1. Today's Contents Review Answer of homework
Mean anomaly and eccentric anomaly
parallelogram

2. = 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.

3. 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)

4. Acceleration in Polar Coordinate
(2)
In order to get the acceleration in polar coordinate, we can express (ar, a) by (ax, ay)
(3)
By differentiating Eqs. (2) with respect to t, we have
(4)

5. From Eqs. (3) and (4), we can express the acceleration in polar coordinate

6. (2.5)
(2.7)
(2.11)
In order to find r as a function of , we need to make the substitution u=1/r.
The general solution of this second-order linear differential equation is
(2.15)
where e (an amplitude) and  (a phase) are two constant of integration, substituting back for r we have
(2.14)
which is the general equation of a conic in polar polar coordinates where e is the eccentricity (이심률) and p is the semi-latus rectum given by
(2.16)
Ellipse  0<e<1 p=a(1-e2);

7. In the right figure, a is the semi-major axis, b semi-minor axis, e the eccentricity and  the longitude of pericenter.
Since most objects in our Solar System are closed orbit, we focus on the elliptical motion in this lecture.
Vernal equinox (춘분점)
The angle  is called the true longitude. Eq. (2.19) shows that the maximum and minimum values of the orbit radius are ra=a(1+e) and rp=a(1-e) when =+and =. These points are called apocentre (or aphelion) and pericentre (or perihelion), respectively.
In the case of ellipse, there is a relationship between the quantities a, e, and b
(2.18)
We also have a relation from Eq. (2.15) and p=a(1-e2)
(2.19)

8. 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)

9. 2-4. The Mean & Eccentric Anomalies
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.

10. 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)

11. (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)

12. (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)

13. (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.

14. 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.

15. 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.

16. 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.