Best orbit to launch a Mars Mission from?

keermalec · Oct 16, 2007

Hi, I thought I would share with you my findings concerning the best orbit to launch a Mars Mission from. Assuming: - 30 minutes burn time - aerocapture at Mars to a 250 x 33793 parking orbit (period: 1 martian day) - Hohmann transfer to a Mars orbital radius at 1.52 AU (average orbital radius of Mars): C3 = 8.64 km2/s2 - 0.1 km/s mid-course adjustments - 0.05 km/s Mars orbital adjustments - Mars aeroshell representing 18% of Mars capture mass - LOX/LH2 engine with zero boiloff using active refrigeration From LEO (ISS orbit) ------------------------------------------------ Altitude: 400 km Departure delta-v: 3.57 km/s Gravity losses: 0.28 km/s Necessary acceleration: 0.22 Gs Total propulsive delta-v: 4 km/s Useful payload in Mars orbit: 14% of initial mass From the Gate Orbit for C3 = 8.64 ------------------------------------------------ Altuitude: 85'600 km Departure delta-v: 2.08 km/s Gravity losses: 0.00 km/s Necessary acceleration: 0.12 Gs Total propulsive delta-v: 2.35 km/s Useful payload in Mars orbit: 34% of initial mass From LLO (Low Lunar Orbit) ------------------------------------------------ Altuitude: 50 km above the Moon Departure delta-v: 1.6 km/s Gravity losses: 0.05 km/s Necessary acceleration: 0.09 Gs Total propulsive delta-v: 1.8 km/s Useful payload in Mars orbit: 43% of initial mass Launching from LLO brings two benefits: 1. Highest useful mass to Mars orbit 2. Important savings if lunar oxygen is used 34% of the LLO-based mission is propellant, of which 7/8ths is Oxygen. That means if our Mars ship were 100 tons, 30 tons would be liquid oxygen. If this were mined from the Moon, it would save the mission the cost of lifting 30 tons of Oxygen from the Earth (ie some 300'000'000 USD + transport to LLO - Lift from Moon...). Fast transfer <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

nexium · Nov 1, 2007

Sorry, I don't understand the math, some of the technology labels nor the assuptions. I presume the position of Mars in it's orbit (relative to Earth) changes the numbers significantly. Neil

gunsandrockets · Nov 2, 2007

How about a gravity assist trajectory, such as used by the Nozomi mission to Mars. http://www.astronautix.com/craft/nozomi.htm http://adsabs.harvard.edu/abs/1991ads..confR....V

qso1 · Nov 2, 2007

Ther have been proposals that suggest a Venus flyby makes for a better trajectory to mars. This however, requires the favorable alignment of all three planets at either MOI (Mars Orbit Insertion) or EOR (Earth Orbit Return) which wont always be available. <div class="Discussion_UserSignature"> My borrowed quote for the time being:There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened. </div>

scottb50 · Nov 2, 2007

Just for fun what would be the requirements from LEO to LLO for a representative payload. Would the difference of bringing LOX to LLO offset the need to bring it from Earth to begin with? Then you have to deal with the equipment on the moon to extract and launch the LOX to LLO. The same would hold true with launching from the Gate Orbit, would the energy needed to get there to begin with offset the energy needed to get to LEO and depart from there? I would assume reaching 85+km as opposed to 400km would require a considerable amount of additional delta-v. <div class="Discussion_UserSignature"> </div>

keermalec · Nov 2, 2007

<blockquote>In reply to:<hr />I presume the position of Mars in it's orbit (relative to Earth) changes the numbers significantly. Neil <hr /></blockquote> Yes, the position of Mars changes the numbers significantly. The numbers given are for Hohmann transfers, when Mars is opposite the Earth in its orbit. It is the most favourable position involving the lowest delta-v. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

keermalec · Nov 2, 2007

<blockquote>In reply to:<hr />How about a gravity assist trajectory<hr /></blockquote> A gravity assist would further decrease the delta-v deeded at Earth departure, therby increasing the useful payload at Mars. However, the best orbit to launch from is LLO, and a lunar gravity assist is impossible here, so this launch delta-v cannot be optimised using lunar gravity assist. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

