Today's Contents Review Elliptic expansion parallelogram

= The motion of m2 with respect to m1 and using Eqs. (2.1) We obtain (2.5) where =G(m1+ m2). Let’s consider the vector product (외적) of r, we have (2.6) angular momentum integral.

We transform to an arbitrary reference frame to a ‘polar coordinate system (극좌표)’ (r, ) referred to an origin centered on the mass m1 and an arbitrary reference line corresponding to  =0. er e r =0  m1 m2 (2.7)

It is convenient to refer the angular coordinate to the pericentre than to the arbitrary reference line. Therefore, we usually use the true anomaly defined as Hence Eq. (2.19) can be written (2.20)

Ideally we want to use an angle, which is linear function of the time. We can determine the orbital radius and velocity as a function of the true anomaly f, when a and e are given. However, we do want to know the orbital radius and velocity as a function of time t instead of true anomaly f! Ideally we want to use an angle, which is linear function of the time. Using the definition of the mean motion, n=2/T, we can define the mean anomaly M by (2.39) where  is the time of pericentre passage. M increases linearly with time at a constant rate, even though it has no simple geometrical interpretation. It is, however, clear that when t= the object is at the pericentre and when t= it is at the apocentre.

M has no simple geometry. Hence we relate it to an angle that does. Consider a circumscribed circle below, which is concentric with an elliptical orbit. We define E (the eccentric anomaly) to be the angle between the major axis of the ellipse and the radius form the centre to the intersection point on the circumscribed circle. E=0  f=0 E=  f=  The equation of a centered ellipse in rectangular coordinates is (2.40) From the figure we have (2.41) (2.41) then (2.42) (2.43)

(2.42) (2.43) These two equations give r and f uniquely when we know E. We can derive a simple relation between E and f (2.44) and From Eq. (2.44) and the standard double angle formulae , we have (2.45) and hence (2.46) Using and Eqs. (2.32) and (2.34), we have (2.47)

(2.47) (2.48) From Eq. (2.42) , then , which is compared with Eq. (2.48) (2.50) (2.49) Eq. (2.50) can be integrated (2.51) Because E=0t= From Eq. (2.39) , finally we can relate E to M (2.52)

(2.52) This is Kepler’s equation and its solution is fundamental to the problem of finding the orbital position at a given time. At given time, we can obtain M from Eq. (2.39) Solve Eq. (2.52) for E Use Eq. (2.43) to obtain f and Eq. (2.20) to determine r Kepler’s equation cannot be solved directly because it is transcendental (cannot be expressed by polynomial) in E and therefore, it is impossible to express E as a simple function of M. To solve the Kepler’s equation, there are two iterative technique: one producing a series solution and the other a numerical solution.

We can derive a series solution by an iterative method of the form (2.54) Using Maclaurin series, we have and the formula We take E0=M as first approximation. And then we obtain (2.55) This equation suggests that we can express E-M as a Fourier since series in M. We will derive bs(e) later.

Kepler equation can be solved numerically. By writing Eq. (2.52) as we can use the Newton-Raphson method to ring the root of the nonlinear equation f(E)=0. The idea of the Newton-Raphson is as follows: (1) You starts with an initial guess which is reasonably close to the true root. (2) The function is approximated by its tangent line, and one computes the x-intercept of this tangent line. (3) This x-intercept will typically be a better approximation to the function's root than the original guess, and the method can be iterated.

Write C-program to solve: f(x)=x2-x-6 void main() { double x0 = 0; // x0 is initial guess double x1, df, f; double eps; do { f = x0*x0 - x0 - 6; df = 2*x0 - 1; // df/dx x1 = x0 -f/df; // x intercept eps = fabs( 1-x0/x1 ); x0 = x1; // substitute x0 by x1 printf("x1 = %lf,eps = %lf\n",x1,eps); } while(eps>1e-6); //repeat the loop unless x converge printf("x1 = %f\n",x1); } fabs() functions compute the absolute value of a floating-point num-ber x.

2-5. Elliptic Expansions There are few integrable problems in solar system dynamics. Thus we have to resort to approximation in order to archive a practical solution to a particular problem. In this section we derive a fundamental expansions that is useful for the solar system dynamics. In the previous section we saw how it was possible to derive a simple series solution for E in terms of M. We left how to deduce bs(e). We then continue to expand the Keplar’s solution. (2.52) (2.55) From Eqs. (2.52) and (2.55), we can write E-M as an odd periodic function (2.72)

Hence let us review the Fourier series. (2.72) Hence let us review the Fourier series. We define that f(x) denotes a function of the real variable x. This function is supposed to be taken to be periodic, of period 2π, that is, f(x+2π) = f(x), for all real numbers x. In this case, f(x) can be written as, and Fourier coefficients are given by Since in Eq. (2.72) Integral by part (2.73) (2.74)

(2.74) (2.75) =M =Js(se) This integral can be written in terms of a standard function ‘Bessel function’ of the first kind. We can write (2.76) (2.77) Bessel function For positive value of s, we can write (2.78) The series for Js(x) for s=1,2,,5 including terms up to O(x5) are given below: (2.79)

Consequently, we can write the solution of Kepler’s equation as (2.72) Consequently, we can write the solution of Kepler’s equation as (2.80)

In addition to the series solution of Kepler’s equation we need a number of other series expansion. The following are series for r/a, cos E, (a/r)3, sin f, cos f, and f-M (Brouwer & Clemence, 1961, Methods of Celestial Mechanics).

Lagrange Inversion Theorem Let  be defined as a function of z in terms of a parameter e by Then any function of  can be expressed as a power series in e which converges for sufficiently small e, and has the form, (2.89) (2.90) Using this theorem, we show f in terms of M, as shown in Eq. (2.88). (2.91) (2.92)

(2.92) (2.93)

Lagrange’s inverse theorem then gives (2.93) This can be written as (2.94) Lagrange’s inverse theorem then gives (2.95) consistent with Eq. (2.88).