Orbital Longitude by Newton's Cycloid Method

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    We have before us in our imaginations a body moving on an elliptical orbit and we want to determine the longitude on that orbit that the body occupies at a given elapse of time after the body passes through the orbit's peritelion, the point at which the body comes closest to the orbit's prime focus. Using his weird dynamic geometry, Isaac Newton gave us a method for solving that problem. Now I want to look at that method and Newton's proof of its validity with the intent of using it. So on the drawing board of your imagination construct an ellipse of eccentricity e and construct around it the diagram:

    We have the ellipse APB with center at O and primary focus at S. The semimajor axis OA has length a, so the line OS has length ea. Using O as the center, construct circles AQB (which circumscribes the ellipse) and GFC with radius r=a/e; that is, line OG has the same proportion to line OA that OA has to line OS. Treating the ellipse and the circles as if we had attached them rigidly to one another, imagine letting the circle GFC roll along the line GKH: as the circle rolls, the point A will trace out the cycloid ALI.

    By Newton's method we so construct the line segment GK that its length has the same proportion to the circumference of the circle GFC (=2πxOG) that the time elapsed since the body passed through its peritelion (the point A) bears to the period of the orbit. Through the point K construct a straight line perpendicular to the line GK and extend it to cross the cycloid at L. Through the point L construct a straight line parallel to GK. That line crosses the ellipse at the point P, the point occupied by the orbiting body at the described time, and crosses the small circle at Q. From that information we can derive the orbital longitude of the body's position.

Proof and Verification

    Construct a straight line through the points O and Q and extend it in both directions. That line crosses the large circle at F. From the point S construct a straight line perpendicular to OQ, meeting OQ extended at R and from the point A construct the straight line AT parallel to RS. Now we proceed by the following steps:

    1. The area of the elliptic sector ASP, the area swept out by the body's radius vector as it evolves from SA to SP (not shown), has the same proportion to the total area enclosed in the ellipse as the time elapsed since the body passed through peritelion (which time we represent with t) bears to the total period of the orbit. We accept that statement as true to geometry by our assent to the truth of the law of conservation of angular momentum expressed in the form of Kepler's second law of planetary motion.

    2. The area of the circular sector ASQ has to the area of the elliptic sector ASP the proportion . We accept that statement as true to geometry by our knowledge that an ellipse shows us a perspective-affine image of a circle; that is, if we take a circle of radius a and shrink it uniformly in one direction by the ratio , we obtain an ellipse of eccentricity e and semi-major axis a.

    3. The area of ASQ equals the area of the circular sector AOQ minus the area of the triangle SOQ. We accept that statement as true to geometry by inspection of the diagram.

    4. The area of the triangle SOQ equals the product of SR and OQ. We accept that statement as true to geometry by reference to Euclid's "Elements", Propositions I: 37-40, with OQ as the triangle's base and SR as the triangle's altitude.

    5. The area of the circular sector AOQ equals the product of OQ and the length of the arc AQ. We accept that statement as true to geometry by Newtonian integration: that is, we inscribe N triangles, each subtending an angle AQ/N, into the sector AOQ, add up their areas, and then let N go toward infinity, taking the limit of the sum of the areas of the minuscule triangles as coming infinitely close to the true area of the sector as the sum of the lengths of the bases of the triangles comes infinitely close to the length of the arc AQ.

    6. The area of ASQ = OqxAQ - OqxSR = OQ(AQ - SR). We accept the truth of that statement to geometry by reference to Statements 3, 4, and 5 above.

    7. We have . We accept that statement as true to geometry by reference to Statements 2 and 6 above.

    8. We have GF - AT = (AQ - SR)/e. We accept that statement as true to geometry because GF = AQ/e by construction (the arcs GF and AQ subtend the same angle on radii OA and OG = OA/e) and because AT = SR/e by construction (both line segments comprise corresponding sides of similar right triangles whose hypotenuses, OS and OA, have the ratio OA = OS/e).

    9. Rolling the large circle a distance GF puts the point A a distance GK = GF - AT from its starting point along the line GH and a distance OF+OT above the line GH. We accept that statement as true to geometry by inspection of the diagram, noting that when the point F coincides with the point H, the line OF lies perpendicular to the line GH.

    10. OF+OT = the altitude of Q above GH before we roll the circle, because OT = QD (by OT and QD being corresponding sides of similar right triangles whose hypotenuses are equal (OQ = OA by construction)) and OF = OG (by construction), so OF+OT = OG+QD, the altitude of the point L.

    11. The area of ASP (= OQ(AQ-SR), by Statements 2 and 6) has to the area of the ellipse (=π(OA)2) the ratio OA(eGK): π(OA)2 (by Statements 7, 8, and 9). That ratio equals the ratio eGK:2πOA :: GK:2πOG :: t:orbital period. Q.E.D.

