I wrote a simulation program in Python to calculate Earth <-> Mars trajectories. Now I would like to test it agains well known trajectories. I used NASA Ames Research Center Trajectory Browser to get the following trajectory:

  • SPK ID: 499
  • Name: Mars
  • Departure: Aug-03-2020
  • Arrival: Feb-27-2021
  • Duration: 208 days
  • Injection C3 (km2/s2): 14.7
  • Abs DLA: 4°
  • Injection ΔV (km/s): 3.87

In order to start the simulation I place my space ship into 200km LEO with the V=7.784 and add 3.87 km/s ("instant" prograde burn for simplicity) to the speed to get to the transmars orbit.

My question: at what exactly position in the LEO should I start the Burn?

Let's take the most distant LEO-point from Sun as angle 0° (also the darkes point). Then going back the LEO trajectory, at which angle should I start the burn?

I ran an optimizer on this question and got out 79°. According to this (being on LEO) I should fire up the engines seconds after transitioning from day to night. Very stange for me.

EDIT (1):

Jupyter Notebook - (Alpha version) of my simulation is now published at GitHub

  • Open in Google Colab to play with the simulation and ODE solver
  • Skip all Test-sections and to long code fragments
  • Look for text explanations and comments in the code
  • Create issues or pull request on GitHub
  • Current version of the file: newest version in master branch

EDIT (2):

Maximizing apogee tweaking only the angle0 parameter I got 60.2369041443°.

Optimizer output:

final_simplex: (array([[-60.2369041443],
       [-60.2369041443]]), array([-2.413841476e+08, -2.413841476e+08]))
           fun: -241384147.60416117
       message: 'Optimization terminated successfully.'
          nfev: 139
           nit: 57
        status: 0
       success: True
             x: array([-60.2369041443])

Source code: Notebook on GitHub

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    $\begingroup$ That seems about right, actually. Mars is sufficiently close enough to Earth that the delta-V requirements in LEO aren't that much above the delta-V requirements for Escape velocity, so your departure trajectory would be a fairly low-eccentricity hyperbola. For a parabolic (exactly escape velocity) trajectory that leaves in the direction of the Earth's motion about the sun, you'd be leaving at 90°. $\endgroup$
    – notovny
    Jul 9, 2020 at 10:52
  • $\begingroup$ @notovny Great news! This means, that my simulation works. I plan to make it public later on. It's a Jupyter notebook. Would you post your comment as an answer, so I can accept it? $\endgroup$ Jul 9, 2020 at 12:56
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    $\begingroup$ Unfortunately, all i can say is that it looks about right. I'm making my way through the orbital equations, and it might be the weekend before I have time to write a full answer. $\endgroup$
    – notovny
    Jul 9, 2020 at 22:23
  • 1
    $\begingroup$ Going to Mars, you need to raise your apoapsis relative to the Sun, and doing so by burning mainly towards the Sun seems strange. Burning at your 0° would be prograde relative to the Sun, which kind of makes more sense to me. But that's just my instinct and orbital mechanics always defy intuition. $\endgroup$ Jul 10, 2020 at 7:21
  • $\begingroup$ @notovny I didn't ment to require such an effort from you. The "looks about right" is exactly the answer I was hoping for. My intuition was, that it should be around 0°, so I thought, my simulation has a bug. The theoretical side would be awesome, but wasn't intended by the question. Should I rephrase my question in order to make your "looks about right" more sutable answer? $\endgroup$ Jul 10, 2020 at 7:34

1 Answer 1


Based on the following calculations, the departure burn angle is about 53.5°

I'm going to be using the following assumptions:

  • The spacecraft is going to depart from Earth moving in the direction of Earth's Travel, to get the maximum benefit from the Earth's path around the sun.

Using the following parameters.

  • Earth's Standard Gravitational Parameter: $\mu_E= 3.97\times10^{14} \mathrm{m^3/s^2}$
  • Earth's Radius $r_E=6.380\times10^6\mathrm{m}$
  • Desired LEO orbital Radius $r_0=6.580\times10^6\mathrm{m}$
  • Injection Delta-V $\Delta v= 3.87 \times 10^3 \mathrm{m/s}$

We can then calculate Circular orbit velocity at the LEO orbit $v_{circ}$: $$v_{circ}=\sqrt{\frac{\mu_E}{r_0}}=7.77\times10^3\mathrm{m/s} $$

Departure velocity at the time of the burn $v_0$: $$v_0=v_{circ}+\Delta v = 1.16 \times 10^4 \mathrm{m/s}$$

From there, we can calculate Specific Orbital Energy of the departure hyperbola $\epsilon$: $$\epsilon=\frac{v_0^2}{r_0} - \frac{\mu_E}{r_0}=7.38 \times10^6\mathrm{J/kg}$$

And the hyperbolic semimajor axis, which we'll use later in the polar equation for the departure trajectory: $$a=-\frac{\mu_E}{2\epsilon}=-2.69 \times 10^7\mathrm{m}$$

Specific Relative Angular Momentum is the cross product of the radial vector and the velocity vector. We just need the magnitude of that vector, $h$ Since at departure, the radial distance vector is perpendicular to the velocity vector, we can just multiply the departure radial distance and the departure velocity. $$h= \|\overrightarrow{r_0}\times\overrightarrow{v_0}\|= r_0v_0\sin\theta=r_0v_0=7.66\times10^{10}\mathrm{m^2/s}$$

And with that, we can calculate Orbital Eccentricity $e$: $$e=\sqrt{1+\frac{2\epsilon h^2}{\mu_E^2}}=1.24$$

This is a bit higher than my intuition assumed when I initially commented.

With Orbital eccentricity, we can use wikipedia's Hyperbolic Trajectory equations get the angle between the asymptotes and the conjugate axis,which I'll call $\theta_0$, listed below in radians, then degrees. $$\theta_0= \frac{2\arcsin(1/e) - \pi}{2} = -1.27 = -36.5^\circ$$

Using the standard polar equation for a hyperbola, that angle $\theta_0$ is the angle we'd have to rotate it to put an asymptote parallel to the X-axis, using the equation below. $$r=\frac{a(1-e^2)}{1-e\cos(\theta+\theta_0)}$$

With the above parameters, the graph below is produced. (I think I probably need to find a better online graphing calculator than Desmos; it's not very good at exporting images. Click on the link for a more comfortable view)

Desmos Graph: Hyperbolic Departure trajectory for a Spacecraft leaving from a 200km parking orbit with a $\Delta v$ of 3.87 km/s

Hyperbolic trajectory

  • Numbers on the graph are in meters.
  • The Sun is in the direction of the positive Y-Axis.
  • The direction of the Earth's motion around the Sun, and the departure asymptote are in the direction of the positive X-axis.
  • The blue circle is Earth. The Red dotted line is the 200km LEO parking orbit.
  • The dotted black line indicates the departure burn point, and is drawn along the major axis of the hyperbola

And to get the angle Boris asked for, between the negative Y-Axis and the hyperbola's major axis, in radians and degrees: $$\phi_{burn}=\frac{\pi}{2}+\theta_0 = 0.93 = 53.5^\circ$$

  • $\begingroup$ Amazing answer! What I would really like to know is where the discrepancy between 53.5° and 60.2° is coming from and what's wrong with my simulation... :) $\endgroup$ Jul 13, 2020 at 4:51
  • $\begingroup$ In my simulation I tweak the angle maximizing apogee of the resulting orbit around the Sun. Should I get the predicted by you 53.5°? $\endgroup$ Jul 13, 2020 at 8:35

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