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$$

r_1 = R_{\earth} (0.5\mathbf{i} + 0.6\mathbf{j} + 0.7\mathbf{k}) ,\quad

r_2 = R_{\earth} (0\mathbf{i} + 1.1\mathbf{j} + 0\mathbf{k})

$$

over an elapsed time of 13 minutes. Determine the semi-major axis $a$ of the orbit for this object's trajectory; is it an elliptical, parabolic or hyperbolic?

\noindent From $\mathbf{r}_1$ and $\mathbf{r}_2$, we can determine $c$ and then $s$.

Additionally, we already have enough to find $\Delta\nu$.

\begin{alignat}{7}

r_1 & = & \quad 6681.96 & \quad &r_2 & = & 7008.1\\

c & = & \sqrt{\lVert \mathbf{r}_2 - \mathbf{r}_1\rVert} & & s & = & \frac{r_1 + r_2 +c}{2}\\

& = & 6339.06 & & & = & 10014.6\\

\Delta t & = & 13\cdot 60 & & \Delta\nu & = & \arccos\left(\frac{\mathbf{r}_1\cdot\mathbf{r}_2}{r_1r_2}\right)\\

& = & 780 & & & = & 55.1048^{\circ}

\end{alignat}

With these relationships, we can solve for $a_m$, $\beta_m$, and then $t_m$. Since $\Delta\nu < \pi$, we can use the exact value of $\beta_m$ that is returned. Moreover, $\alpha_m = \pi$ for the minimum semi-major axis.

\begin{alignat} {7}

a_m & = & \frac{s}{2} & \quad & \beta_m & = & 2\arcsin\left(\sqrt{\frac{s - c}{s}}\right)\\

& = & 5007.28 & & & = & 74.5754^{\circ}

\end{alignat}

Finally, for $t_m$, we need to solve numerically.

Code:

```
In[265]:= Solve[
Sqrt[\[Mu]]*tm == Sqrt[s^3/8]*(\[Pi] - \[Beta]m + Sin[\[Beta]m]), tm]
Out[265]= {{tm -> 1573.66}}
```

I need to find $\alpha$ and $\beta$ in order to solve

$$

\sqrt{\frac{\mu}{a^3}}\Delta t = \alpha - \beta -(\sin(\alpha) - \sin(\beta)).

$$

Formulas I have are

$$

\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{r_1 + r_2 + c}{4a}} = \sqrt{\frac{s}{2a}}

$$

and

$$

\sin\left(\frac{\beta}{2}\right) = \sqrt{\frac{r_1 + r_2 - c}{2a}} = \sqrt{\frac{s-c}{2a}}

$$

but those equations are dependent on $a$ which I need to find.

Also,

$$

\alpha - \beta = 2E_m

$$

$E_m$ is the mean anomlay.