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".

Thanks in advance.


1 Answer 1


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.

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

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.


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