# How to find the hyperbolic angle given the mean anomaly?

I'm modelling a hyperbolic gravity assist trajectory around Jupiter and trying to calculate the coordinates for each hour interval before/after passing periapsis. I've calculated $$M_h = 0.0176$$ is the mean anomaly 1 hour from periapsis, but how can I determine the corresponding hyperbolic angle, i.e solve this equation for H, given e = 1.3893:

$$M_h = 0.0176 = e~\rm sinh\it(H) - H$$

Using iterative calculations I know the answer is approximately $$H = 0.04$$, but I'm hoping to solve the equation above "precisely".

Like the corresponding eccentric anomaly for elliptical orbits, there is no closed-form formula for going from mean anomaly to hyperbolic anomaly.

You're going to have to use some sort of numerical method to go in that direction. Newton-Raphson tends to converge quickly enough.

$$x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}}$$

With Mean anomaly at the chosen time as $$M$$, we'll look for zeroes on the function: $$f(H_n) = e \sinh H_n - H_n -M$$ And we'll need its first derivative: $$f'(H_n) = e \cosh H_n - 1$$

And we'll iterate with:

$$H_{n+1} = H_n - \frac{e \sinh H_n - H_n -M}{e \cosh H_n -1}$$

In almost every case I've used it, it's been useful to set the initial guess of the hyperbolic anomaly $$H_0$$ equal to the Mean Anomaly $$M$$. Given your chosen parameters of eccentricity $$e = 1.3893$$ and mean anomaly $$M = 0.0176$$, these are the values pulled up from a quick Google Sheets Spreadsheet:

Iteration Hyperbolic Anomaly
$$H_0$$ $$\underline{0.0}176$$
$$H_1$$ $$\underline{0.0451}9085695$$
$$H_2$$ $$\underline{0.045154584}33$$
$$H_3$$ $$\underline{0.04515458422}$$
... ...

Newton-Raphson's convergence is typically quadratic, resulting in roughly doubling the number of correct digits each iteration. We're at the three significant figures of your mean anomaly value by $$H_1$$, and by $$H_3$$ the iterated value doesn't change under the floating-point precision Google Sheets can handle.

One more thing: The convergence of using the Newton-Raphson method above with Kepler's equations gets slower as orbital eccentricity approaches $$e=1$$. If Orbital eccentricity was $$e=1.01$$, it would take until $$H_7$$ to get three significant figures stable from the specified Mean Anomaly. At $$e=1.001, H_{21}$$, and $$e=1.0001, H_{74}$$, and so forth. If your hyperbolas are extremely near-parabolic, you may need to look into an alternate method to calculate position as a function of time.