Elliptic orbits

Let us determine the radial and angular coordinates, $r$ and $\theta $, respectively, of a planet in an elliptical orbit about the Sun as a function of time. Suppose that the planet passes through its perihelion point, $r=r_p$ and $\theta=0$, at $t=\tau$. The constant $\tau$ is termed the time of perihelion passage. It follows from the previous analysis that

$\displaystyle r = \frac{r_p\,(1+e)}{1+e\,\cos\theta},$ (4.51)


$\displaystyle {\cal E} = \frac{\skew{3}\dot{r}^{\,2}}{2} + \frac{h^{\,2}}{2\,r^{\,2}} - \frac{G\,M}{r},$ (4.52)

where $e$, $h =\! \sqrt{G\,M\,r_p\,(1+e)}$, and ${\cal E} = G\,M\,(e-1)/(2\,r_p)$ are the orbital eccentricity, angular momentum per unit mass, and energy per unit mass, respectively. The preceding equation can be rearranged to give

$\displaystyle \skew{3}\dot{r}^{\,2} = (e-1)\,\frac{G\,M}{r_p} - (e+1)\,\frac{r_p\,G\,M}{r^{\,2}}
+ \frac{2\,G\,M}{r}.$ (4.53)

Taking the square root, and integrating, we obtain

$\displaystyle \int_{r_p}^r\frac{r\,dr}{[2\,r + (e-1)\,r^{\,2}/r_p - (e+1)\,r_p]^{1/2}} =
\sqrt{G\,M}\,\,(t-\tau).$ (4.54)

Consider an elliptical orbit characterized by $0<e < 1$. Let us write

$\displaystyle r = \frac{r_p}{1-e}\,(1-e\,\cos E),$ (4.55)

where $E$ is termed the eccentric anomaly. In fact, $E$ is an angle that varies between $-\pi$ and $+\pi$. Moreover, the perihelion point corresponds to $E=0$, and the aphelion point to $E=\pi$. Now,

$\displaystyle dr = \frac{r_p}{1-e}\,e\,\sin E\,dE,$ (4.56)


$\displaystyle 2\,r + (e-1)\,\frac{r^{\,2}}{r_p}- (e+1)\,r_p = \frac{r_p}{1-e}\,e^{\,2}\,(1-\cos^2 E)= \frac{r_p}{1-e}\,e^{\,2}\,\sin^2E.$ (4.57)

Thus, Equation (4.54) reduces to

$\displaystyle \int_0^E (1-e\,\cos E)\,dE = \left(\frac{G\,M}{a^{\,3}}\right)^{1/2} (t-\tau),$ (4.58)

where $a = r_p/(1-e)$. This equation can immediately be integrated to give

$\displaystyle E - e\,\sin E = {\cal M}.$ (4.59)


$\displaystyle {\cal M} = n\,(t-\tau)$ (4.60)

is termed the mean anomaly, $n=2\pi/T$ is the mean orbital angular velocity, and $T= 2\pi\,(a^{\,3}/GM)^{1/2}$ the orbital period. The mean anomaly is an angle that increases uniformly in time at the rate of $2\pi$ radians every orbital period. Moreover, the perihelion point corresponds to ${\cal M}=0$, and the aphelion point to ${\cal M} = \pi$. Incidentally, the angle $\theta $, which determines the true angular location of the planet relative to its perihelion point, is called the true anomaly. Equation (4.59), which is known as Kepler's equation, is a transcendental equation that does not possess a convenient analytic solution. Fortunately, it is fairly straightforward to solve numerically. For instance, when we use an iterative approach, if $E_n$ is the $n$th guess then

$\displaystyle E_{n+1} = {\cal M}+ e\,\sin E_n;$ (4.61)

this iteration scheme converges very rapidly when $0\leq e\ll 1$ (as is the case for planetary orbits).

Figure 4.5: Eccentric anomaly.

Equations (4.51) and (4.55) can be combined to give

$\displaystyle r\,\cos\theta = a\,(\cos E-e).$ (4.62)

This expression allows us to give a simple geometric interpretation of the eccentric anomaly, $E$. Consider Figure 4.5. Let $PGA$ represent the elliptical orbit of a planet, $G$, about the Sun, $S$. Let $ACP$ be the major axis of the orbit, where $P$ is the perihelion point, $A$ the aphelion point, and $C$ the geometric center. It follows that $CA=CP=a$ and $CS=e\,a$ (see Section A.9), where $a$ is the orbital major radius and $e$ the eccentricity. Moreover, the distance $SG$ and the angle $GSQ$ correspond to the radial distance, $r$, and the true anomaly, $\theta $, respectively. Let $PRA$ be a circle of radius $a$ centered on $C$. It follows that $AP$ is a diameter of this circle. Let $RGQ$ be a line, perpendicular to $AP$, that passes through $G$ and joins the circle to the diameter. It follows that $CR=a$. Let us denote the angle $RCS$ as $E$. Simple trigonometry reveals that $SQ=r\,\cos\theta$ and $CQ=a\,\cos E$. But, $CQ=CS+SQ$, or $a\,\cos E = e\,a+r\,\cos\theta$, which can be rearranged to give $r\,\cos\theta = a\,(\cos E-e)$, which is identical to Equation (4.62). We, thus, conclude that the eccentric anomaly, $E$, can be identified with the angle $RCS$ in Figure 4.5.

Equations (4.51) and (4.55) can be combined to give

$\displaystyle \cos\theta = \frac{\cos E - e}{1-e\,\cos E}.$ (4.63)


$\displaystyle 1+\cos\theta = 2\,\cos^2(\theta/2) = \frac{2\,(1-e)\,\cos^2( E/2)}{1-e\,\cos E},$ (4.64)


$\displaystyle 1-\cos\theta = 2\,\sin^2(\theta/2) = \frac{2\,(1+e)\,\sin^2 (E/2)}{1-e\,\cos E}.$ (4.65)

The previous two equations imply that

$\displaystyle \tan (\theta/2) = \left(\frac{1+e}{1-e}\right)^{1/2}\tan (E/2).$ (4.66)

The eccentric anomaly, $E$, and the true anomaly, $\theta $, always lie in the same quadrant (i.e., if $0\leq E\leq \pi/2$ then $0\leq\theta\leq\pi/2$, etc.) We conclude that, in the case of a planet in an elliptical orbit around the Sun, the radial distance, $r$, and the true anomaly, $\theta $, are specified as functions of time via the solution of the following set of equations:

$\displaystyle {\cal M}$ $\displaystyle = n\,(t-\tau),$ (4.67)
$\displaystyle E - e\,\sin E$ $\displaystyle = {\cal M},$ (4.68)
$\displaystyle r$ $\displaystyle =a\,(1-e\,\cos E),,$ (4.69)
$\displaystyle \tan(\theta/2)$ $\displaystyle =\left(\frac{1+e}{1-e}\right)^{1/2} \tan (E/2).$ (4.70)

Here, $n=2\pi/T$, $T=2\pi\,(a^{\,3}/G\,M)^{1/2}$, and $a = r_p/(1-e)$. Incidentally, it is clear that if $t\rightarrow t+T$ then ${\cal M}\rightarrow {\cal M}+2\pi$, $E\rightarrow E + 2\pi$, and $\theta\rightarrow
\theta+2\pi$. In other words, the motion is periodic with period $T$.