The Kepler Problem

In a nutshell, the so-called Kepler problem consists of determining the radial and angular coordinates, $r$ and $\theta$, respectively, of an object in a Keplerian orbit about the Sun as a function of time.

Consider an object in a general Keplerian orbit about the Sun which passes through its perihelion point, $r=r_p$ and $\theta=0$, at $t=0$. It follows from the analysis in the previous subsections that

$$r = \frac{r_p\,(1+e)}{1+e\,\cos\theta},$$ (347)

and

$${\cal E} = \frac{\dot{r}^{\,2}}{2} + \frac{h^2}{2\,r^2} - \frac{G\,M}{r},$$ (348)

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 above equation can be rearranged to give

$$\dot{r}^{\,2} = (e-1)\,\frac{G\,M}{r_p} - (e+1)\,\frac{r_p\,G\,M}{r^2} + \frac{2\,G\,M}{r}.$$ (349)

Taking the square-root, and integrating, we obtain

$$\int_{r_p}^r\frac{r\,dr}{[2\,r + (e+1)\,r^2/r_p - (e+1)\,r_p]^{1/2}} = \sqrt{G\,M}\,\,t.$$ (350)

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

$$r = \frac{r_p}{1-e}\,(1-r\,\cos E),$$ (351)

where $E$ is termed the elliptic anomaly. $E$ is an angle which varies between $-\pi$ and $+\pi$. The perihelion point corresponds to $E=0$, and the aphelion point to $E=\pi$. Now,

$$dr = \frac{r_p}{1-e}\,e\,\sin E,$$ (352)

whereas

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

$$= \frac{r_p}{1-e}\,e^2\,\sin^2E.$$ (353)

Thus, Eq. (350) reduces to

$$\int_0^E (1-e\,\cos E)\,dE = \left(\frac{G\,M}{a^3}\right)^{1/2} t,$$ (354)

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

$$E - e\,\sin E = 2\pi\left(\frac{t}{T}\right),$$ (355)

where $T= 2\pi\,(a^3/GM)^{1/2}$ is the orbital period. Equation (355), which is known as Kepler's equation, is a transcendental equation with no known analytic solution. Fortunately, it is fairly straightforward to solve numerically. For instance, using an iterative approach, if $E_n$ is the $n$th guess, then

$$E_{n+1} = 2\pi\left(\frac{t}{T}\right) + e\,\sin E_n.$$ (356)

The above iteration scheme converges very rapidly (except in the limit as $e\rightarrow 1$).

Equations (347) and (351) can be combined to give

$$\cos\theta = \frac{\cos E - e}{1-e\,\cos E}.$$ (357)

Thus,

$$1+\cos\theta = 2\,\cos^2(\theta/2) = \frac{2\,(1-e)\,\cos^2( E/2)}{1-e\,\cos E},$$ (358)

and

$$1-\cos\theta = 2\,\sin^2(\theta/2) = \frac{2\,(1+e)\,\sin^2 (E/2)}{1-e\,\cos E}.$$ (359)

The previous two equations imply that

$$\tan (\theta/2) = \left(\frac{1+e}{1-e}\right)^{1/2}\tan (E/2).$$ (360)

We conclude that, in the case of an elliptical orbit, the solution of the Kepler problem reduces to the following three equations:

$$E - e\,\sin E = 2\pi\left(\frac{t}{T}\right),$$ (361)

$$r = a\,(1-e\,\cos E),$$ (362)

$$\tan(\theta/2) = \left(\frac{1+e}{1-e}\right)^{1/2} \tan (E/2).$$ (363)

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

For the case of a parabolic orbit, characterized by $e=1$, similar analysis to the above yields:

$$P + \frac{P^3}{3} = \left(\frac{G\,M}{2\,r_p^{\,3}}\right)^{1/2} \vert t\vert,$$ (364)

$$r = r_p\,(1+P^2),$$ (365)

$$\tan(\theta/2) = P.$$ (366)

Here, $P$ is termed the parabolic anomaly, and varies between $-\infty$ and $+\infty$, with the perihelion point corresponding to $P=0$. Note that Eq. (364) is a cubic equation possessing a single real root which can, in principle, be solved analytically. However, a numerical solution is generally more convenient.

Finally, for the case of a hyperbolic orbit, characterized by $e>1$, we obtain:

$$e\,\sinh H - H = \left(\frac{G\,M}{a^3}\right)^{1/2} t,$$ (367)

$$r = a\,(e\,\cosh H - 1),$$ (368)

$$\tan(\theta/2) = \left(\frac{e+1}{e-1}\right)^{1/2} \tanh (H/2).$$ (369)

Here, $H$ is termed the hyperbolic anomaly, and varies between $-\infty$ and $+\infty$, with the perihelion point corresponding to $H=0$. Moreover, $a=r_p/(e-1)$. As in the elliptical case, Eq. (367) is a transcendental equation which can only be solved numerically.