he goes. The problem is that Kepler's 3rd law does not tell you how far Neptune is from the Sun, it only tells you how much farther from the Sun it is than the Earth. The real trick to to find out how far the Earth is from the Sun. Actually, knowing the absolute between any two planets would be enough information, because then Kepler's 3rd law could then be used to find all the other absolute distances from the ratios.
384805
The First law of Kepler is: \nPlanets move along elliptical orbits with the Sun\nat a focus.\n\nSo, all i can say is that the sun is relatively stagnant in our immediate universe.\nThe earth rotates along an elliptic orbit around the sun.\nThe moon rotates along an elliptic orbit around the earth.\n\nWell, all "movements" are relative to some reference point.
329537
5 billion years or so.\n\nhttp://www.enchantedlearning.com/subjects/astronomy/sun/sundeath.shtml
177995
Earth and Neptune Compared :\n1) Distance from Sun , Earth- 147,000,000 Kilometers,   Neptune- 4537,000,000 Kilometers.           (Both minimum)\n2) Diameter at the equator,  Earth-12,756 Kilometers.    Neptune- 50,538 Kilometers.               (It is bigger than earth.)\n 3) Surface Temperature,  Earth – 22 Degree Celsius , Neptune – Minus 220 degree Celsius.(-220)( Both average.)\n 4) Atmosphere,   Earth- Mainly Nitrogen and Oxygen. Neptune- Mainly Hydrogen and Helium.\n 5) Number of Moons  Earth – Only One   Neptune -- Eight.
823309
If you write the number sequence 0,3, 6, 12, 24, 48, 96, and add 4 to each number (4, 7, 10, etc) then divide by ten, you get the distance of each planet from the Sun, measured in astronomical units (1 a.u.=93 million miles=150 million kilometers).  This works out as Mercury 0.4, Venus 0.7, Earth 1.0, Mars 1.6, --- 2.8, Jupiter 5.2, and Saturn 10.0  This works out pretty accurately, but there was nothing at 2.8. Herschel discovered Uranus in 1781 at 19.6, right where another planet was supposed to go according to the Titius rule.   When Ceres was discovered in 1801, it was at 2.72, filling the gap.  Neptune, however is not near 38.8, but at 30.  Astronomers now believe the "law" is either a coincidence, or the result of billions of years of gravitational interactions.
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)
392987
How far is Neptune from the Earth?\nYour answer: 2 billion miles\nCORRECT!
725034
An interesting question, which requires thinking carefully about how our common measure of time is related to certain astronomical (solar) positional recurrences.  A lot of people (though not all) have answered you as though the year were exactly 365.25 days long on average, but it's more subtle than that.  \n\nThere are in fact two particular "years" of interest:\n\n1. The sidereal year (period of the Earth's revolution about the Sun with respect to the stars, i.e. operationally, the time for the Sun to apparently "return" to the same position against the stars, after a "year" [crudely 365+ days], as viewed from Earth):\n\n365.256366... days.\n\n2. The tropical year (time for the Sun to cross the celestial equator [the projection of the earth's equator in space] again [e.g. from south to north] after a "year" [crudely 365+ days]).  Alternatively, "the period of revolution of the Earth about the Sun with respect to the Vernal Equinox":\n\n365.242199... days.\n\n(Both figures are quoted to an accuracy sufficient for our present purposes.)\n\nThese two different "years" differ by just ~20.4 minutes, owing to the precession of the Earth.  (One full precession of the Earth's spin axis in space takes ~25781 years.  This long, latter number of years is in fact equal to 365.25 or so days divided by the (essentially observed) "motion of the Vernal Equinox along the celestial equator," at the rate of ~20.4 minutes per year as just mentioned --- check it out.  [Sorry about the technical language, but this is technical stuff.]) \n\nIt's actually the tropical year that determines the rate at which the observed seasons recur, and since we want them to be "at the same time of the year," our long-term timekeeping is geared to the tropical, not the sidereal year.\n\nNotice that at 365.242199 days, this "tropical year" is just a tad short of 365.25 days.  That's why, instead of the simplistic idea of a leap year occurring every 4 years (and having an extra day that accounts "exactly" for 4 x the extra 0.25 days), there will come a time when we've overcounted a bit.  \n\nFor example, after 100 years, we should have had 100 x 0.242199 "extra days," or 24.2199 days.  The "one more day every 4th year" rule would give us 25 extra days per century, ~ 0.78 days more than needed.  If we could drop just 1 day then, that would be better --- but then we'd be under by ~ 0.22 days --- aha! --- close to a quarter of a day!  That's why the revised rule comes in that complete century years are NOT leap years --- UNLESS they're divisible by 400!  (Figure out for yourself from the 365.242199 figure how well this works in the long run.  I think you'll find that the calendar, on average, doesn't get a full day out of step with these revised rules until something like 3,300 years have gone by!) \n\nO.K., so maybe this is much more than you wanted to know.  However, with the right understanding of what the length of the year we usually measure is, 365.242199... days, then 1 million days is 10^6 divided by this, or 2737.90926... years, or 2737 years and 332.10... days.***\n\n*** Remember, this is a mathematically average result, for essentially perfect intervening TROPICAL years (to the accuracy given).  If you want to know it in CALENDAR years and a specific whole number of days, the answer will vary slightly about the preceding result, according to when you start counting. This variability is not just because of the usual "every fourth year is a leap year"  pattern, but also because of the way that century years aren't all treated in the same manner.  It's awkward to give a general rule!  \n\nLive long and prosper.
