Talk:Hohmann transfer orbit

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The G in the formula should be changed to a nice greek Gamma in my eyes — Preceding unsigned comment added by 217.85.247.32 (talk) 22:15, 4 August 2004 (UTC)[reply]

It shouldn't. The G you refer to is the Gravitational constant and is very standard practice Cole.christensen (talk) 09:00, 30 October 2008 (UTC)[reply]

Where it shows the vis viva equation, the equation has an M but the meanings for the parameters only mention a mu (greek letter). Which is? — Preceding unsigned comment added by Gonzalo Diethelm (talk • contribs) 03:17, 12 December 2004 (UTC)[reply]

mu is the Standard gravitational parameter, M is the mass of the larger object Cole.christensen (talk) 09:00, 30 October 2008 (UTC)[reply]

This is confusing:

At the other end, the spacecraft will need a certain velocity to orbit Mars, which will actually be less than the velocity needed to continue orbiting the Sun in the transfer orbit, let alone attempting to orbit the Sun in an Mars-like orbit. Therefore, the spacecraft will have to decelerate and allow Mars' gravity to capture it.

A spacecraft that has Hohmann transferred from Earth's orbit to Mars's will have a linear velocity of about 21.4 km/s, where Mars is going around the Sun at about 24.1 km/s.

That's not a deceleration. What will happen in practice is that the craft will be slightly ahead of Mars in the orbit, and let Mars "catch up with it."

Perhaps what the article meant to say is that the craft will decelerate relative to Mars. — Preceding unsigned comment added by 71.71.238.231 (talk) 19:57, 20 January 2006 (UTC)[reply]

Not all transfer orbits are Hohmann transfers. In practice, you seldom get a perfect minimum energy transfer orbit. Instead, you must use an affordable transfer orbit - i.e., one with a relatively low fuel cost.

The reason the Hohmann transfer orbit mathematics are simple is that both the departure and the arrival occur at (opposite) apsides of the transfer ellipse. A more general sort of transfer involves only one of the end points of the intended trajectory - either departure or arrival, but not both - occuring at one of the transfer orbit's apsides. The transfer path itself is a segment along the transfer orbit that might, or might not, include the other apside.

I've prepared a text based treatment of this more general transfer orbit solution for the "physics forums" website. If someone wants to construct a version of my treatment for Wikipedia, giving me credit for the original work, I'd have no objection.

The URL is

http://www.physicsforums.com/showthread.php?t=29524

Incidentally, I have another essay posted on "Physics Forums" that relates to the determination of the elements of an asteroid's orbit using the Method of Gauss on telescope observations from Earth at three different times. The URL for that essay is

http://www.physicsforums.com/showthread.php?t=36657

Credit in that case would be to me and to the Russian astronomer A. D. Dubyago.

Jerry Abbott — Preceding unsigned comment added by 170.215.171.179 (talk) 00:38, 27 February 2006 (UTC)[reply]

This statement seems backwards, or at least ambiguous:

"Hohmann transfer orbits also work to bring a spacecraft from a higher orbit into a lower one – in this case, the spacecraft's engine is fired in the opposite direction to its current path"

To go into a lower orbit, you would fire the engine in the same direction as the current path, that is to say, the rocket engine's exhaust would be emitted in that direction.

Perhaps less ambiguous would be something like "the spacecraft's engine is fired to create a thrust in the opposite direction to its current path".

Spope3 (talk) 23:49, 9 November 2008 (UTC)[reply]

How is "Hohmann" pronounced? — Preceding unsigned comment added by 69.132.53.190 (talk) 01:36, 2 June 2007 (UTC)[reply]

hōmân -- emphasis on the first syllable. First syllable rhymes with "Joe". Second syllable rhymes with "John" (kind of). Karl Hahn (T) (C) 02:12, 2 June 2007 (UTC)[reply]

It doesn't rhyme with "John", it's more like "Bro hun" or "No gun" 79.199.56.26 (talk) 20:03, 19 July 2010 (UTC)[reply]

Karl Hahn is correct. The comment directly above is wrong. — Preceding unsigned comment added by 142.136.123.62 (talk) 04:48, 22 February 2013 (UTC)[reply]

The velocities listed as delta V's are not the change in velocity required to complete the maneuver, but rather they are the total velocity at the point of maneuvering.

