10
$\begingroup$

The Hohmann transfer orbit is the most propellant-saving transfer orbit to Mars. When using this kind of transfer, the rover/rocket/etc. starts from Earth when Earth and Mars are in conjunction.

But now, in 2020, many Mars missions will be launched, but Earth and Mars are in opposition. I do not understand why they do that, since the Hohmann transfer orbit saves the most propellant?

If somebody would explain me the reasons, I would be very glad!

$\endgroup$
  • 2
    $\begingroup$ You probably misunderstood something about conjunction and opposition. Mars is now in opposition to the Sun, which means that the Sun, Earth and Mars are almost on a straight line with Earth approaching its closest position to Mars. Now it is the best time to launch rockets to Mars along the Hohmann transfer orbit, just look at the animation in Wiki article you gave a link to, when the rocket is launched in the animation it shows exactly the position of Earth and Mars right now. $\endgroup$ – Yellow Sky Jul 25 '20 at 10:57
  • $\begingroup$ @YellowSky apparently I misunderstood something, but not about conjunction and opposition ;-) You can read my comments in the answer below. $\endgroup$ – Meiki Jul 25 '20 at 16:07
  • $\begingroup$ A Hohmann transfer can only occur between two coplanar circular orbits. As neither the Earth's nor Mars's orbit about the Sun is circular, and as Earth's orbit and Mars's orbits about the Sun are not coplanar. a Hohmann transfer orbit to Mars does not exist. $\endgroup$ – David Hammen Jul 26 '20 at 7:51
  • $\begingroup$ The right approach is to use some metric such as delta V or dollars needed to transfer from Earth to Mars, measuring the metric as a function of a departure time $t_1$ and the arrival time $t_2$. The result of such an analysis is called a pork chop plot. Those key words, "pork chop plot" (or sometimes "porkchop plot"), are used in multiple answers at this site and also at the space exploration and physics sister sites. $\endgroup$ – David Hammen Jul 26 '20 at 7:55
23
$\begingroup$

The thing your're missing is that the Hohmann Transfer orbit takes time, and both Mars and Earth are moving around the sun. For the Hohmann Transfer orbit to work, the position of Mars at arrival has to be opposite the point of of Earth at Departure.

The following image depicts Earth's and Mars' orbit as circular, rather than elliptical to simplify calculations, and shows a sample Hohmann transfer from Earth to Mars. Distance is in AU. All objects depicted are moving counterclockwise.

Geogebra Graph of simplified Hohmann Transfer

Corrected Hohmann Transfer Image A Hohmann Transfer from Earth to Mars takes about 0.708 years with the aforementioned simplifications. In that time, our depicted Mars (orbital period about 1.88 years) will move approximately 135° around its orbit, and the depicted Earth will move about 255°.

As such, in this simplified view of the transfer, at the time of departure, Mars is near Western Quadrature (Rather than Opposition) to Earth, similar to its current position as of late July, 2020.

Solar System Live image - UTC 2020-07-25 12:53:47

Image is from Solar System Live, UTC 2020-07-25 12:53:47, with Helocentric Longitude set to -55° to rotate the image similar to the graph above. Image uses the actual orbits, rather than simplified circular ones, and the planets are not to scale.

$\endgroup$
  • 8
    $\begingroup$ @Meiki It sounds like Zubrin misunderstood how a Hohmann transfer works. As notovny says, the point of arrival must be opposite the point of departure. That is, the major axis of the transfer ellipse touches the 2 planets' orbits. Calculating Hohmann orbits between Earth & Mars is slightly trickier than shown in the simplified diagram above because you do need to account for the eccentricity of the planets' orbits, especially of Mars, since its eccentricity is 0.0934. See en.wikipedia.org/wiki/Orbit_of_Mars $\endgroup$ – PM 2Ring Jul 25 '20 at 16:07
  • 3
    $\begingroup$ Upvoted because @novotny tells us what he's used for the pretty pictures. $\endgroup$ – Mark Morgan Lloyd Jul 25 '20 at 20:56
  • 7
    $\begingroup$ @PM2Ring That's very interesting. Zubrin, as quoted, is wrong, for precisely the reasons stated in the post. That's a pretty elementary screwup for someone at Zubrin's level, and, having seen him speak in person, I know he has a much better grasp of orbital mechanics than this. Rather than him truly misunderstanding, I think it's much more likely that he tried to simplify the subject for his audience, and went too far. $\endgroup$ – lwr Jul 26 '20 at 6:39
  • 2
    $\begingroup$ @lwr - Or perhaps Zubrin is not at the level he thinks he is at. $\endgroup$ – David Hammen Jul 26 '20 at 8:05
  • 4
    $\begingroup$ @PM2Ring, the correct alignment is when the two planets are in "time-shifted" conjunction: that is, the position of Earth now is on the opposite side of the Sun from where Mars will be when the spacecraft arrives. $\endgroup$ – Mark Jul 26 '20 at 19:50
3
$\begingroup$

Much of engineering is about compromises. One can find an ideal solution, like a Hohmann transfer orbit. Yes, that is the most fuel efficient way to get from earth orbit to Mars orbit. It is like the top of a rounded mountain. That is the peak, but there is a lot of ground near the peak that is almost as high. Maybe you are willing to give up a bit of fuel efficiency for some other objective, like launching this year, getting there faster, or having a radio link for your data when you get there. (Almost) every mission has a number of objectives, which need to be traded off against each other to make a mission design. Fuel efficiency is important, but not the only objective. If launching this year is important, that may cost you one instrument, or degraded performance on one. It may be worth it.

$\endgroup$
  • $\begingroup$ A Hohmann transfer from Earth's orbit to Mars's orbit does not exist. $\endgroup$ – David Hammen Jul 26 '20 at 8:04
  • 3
    $\begingroup$ @DavidHammen: It does in the approximation that the two orbits are coplanar circles. In the real case there is a similar orbit leaving the earth orbit by thrusting in the direction of earth's motion plus a little out of plane, going halfway around the sun, and thrusting in the direction of Mars' orbit plus a little out of plane. It will be more fuel efficient than other, faster, transfers. I think the question is well defined. $\endgroup$ – Ross Millikan Jul 26 '20 at 13:20
  • 1
    $\begingroup$ The inclination to the ecliptic of Mars is only 1.850°, so a coplanar approximation isn't too shabby. But maybe you could incoroporate something about that in your answer. $\endgroup$ – PM 2Ring Jul 26 '20 at 14:02
  • 3
    $\begingroup$ @uhoh: I think the most efficient orbit is close enough to use the term. As I said in the previous comment, you want to thrust in the direction earth is travelling and the direction Mars is traveling, up to the inclination correction. That will be more fuel efficient than any other transfer orbit. The fact that it is not strictly Hohmann is not important, The question is asking why we choose other transfers than the most efficient. I answered in that spirit. $\endgroup$ – Ross Millikan Jul 27 '20 at 13:12
  • $\begingroup$ @RossMillikan I agree, that's why I moved that discussion to a new question. $\endgroup$ – uhoh Jul 27 '20 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.