For an exam I was asked to derive the follow equations for epoch departure $t_1$ and epoch arrival $t_2$. (planet 1 has a smaller orbit, eccentricity/inclination can be assumed 0)

$$t_1 = t_0 + \frac{\theta_2(t_0)-\theta_1(t_0) + n_2T_H - \pi}{n_1 - n_2}$$ $$t_2 = t_1 + T_H$$

Where $T_H$ is the transfer time and $\theta$ denotes the true anomaly. and finally $n$ the mean motion.

I started proving the first equation, firstly by noting that:

$$\Delta \theta_2 = T_H \cdot n_2$$ $$\theta_2(t_1) - \theta_1(t_1) = \pi - \Delta \theta_2$$ Last equations follows from geometry, a hohman transfer is exactly half an orbit ($\pi$), and then a simplistic geometrical sketch should show this equation.

Now using mean motion one can link in the poch time:

$$\theta(t_n) = t_0 + n \cdot t_n$$ So one gets the following equation:

$$\theta_2(t_0) - \theta_1(t_0) + t_1(n_1 - n_2) = \pi - T_2 n_2$$ Reshuffling for $t_1$ then gave me:

$$t_1 = \frac{\pi - T_2 n_2 - \theta_1(t_0) - \theta_2(t_0)}{n_1 - n_2}$$

Close but not quite the expected result. - First of all I have no notion of a $t_0$ term in the equation other than for angles. Secondly I seem to have the complete opposite sign of the equation?

Which assumption/mathematical error did I make?


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For an Hohmann transfer you want the phase angle to be correct, namely you want to aim for where planet 2 is going to be. You cover an angle of $\pi$ radians, starting as planet 1, which takes $T_H$ seconds, so planet 2 will have moved $T_H n_2$ radians, so the phase angle, $\Delta\theta$, between the two planets at the start of the transfer should be,

$$ \Delta\theta = \pi - T_H n_2, $$

So far so good.

The true anomaly for a circular of planet $i$ can be written as,

$$ \theta_i(t) = \theta_i(t_0) + n_i (t - t_0), $$

note taking the difference of the times, such that at $t=t_0$ then $\theta_i(t)$ does equal $\theta_i(t_0)$.

Thus the phase difference between planet 1 and 2 as a function of time can be written as,

$$ \theta_2(t) - \theta_1(t) = \theta_2(t_0) - \theta_1(t_0) + (n_2 - n_1) (t - t_0). $$

By equating this equation to $\Delta\theta$, then solving for $t$ should give you $t_1$.

  • $\begingroup$ While this adds The "$t_0$" to the equation, it is still the "negative (actually before you move t_0 around)" of the equation I have to find. $\endgroup$ – paul23 Jun 20 '15 at 1:49
  • $\begingroup$ @paul23 but the denominator will also have opposite sign $n_2-n_1$ instead of $n_1-n_2$. $\endgroup$ – fibonatic Jun 20 '15 at 1:56

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