keermalec · Nov 2, 2007

<blockquote>In reply to:<hr />Just for fun what would be the requirements from LEO to LLO for a representative payload. Would the difference of bringing LOX to LLO offset the need to bring it from Earth to begin with? Then you have to deal with the equipment on the moon to extract and launch the LOX to LLO.<hr /></blockquote> From LEO to LLO requires about 4 km/s. From Lunar surface to LLO requires about 2.4 km/s. Assuming you can extract lunar oxygen at little cost (...), it is obviously cheaper to lift it for a mere 2.4 km/s compared to the 10-11 km/s you need to launch it from Earth to LEO. <blockquote>In reply to:<hr />The same would hold true with launching from the Gate Orbit, would the energy needed to get there to begin with offset the energy needed to get to LEO and depart from there? I would assume reaching 85+km as opposed to 400km would require a considerable amount of additional delta-v. <hr /></blockquote> In the case of the Gate Orbit, the answer is different: Delta-v required to get there is greater than the delta-v gained, if using high-thrust fast transfers. Its real advantage is in functioning as a refuelling spot. If you tug your ship there using a low thrust continuous engine like the ion drive, and launch from there, then mass savings are substantial. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

scottb50 · Nov 3, 2007

So if it takes 4km/s to get from LEO to LLO and 3.57km/s to get to Mars from LEO where are we saving that much by taking probably twice the mass needed to Mars to the moon first, to produce Oxygen, and then putting it in LLO for departure to Mars, requiring 2.4 km/s? If, we assume, Oxygen can be extracted on Mars it would be much simpler and much more expedient to just take all the resources to LEO and then to Mars and use Mars resources for a return. Which would be stupid, initially. Which means we take everything from the surface to LEO we need for the first roundtrip(s) and work on extracting resources once we get there. While the moon offers a number of reasons to explore and exploit in it's own right, I really question it has much value in going to Mars. <div class="Discussion_UserSignature"> </div>

keermalec · Nov 3, 2007

Good point Scott, I would tend to agree with you: better extract ressources at Mars (meaning martian moons) rather than the Moon. The reason getting to Mars orbit costs less than getting to lunar orbit is because of the possibility of aerocapture at Mars. Lunar capture has to be all-propulsive. For this same reason, getting to Deimos or Phobos actually costs less in terms of delta-v than getting to the Moon. But it does take 9 months whereas getting to the Moon takes 5 days... <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

keermalec · Nov 3, 2007

<blockquote>In reply to:<hr />How about a gravity assist trajectory, such as used by the Nozomi mission to Mars.<hr /></blockquote> I did a quick check on lunar gravity assist. At first glance, the delta-v gain is quite substantial: Launch from LEO at 3.57 km/s for a C3 of 8.64 km2/s2 (Hohmann transfer to Mars in 9 months). Launch from Earth at 3.57 km/s but do a lunar gravity assist with a periapse 50 km above the lunar surface. Resulting C3 = 28 km2/s2... (fast transfer to Mars in 4 months) Using a lunar gravity assist may add 2.4 km/s to final velocity and shorten trip time by as much as 50%! I checked my calculations again and your link says true, gunsandrockets: I get either 12.9% more payload to Mars (be reducing departure delta-v), or 4.5 months less travel time (by increasing final velocity) with lunar grav assist. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · Nov 4, 2007

Resulting C3 = 28 km2/s2... (fast transfer to Mars in 4 months) This may be correct - it's a bit more lunar assist dV than I thought possible - but it doesn't account for the high dV at arrival at Mars. C3 is the square of the excess velocity of your escape orbit - so C3 = 28 km^2/s^2 gives you 5.3 km/s of dV after you've climbed out of Earth's gravity well. I show that as being a reasonable departure dV, but the arrival dV is on the order of 10 km/s (!!!) <div class="Discussion_UserSignature"> </div>

keermalec · Nov 4, 2007

<blockquote>In reply to:<hr />it's a bit more lunar assist dV than I thought possible <hr /></blockquote> Your impression is right. I did a full verification and it is not 5.3 but 3.35 km/s (C3 = 11.2). :-( Lunar gravity assist therefore adds 0.39 km/s (or 13%) to a normal trans-Mars injection velocity. I was assuming the Moon was massive enough to bend the trajectory such that it left in the same direction as the Moon's motion, thereby gaining an extra 1 km/s but it isnt so. The vehicle leaves LEO at 11.4 km/s and an angle of 150Â° wrt to argument of perihelion. It enters the Moon's SOI more or less perpendicular to the Moon's direction of motion. At this distance the vehicle's velocity is 3.27 km/s. Relative to the Moon its velocity is therefore 3.42 km/s. The vehicle grazes the lunar surface at a periapse 50 km high (passing at 4.14 km/s) and leaves the lunar SOI at 3.42 km/s on the other side, having undergone a change in direction of only 18Â° (I initally thought 120Â° was possible). Adding in lunar orbital velocity and using the cosine rule, its final velocity on leaving lunar SOI at 66'500 km (444'000 km from the Earth's surface) is therefore 3.6 km/s. Its velocity at Earth SOI is therefore 3.35 km/s, which equates to a C3 of 11.2 km2/s only. Trip time is therefore not 4 months, but 185 days, or 6 months. Arrival delta-v wrt Mars is 4.65 km/s, which gives 6.77 km/s at atmospheric entry. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

gunsandrockets · Nov 4, 2007

My suggestion to use gravity assist is not so much to shorten travel times but instead to reduce the size of propulsion systems needed for slow cargo missions to Mars. The Nozomi mission swung by the moon twice and Earth once before heading out towards Mars, and in the process added two or three extra months to the travel time. But I imagine that by exploiting gravity assist the Nozomi mission was possible using a launch vehicle much smaller than customary.

