Hohmann Transfer to Mars Reference Thread

[gunsandrockets]

Good grief. I've done nothing but try to play things straight with you. The only person playing 'gotcha' is you. You jumped at me from the start when all I did was point keermalec to a thread discussing the exact question he raised.<br /><br />I'm not your enemy despite your first and last instincts to assume I'm out to get you.

 

[jimfromnsf]

Good, it isn't just me that has trouble with gunsandrockets or spacester. It is the perfect storm, they found each other

 

[keermalec]

Easy guys, no use oiling the fire.<br /><br />Thanks, Spacester, now I get it I think. It is a mistake to accelerate in LEO because the change in angle per unit time is just too high (1 revolution in 1.4 hours!). If I burn for 30 minutes in this orbit I will have done over a third of a revolution, therefore a large angle difference between "horizontal" and direction of flight, which will give me huge gravity losses.<br /><br />I obviously have to start my Mars trip from a high orbit to minimise this.<br /><br />I suppose calculating the exact loss due to gravity involves plotting the hyperbolic escape curve against the circular LEO to determine the change in angle during acceleration. I plotted a straight escape line to simplify things and realised that around 20'000 km altitude gravity losses are reduced to around 1%. So I'm assuming I should go this high if I want to avoid them. <br /><br />This actually reduces my inital burn substantially from 3.10 km/s (at 150 km altitude) to 2.25 km/s at 20'000 km altitude! However, to transfer the ship from 150 to 20'000 km alt. will require 3.48 km/s by Hohmann transfer. therefore, parking and launching the ship from 20'000 km alt. will entail a total delta-v of 5.73 km/s, with negligeable gravity loss, as opposed to 3.10 km/s at 150 km alt. + some 0.2 km/s due to gravity losses (according to Spacester). SO it does still seem better to launch from low orbit, even considering gravity loss.<br /><br />On the other hand, if my ship is launched from a 20'000 km orbit, even though I will in the end use up more delta-v getting the ship there and then to Mars, I can refuel in that high orbit! Thereby substantially augmenting the payload capacity to mars, from 16% for a launch from LEO (150 km) to 22% for a launch from HEO (20'000 km), assuming a measly 309 isp from an N2O4/MMH drive. <br /><br />Unless I missed a beat somewhere, launching from high orbit is obviously the way to go!<br /> <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[keermalec]

OK this is getting exciting: there is an "optimal altitude" at which the delta-v needed to send a ship to Mars (with 2.94 km/s at SOI) on a Hohmann transfer is minimal. IE: the delta-v is more important at lower or higher orbits. I haven't yet worked out why this is so but the "optimal altitude" is:<br /><br />85'800 km above the surface, or about 92'300 km from the Earth's center.<br /><br />At this altitude the necessary delta-v required to get to mars is 2.08 km/s.<br /><br />On Mars the optimal altitude is at 8'900 km above the surface, or 12'300 km from the center. Here the required delta-v to attain 2.64 km/s at MSOI for Hohmann transfer to Earth is only 1.87 km/s.<br /><br />It seems strange but the math checks out right for now. I will delve deeper.<br /><br />At the moment it seems the best location to launch a Mars expedition from is at 92'300. Getting there requires 5.63 km/s delta-v for orbital transfer, but getting from there to Mars requires only 3.95 km/s!!! (!!!) (!!!)<br /><br />This gives us a 31% payload capacity if using a 372s ISP CH4/O2 drive... (assuming the ship is tugged to HEO and refuelled there by another ship).<br /><br />Just imagine what we could do with the extra delta-v or capacity:<br /><br />1. make the mission cheaper by making the ship smaller<br />2. make the mission safer by launching 2 ships<br />3. get there faster by choosing faster-than-Hohmann transfers, thereby also making the mission safer<br /><br />I will post the details of calculations for checking. <br /><br />If this does prove to be true, I suggest we call these altitudes the Interplanetary Gate Orbits ;-) <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[keermalec]

Here are my calculations, showing optimum delta-v requirements at the Earth IGO @ 92'300 km:<br /><br />Earth delta-v required for hohmann transfer to mars: vh = 2.94<br /><br />Orbital speed at 92'300 km from Earth: vo = sqrt(GM/92300) = 2.08 km/s<br /><br />Speed at 92'300 km necessary to escape the Earth and attain +2.94 km/s at SOI: ve = sqrt(vh^2 + 2*GM/92300) = 4.16 km/s<br /><br />Therefore delta-v at 92'300 km = 4.16 - 2.08 = 2.08 km/s. <br /><br />.......................<br /><br />And the same for the Mars IGO @ 12'300 km:<br /><br />Earth delta-v required for Hohmann transfer to Earth: vh = 2.64<br /><br />Orbital speed at 12'300 km from Mars: vo = sqrt(GM/12300) = 1.87 km/s<br /><br />Speed at 12'300 km necessary to escape Mars and attain +2.64 km/s at SOI: ve = sqrt(vh^2 + 2*GM/12300) = 3.73 km/s<br /><br />Therefore delta-v at 12'300 km = 3.73 - 1.87 = 1.87 km/s. <br /><br /><br />The altitude of the IGO changes with required interplanetary velocity. The ideal orbit is lowered with increased interplanetary delta-v. But for a given planet, there is a given IGO at Earth. For example, the Earth IGO for minimum delta-v Hohmann transfer to Venus is @ 106'200 km from the center of the Earth. <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[spacester]

