MHB Solving for time in a hyperbolic trajectory

[Dustinsfl]

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*}

 

[member38644]

This solution is correct and provides the elapsed time interval between the two positions in the spacecraft's hyperbolic trajectory. If you are looking for a simpler or alternative method, you can also use Kepler's third law which states that the square of the orbital period is proportional to the cube of the semimajor axis. Using this, you can solve for the orbital period and then find the elapsed time interval between the two positions. However, this method may not work for all hyperbolic trajectories as they do not follow the same pattern as elliptical orbits.