To get the change in velocity one must subtract the current velocity in the orbit.

I am changing the equations to reflect that these are total velocities at periapsis and apoapsis, and not delta V's.

I would like to add equations that reflects total delta V for both impulses, by subtracting the velocity of a circular orbit at that altitude, but since I'm at work I can't do that immediately, so anybody else feel free to add that if they want! Mattski (talk) 23:28, 17 January 2008 (UTC)[reply]

Your change does not seem correct, take e.g. r1=r2. The subtraction is already in the formulas.--Patrick (talk) 01:45, 18 January 2008 (UTC)[reply]

Sorry for the bad edit. I was using the formulas on the page and getting incorrect answers because I was using altitudes instead of radii. I redid the derivation and these formulas are indeed correct delta-v's. 65.160.147.253 (talk) 21:30, 18 January 2008 (UTC)[reply]

It's not clear how you get from the Vis-Viva equation to the delta velocity equations.. — Preceding unsigned comment added by 87.55.107.211 (talk) 05:27, 17 June 2013 (UTC)[reply]

Is there some way to make clear in the lead paragraph of the article that the two engine burns are assumed to take place in zero time, i.e. they each create an instantaneous change in delta-v? The lead sentence is already a bit lengthy.... (sdsds - talk) 03:37, 28 May 2008 (UTC)[reply]

Reference 1 appears to be broken. --Coosbane (talk) 21:01, 7 February 2009 (UTC)[reply]

The article had a mistake in attributing the first delta-v to periapsis; that's clearly only so when transferring from a lower into a higher orbit. I've corrected the mistake, generalizing the article to transfers from a higher into lower orbit as well. The edit was reverted, on the grounds of "unencyclopedic tone." Could you please elaborate on that? I guess you (i) agree with the need for the edit but (ii) doesn't agree with the particular wording. If you don't care to reword rather than revert, correctness should take precedence over tone. 128.138.43.113 (talk) 02:41, 9 March 2009 (UTC)[reply]

Sorry, I think I jumped the gun a little too fast there. The revert stating "unencyclopedic tone" did not refer to the edit that I was concerned with, which remains in effect. So I take back my protest. 128.138.43.113 (talk) 04:08, 9 March 2009 (UTC)[reply]

At the end of the "Maximum delta-V" section, I added a fact and removed the citation needed tag. It's simple enough that I don't think a citation is needed, but if someone wants me to work through the math of it here, I can. I felt it would simply clutter the main article to include it there.Gmalivuk (talk) 21:13, 27 May 2009 (UTC)[reply]

As far as I can tell, *all* of the citation needed marks on this page don't need to be there; a simple check of the math already provided on the page with demonstrate the points being made without any real need to call on an external source. 96.30.136.159 (talk) 20:28, 14 March 2012 (UTC)[reply]

Agreed some bozo put it back. I shall remove it again. Montestruc (talk) 02:12, 29 September 2015 (UTC)[reply]

I haven't confirmed the numbers leading up to this, but when it gets to the extra delta-v needed to reach escape velocity, rather than enter the transfer orbit, I think the difference is 0.78:

10.93 - 10.15

or 3.20 - 2.42

Jmichael ll (talk) 02:26, 21 July 2009 (UTC)[reply]

The article makes the simple mistake of stating that the potential energy at the farthest point (apoapsis) is given by -GMm/a. The correct formula is -GMm/r_a where r_a is the distance from the cental body to the apoapsis. Dauto (talk) 20:21, 8 August 2009 (UTC)[reply]

