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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*}

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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