309054
given the figures, another way to "imagine" how far it is\n IS that Light (at 186,000miles/sec)takes about 8.3 minutes to get to earth from the sun and takes 5.5 hours to get to pluto\n\nPluto is 40 times the distance away from the sun as the earth \n\nOver 11,000 earths will fit between the sun and earth if lined up so about 460,000 earths fit between the sun and pluto.\n\nAlso if the Sun was a beach ball, the earth would be a pea and Pluto would be A LARGE grain of sand.\n\n\n\nThe numbers are often useless unless we "translate them" into "concrete" terms because we have peas er I mean grains of sand for brains ;)))).
354744
This is a great question, actually, and one that bothered astronomers for centuries.  In short, there’s no simple way to measure the distances to planets—at least not without really good telescopes (and a bit of luck).\n\nIn the early 17th Century, Kepler devised his Third Law, which releated the distance of a planet from the Sun and the time it took to revolve around the Sun.  So if you know one interplanetary distance, you can figure out the rest.  Which is all well and good, except that there were no good distances at the time.  (A rough sense of the distance between the Sun and Earth constrained the numbers somewhat, but the estimates were pretty far off—Kepler thought the Solar Syatem was about one seventh its actual size.)\n\nThe first accurate means of determining the distances was by observing transits of Venus across the face of the Sun.  Through relatively simple trigonometry, one can calculate the actual distances between the Sun, Venus, and Earth.  Unfortunately, transits don’t occur very frequently, so it was only in the 19th Century that reasonable results were achieved.\n\nNowadays, the distances to planets are determined by bouncing radar off them.  Because we know how fast light travels, we can simply measure the time between emitting a radio pulse and its return back to Earth, which gives us the distance!  Easy as pie…\n\n(In the case of the Moon, we actually bounce lasers off a reflector left there by Apollo astronauts.  Because of the phase information of the laser light, we can get extremely accurate distances to our nearest neighbor.  That’s how we confirmed that the Moon is moving away from us at the rate of about 3.8 centimters per year.)\n\n\nRyan Wyatt\nRose Center for Earth & Space\nNew York, New York
351394
Pluto.\n\nPluto\nTiny Pluto is the last planet in our family. The dark, cold planet's average distance from the Sun is a little over 3.5 billion (3,500,000,000) miles. If you tried to step off this distance, you would be over two miles away from your Sun when you reached the location of Pluto. If you got on a jet and flew at 600 miles per hour from the Sun to Pluto, it would take 675 years. If you could drive the same distance, it would take over 5,700 years.
456292
Pluto is 6 Billion kilometers away from the sun.  Light takes about 5.5 hours to reach Pluto.  Which means Pluto is only 6 light HOURS away from the sun.