I see on the Russian Academy of Sciences site, some of Tsiolkovsky's pre-revolutionary notes (around 1911) show elliptical transfer orbits in diagrams. Some Soviet literature attributes the invention of the fully worked out math to Vetchinkin, others (the Keldysh Institute's reports) attribute it to Tsander. In my own historical research, it is unclear who did it first, as Tsander and Vetchinkin worked closely together on these problems. It is clear that these Russians developed and communicated the idea before Hohmann. In fact, Vetchinkin wrote a famous letter expressing his belief that Hohmann got the idea from the Russians, but this is unprovable. Personally, I don't think it is a difficult concept. Anyone who asks the question of how to transfer from on planet to another will quickly reach this answer. So I think it is believable that Hohmann arrived at the result independently. DonPMitchell (talk) 22:07, 31 March 2010 (UTC)[reply]

The name Kurd Laßwitz is spelled with the German letter ß (eszett = "sz"). This is not a letter in the English language. It's use in German is, according to the Wikipedia german alphabet article, not universal. It is only used in lowercase (uppercase uses "SS") and regional (Swiss German replaces both upper and lower case with "ss"). Since this is the English the English equivalent should be used.154.5.32.113 (talk) 05:19, 15 June 2011 (UTC)[reply]

Since nothing in the Solar System has a truly circular orbit (moreover, no two planetary orbits are exactly coplanar), the current first paragraph defines something which cannot exist in the System - although it may be what Hohmamm actually described.

The paragraph, and the article as a whole, should be rewritten on the basis that the orbits are at most approximately coplanar and circular.

Contributors should check for a definition formally recognised by an international body. Otherwise, I suggest that a Hohmann is an orbit tangential to the two other orbits. If one wants, as one must, to allow one truly Hohmann orbit from Earth to Mars for each opposition, one must be careful about calling it minimum-energy, as the least energy depends on the circumstances of the opposition.

There remains the question of what term might be applied to an orbit in a field which is not inverse-square from a single centre - from low Earth orbit to Earth-Moon L1, for example.

94.30.84.71 (talk) 10:43, 11 June 2012 (UTC)[reply]

Thank you for that observation; I had (and have) the same question. "Circular" was omitted until 10 April 2006, when it was added along with "under standard conditions". Somewhere further along the line, the "standard conditions" part was dropped.

circular-to-circular.Captain Puget (talk) 06:02, 9 December 2012 (UTC)[reply]

The Application section reeks of astroturfing. It adds nothing useful, is out of context and reads like PR talk. Surely the vast majority of Mars missions, if not all, has employed a Hohmann Transfer Orbit. 85.139.13.107 (talk) 00:57, 17 November 2014 (UTC)[reply]

Should the location of the picture be changed? The section titled "Explaination" describes the drawing, but when its being read, the picture itself is off screen. Should this be lowered into that section? Leobold111 (talk) 20:16, 30 March 2015 (UTC)[reply]

I is mathematically obvious to anyone who can do the math. Montestruc (talk) 02:10, 29 September 2015 (UTC)[reply]

The Diagram shows the arc labelled 2 - wouldn't that be a straight line rather than an arc? An arc would require continuous fuel burn, no? — Preceding unsigned comment added by 2601:1C0:8100:1D:9476:948D:388D:90A0 (talk) 14:57, 3 October 2015 (UTC)[reply]

Probably a bit late to clarify for you, but Earth is constantly exerting a gravitational force on any object with mass in its vicinity. this force is always radial towards the earth, so once a burn stops its trajectory is still being changed by gravity. These trajectories can be approximated as conic sections (e.g. circles, ellipses, hyperbolae). It would take a constant burn to move in a straight line to counteract gravitational forces, but not necessarily to move on an arc. You can look at orbital mechanics for more. Hope I cleared this up for you and anyone else thinking the same thing.--Cincotta1 (talk) 16:29, 24 March 2016 (UTC)[reply]

Section doesn't seem to mention where you would aim for. I could be being silly, but if travelling to higher orbiting planet say Mars and trying to minimize delta-v would you aim for the planets orbital distance or somewhere just above (or even below) the Mars L1 orbit and then let the planets gravity pull you into an orbit rather than needing much of a delta-v when you arrive at Mars orbital distance? Maybe this is just insignificant? Seems to me like this could be more efficient than requiring a slow down delta-v to achieve capture? Perhaps too far from concept of Hohmann transfer orbit? crandles (talk) 20:14, 9 May 2016 (UTC)[reply]

