Why is the bend angle of a hyperbolic trajectory giving different results?

Question

The bend angle ($\delta$) is the reflex angle between the two asymptotes of a hyperbolic trajectory.

From this Stack Exchange answer, the formula to calculate the bend angle is: $$ \delta = 2 \sin^{-1}\bigg( \frac{1}{e} \bigg)$$

Where $e$ is eccentricity and is calculated by: $$e = \frac{rv_\infty^2}{\mu}+1$$

Where $r$ is the distance of the spacecraft from the body during the closest approach (periapsis), and $v_\infty$ is the velocity as if the gravitational body wasn't there.

I put an arbitrary value for $v_\infty$ which is 21,000 m/s, and an arbitrary altitude of the orbit which is 60,000 km from the centre of the planet (which has a radius of 6,371 km) to give the total value of $r = 6.6371 \times 10^7$ meters. The GM of the planet is also made up, so $\mu = 1.47 \times 10^{14}$. This gives me $e = 200.113$. And when I substitute $e$ into equation 1, I get $\delta = 0.573$.

I plotted this hyperbolic trajectory on Desmos and here is the graph for reference (the dotted lines are the asymptotes).

Note, a = 500, and b = 100,055.25 because this yields an eccentricity of 200.113, using the equation for eccentricity of a hyperbola $e = \frac{\sqrt{a^2+b^2}}{a}$. What I basically did was choose an arbitrary $a$ value and solved for $b$ to obtain an eccentricity of 200.113.

Question: The image above shows that this hyperbolic trajectory clearly has a very big bend angle, about $180^o$ (less actually but to the eye it looks 180 deg). But formula 1 gave a result of $\delta = 0.573$. This can't be degrees. But it's also not radians because $0.573 rad = 32^o$. It can't be arc mins or arc sec. I then though that it must be revolutions. But a revolution greater than 0.5 yields a degree value of greater than 180 degrees, which the bend angle is less than. What's going on here? Am I misunderstanding something?

Answer 1

There are two major issues that I can see.

Whatever you're using to calculate the $\arcsin$ is indeed giving you a value in degrees.

The parameters you've provided are basically that of a satellite racing by an object at a moderate distance, moving far above escape velocity all along its trajectory. It neither gets close enough, nor hangs around long enough to get a significant trajectory bend, so yes, the deflection angle is roughly 0.57°.

Also, your chosen values to produce an eccentricity of 200 are off. $\frac{\sqrt{500^2 + 4444^2}}{500} = 8.944 $

To expand further, I generally use the Polar conic sections for orbits; they're easier to deal with when working with Kepler's equations.

You've got the periapsis distance and the eccentricity: (edit: Initially had the periapse 10 times too large) $$r_p = 6.67 \times 10^7m$$ $$e = 200$$

From there, you can calculate the semimajor axis $a$ as :

$$ a = \frac{r_p}{1-e} = -3.34 \times 10^5 m$$

And the polar equation of the resulting hyperbola is:

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

And your flyby of the world you've chosen looks like this:

GeoGebra Graph - Hyperbolic Flyby

The purple circle at the origin is the central body on the flyby. I've drawn both lobes of the Hyperbola here, the green one is the actual flyby trajectory, and the red dotted line is the other, non-trajectory lobe.

A single lobe of an eccentricity-200 hyperbola is difficult to distinguish from a straight line, by eye.

Answer 2

Let's go through this one thing at a time.

• First off, your value for $\delta$ is clearly in degrees. With an eccentricity of $200$, the reciprocal $1/e$ is so small it's nearly equal to it's own inverse sine -- in radians. $\delta$ is basically $2/200$ radians $=0.573°$.

• Next, realize that geometrically, the eccentricity of a hyperbola equals the center to focus distance divided by the center to vertex distance. With your form for the hyperbola, having vertices on the $y$ axis, the former distance is $\sqrt{a^2+b^2}$ and the latter distance is $b$ (not $a$).

• For the high eccentricity you render you have to have $a$ bigger than $b$, not the other way around, and moreover the ratio between them has to be much higher than the $4444/500$ supposed here. When the eccentricity of the hyperbola described above is greater than $10$, the $b^2$ term under the radical is very small so the eccentricity is only a little different from just $a/b$. For an eccentricity of $200$, then, you basically need $a$ to be $200$ times $b$, not the $8.888$ ratio given here.

Put numbers in according to the above and your hyperbolic curve should conform with the formulas.