1005688
This is actually a difficult question to answer because the planets are moving in their orbits all the time, and so the distance from each planet to Earth is constantly changing. When both planets are on the same side of the Sun, and form a line with the Sun, then they are closest to each other. When two planets are on opposite sides of the Sun, and form a line with the Sun, then they are farthest away. For example, the planet Venus is about 42 million kilometers (26 million miles) from Earth at its closest point and about 258 million kilometers (160 million miles) away at its most distant point. \n\nUsually when people ask this question, what they mean is "What is the distance between the orbit of Earth and the orbit of each planet?" or "What is the closest that each planet comes to Earth?" (These are essentially the same question, right, because the planets can't get any closer than their orbital spacing allows.) You can compute this in a rough way by assuming that the orbits are circular, and looking at the planet-to-Sun distance for each planet. Since the distances are so large, we usually express them in Astronomical Units (AU). (An AU is the distance from Earth to the Sun, about 150 million kilometers or 93 million miles). The list below shows the distance of each planet from the Sun in AU. \n\nPlanet Distance from Sun in AU \nMercury 0.39 \nVenus 0.72 \nEarth 1.0 \nMars 1.5 \nJupiter 5.2 \nSaturn 9.5 \nUranus 19.2 \nNeptune 30.1 \nPluto 39.5 \n\nTo find the distance between the orbits of two planets in AU, subtract the two planet-Sun distances. For example, Earth orbits at 1 AU from the Sun and Venus orbits at 0.72 AU from the Sun. The difference between these two distances is 1 - 0.72 = 0.28 AU. To get the number in kilometers, multiply by the conversion above: 0.28 x 150 = 42 million kilometers. \n\nThe numbers above are generally what's used in introductory astronomy classes.
647693
lots :-)  Let me figure it out...\n\nFirst the number is spelled 'googol' and is equal to 10^100 (1 followed by 100 zeroes).\n\nA year nominally has 31,556,926 seconds (365.242199 days x 24 hours x 60 minutes x 60 seconds).  I used the number 365.242199 because that is how long it takes for the earth to go once around the sun... we round this to 365 and add an additional day every 4 years (with some exceptions 3 out of every 4 centuries, blah, blah, blah)...\n\nAnyway, I digress.  Back to the problem at hand.\n\n10 x 10^99 / 3.1556926 x 10^7 = (10/3.1556926) x 10^92\n\n3.16887646 × 10^92 years.  There is no English name for this word but it is *lots* like I said at the beginning.  :-)\n\nThe closest I can get to a name would be to use vigintillion (10^63) as a base and go from there...\n\n3.16887646 x 10^29 x 10^63 or\n\n316 x 10^27 x 10^63 +\n887 x 10^24 x 10^63 +\n646 x 10^21 x 10^63\n\nSo that would be:\n316 octillion-vigintillion, 887 septillion-vigintillion, 646 sextillion-vigintillion years\n\nOkay, a little more research and I found that my names were close but not exactly correct.  This number should be called:\n  \n316 octovigintillion 887 septenvigintillion 646 sexvigintillion years!
928843
see: http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Kepler.27s_third_law_2
67566
Kepler's 3rd Law answers this.  His law is based on the fact that the distance must be the distance that causes the gravitational attraction to between the sun and planet to be the same as centrifigual forces that tend to fling the planet away from the sun.  If the planet were to close, the gravitational attraction would pull the planet into the sun; if it were too far away, it would fly away from the sun.\n\nOn to Kepler's 3rd Law:\n   Period^2 = 4*(PI^2)*(Distance^3)/G*(Mass)\nAccording to Kepler, Mass was the mass of the sun.  Newton later pointed out that it should be the Mass of the sun plus the mass of the planet, but your planet is so small relative to the sun that it doesn't matter.\n\n    To solve this equation, get everything into base SI units (kg, s, m, N, etc.)\n    775 days = 6.696x10^7 seconds\n    G = 6.672 x 10^-11 N m^2 / kg^2\n\nAnd the equation becomes\n   (6.696x10^7)^2 = 4PI^2(Dist^3) / (6.672*10^-11)(1.99*10^30)\n   Distance = 2.47x10^11 meters\n\nThere is a simpler from of the equation using units of AU and years, if the mass of the sun is our sun (1.989x10^30 kg).  It is simply \n\n    (Period in years)^2 = (Distance in AU)^3\n\nSince 775 days / 365.34 days = 2.122 (in years)\n     2.122^2 = (Distance in AU)^3\n     Distance = 1.65 AU\n\n1.65 AU = 2.47x10^11 meters, which is the same answer as above.
652664
The length of the earth's orbit (orbital circumference) is 924,375,700 kilometres according to wikipedia.