Find the average radius of the orbit of Saturn?
830105
8.8 AU....or 868 million miles from the sun.
471913
The daily path of an object on the celestial sphere, including the possible part below the horizon, has a length proportional to the cosine of the declination. Thus the speed of the diurnal motion of a celestial object is this cosine times 15 °/hr = 15'/min = 15"/s, i.e. (compare angular diameter):\n\nup to a Sun or Moon diameter every two minutes \nca. four seconds for the largest planet \n2000 diameters of the largest stars per second
248549
Mean Equatorial Radius\n71,492 km\n(11.209 of Earth's radius)\n\n142,984 km diam *3.14\n\n\n448,969.76 km is the cirumference\n\nif you search for astronomy pic of the day then search jupiter everyting is there.  Plus some relly awesome pictures
448267
the average orbital velocity of earth is 29.783 km/s.\n\nhttp://en.wikipedia.org/wiki/Earth
106990
Period is proportional to R^3/2.\nThis is an answer I posted to bombay2pac2000 under Astronomy & Space 8 hours previously, and it applies here as well.\nKepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2 \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*(R^3/2)/SQRT(GM)
952581
http://planetmath.org/?op=getobj&from=objects&id=3808
636756
Radius:\nEquatorial 6,378.135 km\nPolar 6,356.750 km\nMean 6,372.795 km
583647
According to the tables in the CRC Handbook of Chemistry & Physics, the surface temperature of Saturn is about 97°K, or -176°C, or -285°F.  Brrrrrrrrrrrrrr!
905767
You are making it needlessly hard on yourself.  The planet has the same mass as the earth, and someone on it weight the same as on earth, therefore the surface of the planet is the same distance from its center as the earth radius, i.e. this planet has the same dimension as earth.\nSo, the radius of the planet is 6,372.797 km, same as earth's.\nYou don't need to compute anything.  The orbital velocity of the planet has nothing to do with it.\nIf the weight was different, you only need to scale it remembering that the weight would vary as an inverse of the square of the radius.
754180
Here's one simple animation: (I'll keep looking).\nhttp://janus.astro.umd.edu/javadir/orbits/ssv.html\n\nHere's a FANCY video:\nhttp://www.bbc.co.uk/science/space/solarsystem/3d_tour.shtml
871968
Here is the math for finding our solar system's barycenter. (It's under the heading "Barycenter").\nhttp://en.wikipedia.org/wiki/Center_of_mass
406753
The average radius of the Earth is about 6378.1 kilometers, or abnout 6378100 meters.\nCompared to this, a few hundred meters doesnt make a significant difference.\n\nFor reference, the mass of the Earth is about 5.9742 E24 kilograms.\nG = 6.67 E-11\n\nThe gravitational acceleration on the surface of Earth (r = r_E = 6378100 m) is,\ng = G*M / r^2 = about 9.795 m/s^2\nIs the object is now 1 km above the surface of Earth (1000 meters higher), the gravitational acceleration is about 9.792 m/s^2, just .003 m/s^2 smaller, barely noticible.
1071951
NASA/JPL lists a detailed ephemeris for this object with celestial coordinates ==> http://neo.jpl.nasa.gov/cgi-bin/db_shm?sstr=2004%20XP14
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.
748523
Saturn has a density less than water.  It's density is 0.6873 g/cm^3.  Water has a density of 1g/cm^3.
251207
Don't have it handy,but it'sjust a little denser than water, from what I remember. Check www.nasa.gov. They have lots of planetary info!
97099
Please take a look at:\n\nhttp://en.wikipedia.org/wiki/Orbital_velocity