How Is A Comet's   Orbit / Course Calculated ?
823944
The basic method is to observe its position over several nights.  The subsequent motion of the comet against the background stars is then fitted to a curve that describes the orbit.  The difficulty with this method is that a elliptical orbit can resemble a parabolic or hyperbolic orbit at certain sections of the orbit.  In the early days of astronomical observations, this closeness made it difficult to determine if a comet was in a closed, periodic orbit or whether it would be on an escape path from the Solar System.
217914
A comet far from the sun would look like a small black rock.  There are many small black rocks out in space.  The difference between a comet and a lot of other objects out there is that comets have large amounts of ice in them.\n\nAs the comet nears the sun these ices turn to gas.  The comet now becomes enveloped in a ball of gas.  This ball of gas, or "coma" scatters sunlight, and the comet becomes much brighter.\n\nStill closer the amount of ices turing to gas increase and the comet grows two tails.  One tail is made up of gas that has been ionized from the UV light the sun puts out.  These gas ions are carried away from the comet by the solar wind.  This tail always points away from the sun.\n\nThe second tail that forms is made up of rocks, dust, and other assorted particulate.  They are blown off of the comet by the rapidly expanding gas being produced by melting ice.  Since these are particles, the behave just like any other particle and try to orbit the Sun.  The trail behind the comet in its orbit.
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)
837333
If you want good data and you are not trying to observe something that is temporary like a comet, I highly recommend skyview. You don't get to steer the telescope and see real time images, but there are some great surveys in many different wavelengths there.\n\nhttp://skyview.gsfc.nasa.gov
289592
You can't-you dont have enough information.  You need to know the mass of the planet as well, to know the gravity.  Orbital velocity is determined based on calculating the gravity based on the mass, assuming all the gravity originates in the exact center of the planet.  Then, if you know how far the surface is from the center (the radius of the planet) you can calculate what the gravity is at this point...gravity decreases as the square of the distace from the center of gravity.  Now, you know the gravity at the surface, and can calculate escape velocity.
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Kepler's Laws (of planetary motion) use Calculus (first quarter/semester if that is what you were wondering about).  You can get a good understanding of them by simplifying them with geometry, trig and logic though I believe.\n\nMost descriptions of the motion of bodies (like planets and comets) use calculus, trig(for angles), and some other basic geometry, as well as good ole logic.
169817
The moon's orbit is nearly circular and traverses 13.2 degrees along this orbit over the course of a day. Its orbital velocity varies from 0.964 to 1.076 (mean 1.023) kilometers/sec (multiply the mean by 86,400 to get the number of km/day traveled. Multiply by 0.62137 to convert to miles). See the provided links for any additional information you may be looking for.
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A continuing effort to identify even near Earth asteroids which are relatively easy to track and have a greater possibility of collision is still incomplete, with only a fraction of their orbits plotted. Since the orbital periods of comets and other Oort cloud entities are in the tens, hundreds, or even thousands of years, many of these objects have never been seen in recorded history, much less their orbits calculated and possibility of collision with Earth evaluated. With this in mind, an event such as this might be possible at any point in the future, although since objects from the outer solar system tend to be rather reflective, chances are that they would be spotted and we would have some advance warning of a likely intersection with Earth orbit presenting a cause for concern. That said, the chance of it happening in any particular year is extremely remote, the likelihood of it happening sometime in the next several million years greater.
97099
Please take a look at:\n\nhttp://en.wikipedia.org/wiki/Orbital_velocity
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I don't know the name, I think it just has a number.\n\nIt is about 1,000,000 km away and is a captured asteroid about 1km across.  It is irregular in shape.  It is in an unstable orbit and will eventually leave the Earth's orbit and enter solar orbit.
881773
A shooting star is a meteorite burning up in our atmosphere.A meteorite is a small piece of space dust usually pea size,maybe a little larger or smaller.Comet are a collection of space debris loosely held together by its own gravity;thus the dust trail.Asteroids are space bodys usually of some size(a few hundred feet to several miles wide).Comets can be quit large as well and the tail is caused by the sun melting the comet.
339322
if any object following a straight or whatever path was to be attracted by some other larger body and put into and orbit then thru mathematical equations the orbit will always be elliptical(considering variable decceleration and velocity) that's why there is a doubt that moon was just a passing object that earth caught up..(coz it's orbit is circular and that it perfectly covers up the sun during eclipses....search for anomalies of the moon on google) so the orbit cannot be circular unless u "put" an object specifically into that shaped path...\n\nregards...\npretty rough explaination ..sorry cud not include the equations :)\n\nvivek
29063
http://www.eskimo.com/~jbm/calculations/traj/traj.html click this link it'll calc the answer for you
538789
The fact that there's pretty much nothing random in orbital mechanics is very helpful along with the fact that we know pretty much everything about what affect all the objects in the solar system have on the spacecraft, that way they can just run the simulations using computers to high precision and take account of pretty much everything.\n\nIt is of course common to use mid-course correction burns to correct for any errors when the craft was first sent off into space.
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The Earth orbits the sun in an ellipse.\n\nEvery ellipse has two foci. It is at these two points where the orbited body is normally located.\n\nPlease see the wiki source that has further information on how to get the numerical data if you need it.
660999
THe orbital angular momentum is determined by the number L (usually a small l, but I want to be clear here). For L=0, you have an s orbital, for L=1 a p orbital. In general, the orbital angular momentum is given by L*hbar.
1090033
There is actually a comet passing by Earth this month.  To calculate where the comet will be visible go to\nhttp://ssd.jpl.nasa.gov/horizons.cgi#top\n\nType in information for comet 73P