You are. 94.30.84.71 (talk) 16:30, 30 September 2017 (UTC)[reply]

Sorry but your aim would not get you to Mars. Your speed relative to Mars at Mars-Sun L1 is so high that gravity of Mars will not pull you to orbit. Jkn2 (talk) 09:13, 29 December 2017 (UTC)[reply]

The speed at apogee of a Hohmann transfer orbit is too slow to maintain circular orbit at the target distance from the sun and without the planet you would be about to start plunging back closer to the sun. I would be surprised if aiming at Mars-Sun L1 would be perfect for minimising delta-v requirements, but I imagine there must be some optimum between reducing delta-v requirement at initiation and minimising delta-v requirement when you arrive at apogee of Hohmann transfer orbit. Are you instead saying the optimum is to aim pretty much at the planet orbital distance and only by going really close to the planet can the planet pull you into orbit. Anything slightly less would mean you are travelling too slowly and are not pulled into orbit and the delta-v requirement to achieve orbit at minimum distance to the target planet is much greater than any saving you can make at initiation of the Hohmann transfer orbit? crandles (talk) 15:01, 10 March 2019 (UTC)[reply]

The present Introduction fails to include a clear statement of the essential benefit of a Hohmann transfer orbit, which is that it requires the least energy and delta-v.

I do not think that the term needs to be restricted to the case of circular planetary orbits.

For that case, I think it would be of sufficient interest to give the initial and final angular separations of the two planets.

94.30.84.71 (talk) 17:23, 30 September 2017 (UTC)[reply]

Definition of the Hohmann transfer orbit is incomplete. It does not mention "both the departure and the arrival occur at (opposite) apsides of the transfer ellipse." (Text taken from Jerry Abbott's comment.)

English translation of 'Die Erreichbarkeit der Himmelskörper' by Walter Hohmann:[1]page 49

"If the vehicle's path is to touch at a distance r_II from the Sun, the orbit of a planet other than the Earth, distance r_I from the Sun (see Figure 25), then the major axis of the ellipse is

a = (r_I+r_II)/2 "

This does match with definition [2]

"A Hohmann Transfer is a two-impulse elliptical transfer between two co-planar circular orbits. The transfer itself consists of an elliptical orbit with a perigee at the inner orbit and an apogee at the outer orbit."

Definition should be changed to generalization of MIT definition:

In orbital mechanics, the Hohmann Transfer is a two-impulse elliptical transfer between two co-planar circular orbits. The transfer itself consists of an elliptical orbit with a periapsis at the inner orbit and an apoapsis at the outer orbit.

or

In orbital mechanics, the Hohmann transfer orbit (/ˈhoʊmən/) is an elliptical orbit used to transfer between two circular orbits of different radii in the same plane. The Hohmann orbit has a periapsis at the inner orbit and an apoapsis at the outer orbit.

Which is better?

This is also good, but not always true (because of Bi-elliptic_transfer)

[3] The German engineer Walter Hohmann showed in 1925 that elliptical orbits tangent to the orbits of both the planet of departure and the target planet require the least fuel and energy.

Jkn2 (talk) 09:27, 30 December 2017 (UTC)[reply]

Magellan (spacecraft) had a 540° trajectory, apparently called a Type-IV. Type II is more than 180°, but 540° is a lot more! [4] Does it belong in this article? Tom Ruen (talk) 07:53, 24 February 2018 (UTC)[reply]

I'm no rocket scientist <pause>, but I've been surfing WP a bit and ran across the assertion here saying: "In general a Hohmann transfer orbit uses the lowest possible amount of energy." just after looking at the Low-energy transfer article. That other article appears to contradict that assertion here. I just thought I would mention this. Wtmitchell (talk) (earlier Boracay Bill) 09:49, 20 July 2019 (UTC)[reply]