spacester · Nov 4, 2007

<img src="/images/icons/laugh.gif" /> Nice correction. Those numbers all look reasonable to me. In fact, 0.4 km/s is what I had on file as the maximum lunar swing-by dV. <div class="Discussion_UserSignature"> </div>

keermalec · Nov 4, 2007

Gunsandrockets, the Nozomi trajectory is very interesting. I must admit though at this time I have no idea how one manages to gain delta-v by swinging twice round the Moon. To be studied for sure. <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

gunsandrockets · Nov 4, 2007

I don't know the details of of the Nozomi mission more than I have already said. I think Nozomi swung by the moon while heading outward yet still didn't have enough velocity to depart the Earth-moon system and so eventually fell back towards Earth. During that long fall towards Earth, Nozomi swings by the moon a second time picking up more speed, and with that higher speed Nozomi is still heading towards Earth thereby providing an opportunity for an Earth gravity assist before heading out towards Mars. From the Nozomi link I provided earlier... "The spacecraft used a lunar swingby on 24 September and another on 18 December 1998 to increase the apogee of its orbit. It swung by Earth on 20 December at a perigee of about 1000 km. The gravitational assist from the swingby coupled with a 7 minute burn of the bipropellant engine put Nozomi into an escape trajectory towards Mars." Multiple flybys of Venus or Earth for gravity assist are sometimes used for higher-energy unmanned missions. So if one considers the Earth-moon system analagous to the Solar System, then multiple flybys of the moon for gravity assist are easy enough to contemplate.

gunsandrockets · Nov 4, 2007

<In fact, 0.4 km/s is what I had on file as the maximum lunar swing-by dV.> Is that for a single lunar or double lunar swing-by?

keermalec · Nov 5, 2007

Single <div class="Discussion_UserSignature"> “An error does not become a mistake until you refuse to correct it.” John F. Kennedy </div>

spacester · Nov 5, 2007

"Single" Yes. Nozomi is very interesting. I remember at the time not being able to explain how they did it. I have a little more understanding now. You cannot use the thing that's establishing your gravity field to do a gravity assist. The final pass by Earth is not technically a gravity assist. What it does is allow you to redirect your course as required to get to Mars. Without that 7 minute burn, it stays in Earth orbit. <div class="Discussion_UserSignature"> </div>

nexium · Nov 5, 2007

Why can't a launch from LLO = low lunar orbit be followed by a gravity assist manuver around Earth? Neil

spacester · Nov 5, 2007

When you are in orbit around something, your motion is dictated by its gravity field. Given an initial velocity vector and position, your elliptical orbit is fixed and the only way to change it is to change your velocity vector. If you can do that, the new velocity will establish a new orbit. A gravity assist works by shifting the direction of your velocity vector. We're cruising along in our orbit . . . lah-de-dah, we know where we are and where we're going . . . then we approach something else orbiting the same thing we are. It's big: it has it's own large gravity field. We time ourselves to miss by a certain distance if we keep going on the same orbit. But the gravity field of our fellow satellite changes our course! It's pulling us in! We are going to get a lot closer to the surface than where we aimed! We are now under the influence of the fellow satellite. It's almost like the body we are both orbiting isn't even there. But we are going too fast to be captured into an elliptical orbit. Still, we are in orbit - it's just called a hyperbolic orbit. A hyperbolic orbit is a brief one. It is the 'swing-by' part of the gravity assist. What happens is, you enter and leave this new gravity field with the same velocity magnitude (speed). It changes the direction of your velocity, but not the magnitude. It does this because the hyperbolic orbit, like all orbits, has to have constant angular momentum. We're getting closer! Going faster and faster! (insert suspense-building text here) But all is well, we missed the surface after all. We are continuing on the hyperbolic orbit . . . the satellite is in our rear-view mirror now. And we're out! We are once again orbiting the original body, we are once again a fellow satellite along with the body we just grazed. But we changed the direction of our velocity. The direction is now more away from the central body than around <div class="Discussion_UserSignature"> </div>

gunsandrockets · Nov 6, 2007

<Without that 7 minute burn, it stays in Earth orbit.> That's one heck of a 7 minute burn to not only escape Earth but also gain the velocity needed to reach Mars!

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