<font color="yellow">Good grief. I've done nothing but try to play things straight with you. The only person playing 'gotcha' is you. You jumped at me from the start when all I did was point keermalec to a thread discussing the exact question he raised. </font><br /><br />You can't ignore all the math I present and then tell me you're playing straight with me. <br /><br />I want to talk math, you want to play politics; that's the problem here from my point of view. You know perfectly well that there was another guy playing 'gotcha' here, he stopped doing it but you chose to bring it back out from a dead thread. SG quit doing it but you're going "AHA! spacester said this here but that there! Got him!" <br /><br />I don't know what your motives are, all I know is what you write and post. If you want to play it straight, talk math. And don't pretend that I made the original conflicting statements when they were made by SG.<br /><br />I don't assume you're my enemy and perhaps I over-reacted but I'm trying to explain this stuff to a hostile audience and I'm done being a pacifist when that happens. I spent years here explaining stuff to an indifferent audience and then it turned hostile and that fact still ticks me off.<br /><br />You're one of my favorite posters here but my perception is that you've been hostile on this particular subject from the beginning. I don't know why you act that way on this subject.<br /><br />The other part that really ticks me off is that we've got a whole society whose eyes glaze over as soon as they see an equation. You're a very knowledgeable, intelligent guy, you are an established leader on a prominent message board, and you refuse to do math. What hope have we as a species to get into space with such an attitude of willful ignorance? These are not abstract concepts of no practical significance. <br /><br />If you want to play it straight, talk math! I cannot believe that high school geometry and algebra is over your head, but if it is, sw <div class="Discussion_UserSignature"> </div>

 

[spacester]

Keermalec (and others), I am spending my energy here on the wrong thing and I apologize. I would much rather focus on the math. I'm way behind now in terms of really carefully reading your posts.<br /><br />Allow me to explain, or at least present, another thing before I do that.<br /><br />The spiral trajectory math, as I understand it, is based on the mathematics of a circular orbit. It is a numerical integration of incremental applications of deltaV to a circular orbit. Making that assumption makes the integration manageable.<br /><br />But the orbit isn't circular, it's elliptical. If you stop thrusting you will be in an elliptical orbit. Not only that but (as I understand it) you will find that the point at which you stopped thrusting was not the periapse. Therefore the thrust you were applying just before you stopped was not coupling the momentum change to the bound energy change. Therefore you were experiencing gravity losses. Therefore you experience gravity losses the entire time you are thrusting on a spiral trajectory.<br /><br />The circular orbit assumption is made to allow you to continually adjust the parameters of the continuously changing elliptical orbit. But the result of that vector math shows that you have experienced gravity losses. The charts allow you to find the gravity losses.<br /><br />I say all that with some trepidation. Keermalek, you have now used those charts more than I have. I'm a brute force engines kind of guy, I've never really gotten into spiral trajectories. So feel free to correct me if that's not ringing true for you.<br /><br />I'm looking very busy for the rest of the week, I don't know when I'll be able to really check out your posts. <br /><br />My initial reaction is that you might possibly be taking gravity losses as provided in my Hohmann tables as a constant, but it is not constant. This might possibly be why you are finding an optimum altitude for interplanetary trajectories. <div class="Discussion_UserSignature"> </div>

 

[keermalec]

Spacester, thanks for your post. I actually started a new thread under Space Science and Astronomy on this new subject which I entitled Interplanetary Gate Orbits.<br /><br />This has nothing to do either with gravitational losses or spiral trajectories. I just happened upon it while investigating gravitational losses.<br /><br />I have found that when leaving the Earth for a specific delat-v at SOI, there is a specific altitude under which AND over which the delta-v is HIGHER. This altitude is 92'300 km for the 2.94 km/s Hohmann transfer to Mars. Above 92'300 km and below it, the delta-v is higher. Check and you will see. I have posted my formulae for this purpose.<br /><br />The difference is not huge, but meaningful. For example, at 4'000 km the delta-v is 3.05 km/s, and at 400'000 km the delat-v is 2.27 km/s. But at 92'300 km it is at a minimum of 2.08 km/s. <br /><br />Here is the new thread.<br /><br />PS: Note also that at these altitudes gravity losses are practically negligeable as 30 minutes burn does not significantly alter a ship's orbital trajectory (2.4° only), which is almost a straight line. I estimate the gravity loss at 0.87 m/s only at this altitude. <div class="Discussion_UserSignature"> <p><em>“An error does not become a mistake until you refuse to correct it.” John F. Kennedy</em></p> </div>

 

[gunsandrockets]

<Good, it isn't just me that has trouble with gunsandrockets or spacester.><br /><br />Hmmm...isn't your list of people you seem to have trouble with about a dozen names short?