The article is assuming that the transfer is from orbit about a gravitationally significant body (i.e. planet) to a different orbit about the same body. Where the transfer of orbit is from one planet to another the gravitation of the second planet changes the physics substantially. — Preceding unsigned comment added by Trewornan (talk • contribs) 21:29, 20 July 2019 (UTC)[reply]

Indeed. I noticed the same thing, and have tried to fix it. But better wording and citations would still be helpful. ★NealMcB★ (talk) 02:38, 30 July 2019 (UTC)[reply]

On the one hand we have: "An ideal Hohmann transfer orbit transfers between two circular orbits in the same plane and traverses exactly 180° around the primary." So presumeably for an orbit to/from Earth, the ideal one would be 6 months?

Yet later we have: "A Hohmann transfer orbit ... for an Earth-Mars journey ... is about 9 months. "

OK, accepting that this particular orbit is 9 months rather than 6 months, the question is ... why? What is it about Earth/Mars that makes the actual Hohman transfer so much longer than the "ideal" one. Is it the [great] eccentricity [of Mars] and the inclinations? How does that actually come into play? — Preceding unsigned comment added by 2001:8003:e422:3c01:a45c:155c:b77a:3dc8 (talk) 08:23, 21 July 2021 (UTC)[reply]

6 months is incorrect. The Earth->Mars transfer orbit has a periapsis at the Earth's orbit, and an apoapsis at Mars' orbit. That makes its period longer than Earth's, and shorter than Mars'. (Although the velocity of the transfer orbit at periapsis is greater than that of Earth, the transfer orbit is "slower" in terms of its period.)

Here's another counterintuitive case: You're in a circular orbit, just behind another object in the same orbit. You want to catch up. Do you speed up or slow down? The answer is, slow down. You will then fall from the circular orbit, pick up speed due to gravity, and when you rise back up to the circular orbit 360° later, you'll be ahead of the other object, because your altered orbit's period was shorter due to its increased speed and shorter path. (But don't forget to increase your speed again once you're back at the circular orbit, or you'll be even farther ahead when you come back around another 360° later. And yes, there are other ways of doing it.)

Historical postscript: The primary purpose of the Project Gemini series of spaceflights was to develop techniques to rendezvous two spacecraft in orbit, in preparation for the Apollo orbiter/lander rendezvous.

BMJ-pdx (talk) 07:01, 8 September 2022 (UTC)[reply]

" Therefore, the spacecraft will have to decelerate in order for the gravity of Mars to capture it. " This is either outright wrong, or misleading. Mars' orbital velocity is about 24 km/sec, the spacecraft is travelling at about 21.7 km/sec. So it has to accelerate.

This is also discussed in several places elsewhere. At apogee - which in this case will be at Mars encounter - there is a requirement for an apogee kick, which is an acceleration. — Preceding unsigned comment added by 2001:8003:e422:3c01:a45c:155c:b77a:3dc8 (talk) 08:40, 21 July 2021 (UTC)[reply]

The wording in the article is bad, but deceleration is not the problem. For efficiency, the kick will be done low in Mars' potential well, changing the hyperbolic trajectory into an elliptic one by reducing the relative velocity from ~6 to ~4 km/s (escape velocity is ~5 km/s).--Rainald62 (talk) 16:08, 15 June 2024 (UTC)[reply]

I just made a series of edits of the "The maneuver uses ..." sentence of the Intro, and I hope it's a lot clearer now. (The "raises the periapsis" of the original, while technically correct, was especially obtuse.) I also split the paragraphs before the sentences about "lowest possible amount of impulse". BMJ-pdx (talk) 06:34, 8 September 2022 (UTC)[reply]

You may also calculate the time it takes to leave one orbital plane and achieve the other when using least impulse using only the radius and the gravitational constant of the Sun.

the square root of (((r1 + r2) divided by .5787) times 2) gives the number of days, with the radii offered in miles (for kilometers one uses a different solar gravitational constant scaled for kilometers.)

Mars at some 142 million, Earth at some 92.956 million. SoftwareThing (talk) 20:19, 17 July 2023 (UTC)[reply]