How did Kepler actually calculate his 3 laws of planetary motion in the 1600s?
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He came to them empirically.  The math needed to work them out was not around yet, but he just fit what he could to his data.  Well Tycho Brahe's adat really.  Kepler essentially analysed the orbit of Mars in particular for almost a decade before he came to his conclusions about the shape of planetary orbits.\n\nOne other thing to point out, Kepler was a real wack-job.  Unlike say Galileo who's writings are very easy to read, and you can see the beginings of modern science in, Kepler's writings are 100's of pages of mysticism, musical notation, and invokations of holy spirit force.  His three laws are buried pretty deeply within his works, it is a challenge for the reader to find them, even if they know what they are looking for.  He also spent quite a bit of time as a lawyer, defending his mither from charges of witchcraft.\n\nAs far as how he came up with law two and law three, they more or less followed from his first law.  Once he became convinced that ellipses were the way to go he was able to see how relationships of time, area, and reletive distance all were interconnected.  Without the calculus of Newton, that is all they were, relationships.  Really that is the brilliance of Newton, he managed to with one fundemental concept explain and unify both the mechanics of Galileo, and the orbits of Kepler.
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By Keplers 3rd Law T² α R³\n\nOr T1²R1³ = T2²R2³\nFor earth R = 1 and T = 1 and for Jupiter R = 5.2\nSo T2²/5.2³ = 1²/1³\nSo T2 = 5.2^(3/2)\n≈11.86 years
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If I understand your question, Earth's velocity around the Sun changes as its distance from the Sun changes; we are farthest from the Sun during summer (northern hemisphere "summer")and our orbital velocity is at its lowest. When our orbit takes us in closer to the Sun in winter, our velocity increases. Kepler determined the principles behind this relationship hundreds of years ago.
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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)
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The only thing I know that is close is called the Johannes law\nand it is Astronomy\n\nI hope this is what you needed
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Perhaps an illustration will help.\nhttp://home.cvc.org/science/kepler.htm
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http://en.wikipedia.org/wiki/Gravitation#Newton.27s_law_of_universal_gravitation
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If you're speaking of the motion of Earth on it's axis or it's rotation about the sun, the answer is the scientific principle of Inertia, read now!\n\nThe principle of inertia is one of the fundamental laws of classical physics which are used to describe the motion of matter and how it is affected by applied forces.Inertia is the property of an object to resist changes in velocity unless acted upon by an outside force.Inertia is dependent upon the mass and shape of the object. The concept of inertia is today most commonly defined using Sir Isaac Newton's First Law of Motion, which states:\n\nEvery body perseveres in its state of being at rest or of moving uniformly straight ahead, except insofar as it is compelled to change its state by forces impressed. [Cohen & Whitman 1999 translation]\n\nThe description of inertia presented by Newton's law is still considered the standard for classical physics. However, it has also been refined and expanded over time to reflect developments in understanding of relativity and quantum physics which have led to somewhat different (and more mathematical) interpretations in some of those fields.\n\nNow read more at the links below:
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1- gravity , elastic forces (as in bows and catapults), Archimedes' push   - In Galileo time was discovered air pressure (Torricelli experiment) \n\n2- Mass \n\n3- Forces add up for the same mass, then acceleration also adds up :   F1 is force from friend 1, A1 is acceleration/thrust  given by him  \nF1= m.A1 \nF2= m.A2\n\nF1 + F2 = m. (A1 + A2)
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Not sure who was the 'first' but the Egyptian Ptolemy left records that show he knew the Earth was a sphere.  He even calculated its circumference fairly accurately.\n\nWhere he was incorrect, of course, was proposing a geocentric view of the universe with the Earth in the middle and fairly complicated system of orbits to describe the movement of the various visible astronomical bodies.  It wouldn't be until Copernicus over a thousand years later, until this particular view was corrected.
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In the 16th century, Nicolaus Copernicus developed the heliocentric model of the solar system, in which the Sun is stationary at the center, and Earth moves around it. This view of the solar system challenged Ptolemy’s geocentric model, which had been the accepted theory since the 2nd century. In Ptolemy’s model, Earth is stationary in the center of the solar system, and the other planets and the Sun move in complex orbits around it. The Copernican model gradually gained acceptance, for it provided a simpler explanation of the planets' motions.
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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)
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http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
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see: http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Kepler.27s_third_law_2
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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.
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three laws of thermodynamics\nthree laws of Kepler\nThree axioms of Euclidean geometry \nplus\nE=mc2\n\nhere's your ten scientific "rules"
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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.