Garmaine Staff asked 2 years ago

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?
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