Hyperbolic Trajectories ($$e > 1$$)#

For the hyperbola, combining Eq. (189) and Eq. (192) results in:

(219)#$M_h = \frac{e\sqrt{e^2 - 1}\sin\nu}{1 + e\cos\nu} - \ln\left[\frac{\sqrt{e + 1} + \sqrt{e - 1}\tan\frac{\nu}{2}}{\sqrt{e + 1} - \sqrt{e - 1}\tan\frac{\nu}{2}}\right]$

where

(220)#$M_h = \frac{\mu^2}{h^3} t \left(e^2 - 1\right)^{3/2}$

The hyperbolic mean anomaly, like the elliptical mean anomaly, is a monotonic function of the true anomaly, as shown in Fig. 55.

Notice that $$\nu$$ cannot exceed $$\nu_{\infty} = \cos^{-1}(-1 / e)$$.

Hyperbolic Eccentric Anomaly#

Similar to the ellipse, we will define an auxiliary angle $$F$$ to simplify the equations. $$F$$ is defined with reference to the hyperbola in Fig. 56.

The ratio $$y/b$$ is the definition of the hyperbolic sine of the angle $$F$$:

(221)#$\sinh F = \frac{y}{b}$

Then, using the hyperbolic trigonometric identity:

(222)#$\cosh^2 c - \sinh^2 c = 1$

we also define

(223)#$\cosh F = \frac{x}{a}$

Note

The hyperbolic angle $$F$$ is weird. The reason we don’t draw it on the figure is because hyperbolic angles aren’t angles per se. Instead, they can be interpreted as half the area between the $$x$$-axis and a line drawn from the origin to the point of interest, bounded by the hyperbola. I think. At least, that’s my best interpretation from what I’ve been able to read. YMMV.

Another way of thinking about this is by analogy to a circle. For a circle, we can draw any two lines from the center of the circle to the perimeter. These two lines will have an angle $$\phi$$ between them, and the area between them will be:

(224)#$A_{\text{circular sector}} = \frac{r^2 \phi}{2}$

If the circle is a unit circle ($$r = 1$$), then the area of the sector will be equal to the angle divided by two. Turned around, the angle is equal to twice the area:

(225)#$\phi = \frac{2A}{1^2}$

Similarly, the hyperbolic angle is defined on the unit hyperbola as twice the area between two lines that start at the origin and touch the hyperbola, called a hyperbolic sector.

However, the circular angle between the $$x$$-axis and the ray from the origin is not the same as the hyperbolic angle.

You can read more about hyperbolic angles on Brilliant and on Wikipedia.

We can relate $$F$$ to the true anomaly $$\nu$$ by plugging in $$y = r\sin\nu$$, and the orbit equation for $$r$$. We also note that $$b = a\sqrt{e^2 - 1}$$. Then:

(226)#$F = \sinh^{-1}\left(\frac{\sin\nu\sqrt{e^2 - 1}}{1 + e\cos\nu}\right) = \ln\left(\frac{\sin\nu\sqrt{e^2 - 1} + \cos\nu + e}{1 + e\cos\nu}\right)$

After some more trigonometry and algebra, we find:

(227)#$F= \ln\left[\frac{\sqrt{e + 1} + \sqrt{e - 1}\tan\frac{\nu}{2}}{\sqrt{e + 1} - \sqrt{e - 1}\tan\frac{\nu}{2}}\right]$

And, with a little more algebra and trigonometry, we find an equation for $$F$$ in terms of $$\nu$$ more directly, analogous to the ellipse:

(228)#$\tanh\frac{F}{2} = \sqrt{\frac{e - 1}{e + 1}}\tan\frac{\nu}{2}$

Kepler’s Equation for the Hyperbola#

Substituting this back into Eq. (219), we find Kepler’s equation for the hyperbola:

(229)#$M_h = e\sinh F - F$

As with the ellipse, Kepler’s equation can be solved easily if $$F$$ is known. However, if time is the known quantity, then Kepler’s equation is transcendental and must be solved numerically. The form of the equation for the Newton solver is $$f(F) = 0$$, or:

(230)#$f(F) = 0 = e\sinh F - F - M_h$

To aid in the numerical solution, the derivative of Kepler’s equation for the hyperbola is:

(231)#$f'(F) = e \cosh F - 1$

In addition, we can estimate an initial value for the guess of $$F$$ from Fig. 59, with a known $$M_h$$ value.

Note that the $$y$$-axis plots the log base 10 of $$M_h$$. To use the graph, take the log base 10 of whatever value you calculate for $$M_h$$ and find that on the graph. For example, assume that $$M_h =$$ 40.69 rad and $$e =$$ 2.5. Then,

$\log_{10}(40.69) = 1.6$

From Fig. 59, we estimate that $$F =$$ 4, and we can use this as the initial guess for a Newton solver.