I am writing an internal assessment for first year university on patched conic approximation. I am having some trouble calculating the departure angle from earth. This is because I have to calculate the eccentricity and parameter of an ellipse which is tangential to both the earth’s and mars’ orbit. The sun is assumed to be at the center of the ellipse. I am also struggling to find any piece of work on this topic which I could reference my results with. Any help in general with the analysis and formula application of the transfer between the earth and mars would be also appreciated.

  • $\begingroup$ Is the "parameter" you're referring to the focal parameter, the semi-major axis, the semi-latus rectum, or some other conic parameter? Also, are you simplifying the orbits of Earth and Mars to be circular and coplanar? $\endgroup$
    – notovny
    Sep 13, 2021 at 21:11
  • $\begingroup$ The orbits are circular and coplanar yes. I wouldn’t know what type of parameter I’m looking for. What do you think? $\endgroup$ Sep 13, 2021 at 21:19
  • $\begingroup$ cos ν1 =( p - r1 ) / ( e *r1 ) cos ν2 =( p - r2 ) / ( e *r2 ) These are the two equations... I know the two radii and I’m trying to find v1 and v2, but I don’t know e and p which are eccentricity and parameter. $\endgroup$ Sep 13, 2021 at 21:23

1 Answer 1


Circular/Coplanar Simplified Hohmann Transfer from Earth to Mars:

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For a Hohmann Transfer from Earth to Mars, using the simplified, circular and coplanar versions of Earth's and Mars' orbits, we'd have the following givens:

  • Semi-major axis of Earth's orbit: $a_E$
  • Semi-major axis of Mars' Orbit: $a_M$

The Hohmann transfer orbit is an ellipse tangent to both circular orbits, with one of its foci at the Sun. The semi-major axis of the Hohmann Transfer ($a_H$ ) is half the sum of Mars' semimajor axis and Earth's semimajor axis:

$$a_H = \frac{a_M + a_E}{2}$$

Orbital Eccentricity ($e_H$) of the Hohmann Transfer orbit is:

$$e_H =1- \frac{a_E}{ a_H}$$

Focal Parameter ($p_H$) of the Hohmann transfer ellipse is then:

$$p_H=\frac{a_H}{e_H} - a_H e_h$$

  • $\begingroup$ Thanks bro. Very helpful $\endgroup$ Sep 16, 2021 at 20:06

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