Calculating the Longitude

    We want to calculate the angle ASP. But that statement sets before us an intractable problem, one that we cannot solve in a finite number of steps. We have several approaches that we can take to the problem, but one obliges us to know the solution before we can solve the problem and the other requires of us that we solve an equation of the form a=bφ+cSinφ for φ, an intractable problem in itself. Fortunately Mister Newton gave us a good approximation method that yields us a solution that we can, with little effort, take arbitrarily close to the true solution.

    We begin by defining an angle beta as the product of our orbit's eccentricity and one radian (β=e x 57.29577951 degrees) and a length W as the quotient of the orbit's semi-major axis by the eccentricity (W = OA/e = OG). We also take the time t, expressed as a fraction of the orbit's period, and define the angle ν=2πt.

    Next we guess at a solution. On our diagram we establish a point p close to where we believe the true solution P lies. From that point p we draw a horizontal line to cross the ellipse's major axis at M and extend that line to cross the circumscribed circle at V. Because most of the ellipses that we will encounter have eccentricities close to zero, we can use the solution that we would get for a circular orbit as our guess, thereby making our estimated angle AOV = 2πt. We now take a series of steps to improve that approximation:

    1. Define the angle δ=βSin(AOV) and calculate ε=(ν-AOV+δ)(W/W-OACosAOV). Because we have set ν=AOV, we can rewrite that equation as ε=(OA/OA-eOACosAOV) = δ(1/(1-eCosAOV)).

    2. Define φ=βSin(AOV+ε) and calculate γ=(-ε+φ)(1/(1-eCos(AOV+ε))).

    3. Define η=βSin(AOV+ε+γ) and calculate


    4. And so on for as many steps as we require for the accuracy that we want.

    Let AOQ=AOV+ε+γ+ι+.... We can calculate the Sine and the Cosine of that angle as if we were going to calculate the sides QE and OE of the right triangle whose hypotenuse is OQ. But we actually want the sides of the triangle EOP: the sine side has length OQSinAOQ and the cosine side has length OQCosAOQ. Now we know that the triangle ESP has a sine side of length OQSinAOQ and a cosine side of length eOA-OQCosAOQ=OQ(e-CosAOQ). Thus we can calculate the angle ESP as the arctangent of SinAOQ/(e-CosAOQ). Finally we have the orbital longitude, the angle ASP=180-ESP (or π-ESP if you use radians).

    For exercise let's calculate the orbital longitude that Earth will occupy on AD2159 May 18. In that year Earth will pass through the perihelion of its orbit at AD2159 Jan 07 PM 09:56. Thus we want to find where Earth will be 131 days later (t=0.3587 year). For our initial estimate we calculate 2.253564 radians (129.1197 degrees). We also calculate β= 0.01671 radian (0.9574 degree). We now follow the steps laid out above:

    1. We have δ=0.742779 degree and ε=0.73503 degree.

    2. We have φ=0.734969 degree and γ=-0.00006044 degree.

    3. We have η=0.73497 degree and ι=0.000000435 degree.

    4. We have AOQ = 129.85467 degrees, SinAOQ =0.76762404, and CosAOQ =-0.640842476.

    5. ESP = arctan (1.167232867) = 49.41244119 degrees.

    6. And finally we have θ= 180 - ESP = 130.5876 degrees.

Final Comment

    I have composed this essay by adapting Proposition 31 (Problem 23) and the following scholium from Section 6, Book I of Isaac Newton's "Philosophiae Naturalis Principia Mathematica", converting Newton's geometric reasoning into the equivalent algebra for convenience of calculation.

    To someone who learned physics, as I did, by way of analytic geometry and the calculus (which Newton invented), it seems strange that Newton founded modern physics on something that looks like a straightforward extension of Greek plane geometry. The Principia looks very much like a sequel to Euclid's "Elements" and Apollonius' "Conics", to which works Newton frequently refers his readers. In that sense, the Principia stands as a continuation of Greek mathematics. Newton even presented the basic ideas of the calculus geometrically, leaving to later generations of natural philosophers the task of translating the work into the algebraic form that we use today.

    I now offer a thought and a challenge. I have said elsewhere that in time others will create alternative versions of the Map of Physics, that my version stands as the first of many and not as the only one. It seems perfectly plausible that someone might combine the material in Euclid's "Elements", Appolonius' "Conics", and Newton's "Principia" into a single treatise and then extend Newton's dynamic geometry onward to create a purely geometric version of the Map of Physics. There you have the challenge, then; to use Newton's extension of the geometry taught to high-school sophomores to create an alternative Map of Physics. Using a geometric approach, instead of our modern algebraic approach, gave Newton a perspective on physics that differs significantly from ours, so we can not even guess at what new insights such a project would give us into the nature of Reality. We shall simply have to wait and see.

    As for me, I remain committed to extending my own version of the Map of Physics, based on Cartesian coordinate geometry and the algebraic version of the calculus.


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