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

Hyperbolic trajectory labelled

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).

enter image description here

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?

  • $\begingroup$ Yeah, something's very wrong. with your image. The eccentricity seems alright. Moving at nearly twice Earth's surface escape velocity for your $v_\infty$ with a periapsis almost 10 earth radii, and a central mass billions of times smaller than Earth should produce a trajectory that's almost a straight line, so a 200+ eccentricity is probably reasonable. $\endgroup$
    – notovny
    Commented Mar 30, 2020 at 23:59
  • $\begingroup$ (Ah, mistook gravitational parameter for planetary mass, and let the edit window lapse. Still, the eccentricity seems okay.) $\endgroup$
    – notovny
    Commented Mar 31, 2020 at 0:08
  • $\begingroup$ I haven't run the numbers, but you say "r is the distance of the spacecraft from the body" and the reference says r is the closest approach radius which sounds like it may be a disconnect. If your plot is labeled correctly r should be 4500 not 6.6x10^8 $\endgroup$ Commented Mar 31, 2020 at 0:12
  • $\begingroup$ I have made a mistake on calculating the b value. It is NOT 4,444, but actually 100,055. I also corrected the formula. $e = \frac{\sqrt{a^2+b^2}}{a}$ $\endgroup$
    – Star Man
    Commented Mar 31, 2020 at 1:16
  • $\begingroup$ @OrganicMarble To the exponent 7. Not 8. Sorry. But Yes. $6.6 \times 10^7$ is the closest approach. (About 60,000 km from the surface of the planet). The purpose of the desmos diagram is to show that the bend angle of the asymptotes don't equal to the value of 0.537. Which is my question. As long as the eccentricity is 200.1, the angle will stay the same. So whether it's $6.6 \times 10^7$ or 4500 is irrelevant as long as the eccentricity is 200.1. $\endgroup$
    – Star Man
    Commented Mar 31, 2020 at 1:24

2 Answers 2


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

Eccentricity 200 hyperbolic flyby - GeoGebra

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.

  • 2
    $\begingroup$ @novotny in the eccentricity formula the denominator is not squared, unless you put the whole fraction under the square root. $\endgroup$ Commented Mar 31, 2020 at 0:53
  • 2
    $\begingroup$ @OscarLanzi Ah. I was taking the formula from the original post, and didn't double-check. Apologies $\endgroup$
    – notovny
    Commented Mar 31, 2020 at 0:55
  • $\begingroup$ @notovny Sorry. I fixed the value of 4,444. It's actually 100,055. This gives an eccentricity of 200.113. And I fixed the formula. The denominator is not squared. $\endgroup$
    – Star Man
    Commented Mar 31, 2020 at 1:30
  • $\begingroup$ This answer has become much improved, but one small clarification request. I would draw the lobes in different colors, perhaps the "real" one blue and the other lobe light gray, to make them easier to distinguish given the shape of the hyperbola and make the actual orbit easier to see. Thank you. $\endgroup$ Commented Mar 31, 2020 at 16:46

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.

  • $\begingroup$ Oops. I don't know how I got 4,444. I have edited the question. The new b value is 100,055 which gives an eccentricity of 200.1. (I also corrected the formula). $\endgroup$
    – Star Man
    Commented Mar 31, 2020 at 1:14
  • 1
    $\begingroup$ Please look again at my second bullet point. I do not agree with $\sqrt{a^2+b^2}/a$ when you draw the hyperbola with vertices at $(0,\pm b)$ as opposed to $(\pm a, 0)$. $\endgroup$ Commented Mar 31, 2020 at 1:31

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