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[SOLVED] Lambert's problem

dwsmith

Well-known member
Feb 1, 2012
1,673
Suppose that an unknown space object is detected and tracked over a short interval of time. Radar measurements give the position vectors as
$$
r_1 = R_{\earth} (0.5\mathbf{i} + 0.6\mathbf{j} + 0.7\mathbf{k}) ,\quad
r_2 = R_{\earth} (0\mathbf{i} + 1.1\mathbf{j} + 0\mathbf{k})
$$
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}}
Now, $\Delta t < t_m$ so $\alpha = \alpha_0$.

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.