# [SOLVED]Solving for time in a hyperbolic trajectory

#### dwsmith

##### Well-known member
A spacecraft is on a hyperbolic orbit relative to the Earth with $a = -35000$ km and an eccentricity of $e = 1.2$.
At some initial time $t_0$, the spacecraft is at a true anomaly of $\nu_0 = 20^{\circ}$.
At some later time $t$, the true anomaly is $\nu = 103^{\circ}$.
What is the elapsed time interval $\Delta t$ between these two positions?

This solution is wrong. The answer should be around an hour. How else can I do this?

Since we are dealing with hyperbolic trajectories, our equations for the eccentric hyperbolic anomaly are
\begin{alignat*}{3}
M_{\text{h}} & = & \frac{\mu_{earth}}{h^3}(e^2 - 1)^{3/2}t\\
M_{\text{h}} & = & e\sinh(F) - F
\end{alignat*}
Therefore, we need the eccentric anomalies $F$ and $F_0$.
\begin{alignat*}{3}
F & = & 2\tanh^{-1}\left[\sqrt{\frac{e - 1}{e + 1}}\cdot\tan\left(\frac{\nu}{2}\right)\right]\\
& = & 41.51866^{\circ}\\
F_0 & = & 2\tanh^{-1}\left[\sqrt{\frac{e - 1}{e + 1}}\cdot\tan\left(\frac{\nu_0}{2}\right)\right]\\
& = & 6.08648^{\circ}
\end{alignat*}
Since we don't know $h$ explicitly, we can solve for $h$,
$$h = \sqrt{a\cdot\mu_{earth}\cdot(1 - e^2)}.$$
Next, we need to find the eccentric anomaly for $F_0$ and $F$.
\begin{alignat*}{3}
M_{F_0} & = & e\sinh(F_0) - F_0\\
& = & -5.95877^{\circ}\\
M_{F} & = & e\sinh(F_0) - F_0\\
& = & -40.5709^{\circ}\\
\end{alignat*}
Finally, we can solve for the time and take the difference to obtain $\Delta t$.
\begin{alignat*}{3}
t & = & M_F\cdot\frac{h^3}{\mu_{earth}\cdot(e^2 - 1)^{3/2}}\\
& = & -420773\text{ s}\\
t_0 & = & M_{F_0}\cdot\frac{h^3}{\mu_{earth}\cdot(e^2 - 1)^{3/2}}\\
& = & -61800.2\text{ s}
\end{alignat*}
Since time isn't negative, we simply take the absolute value of the $t$ and $t_0$.
\begin{alignat*}{3}
\Delta t & = & t - t_0\\
& = & 358973\text{ s}\\
& = & 4.15478\text{ days}
\end{alignat*}

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