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I read somewhere that a lot of the fuel used for Earth-to-Mars trips is burned up in the first few hours -- leaving Earth's gravity well.

Working on the premise that it takes a whole lot of fuel to get out of the gravity well, let's imagine how much fuel and time would be needed to launch from in Earth's gravity well to Mars, keeping 1 G of thrust the entire time (accelerate halfway and decelerate halfway).

Time: arithmetic says it'd take about 35 days.

Fuel: I read somewhere that this would take an astronomical amount, which is why, as of how much bang for the buck we currently get, the best thing to do is accelerate of to some sort of cruising speed, coast for a while (many months), and decelerate.

However, if the ship were to launch from outside of the gravity well -- say, the L2 Earth-Moon-LaGrange Point -- which resource would more likely be saved: more fuel would be saved; or would more time be saved?

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    $\begingroup$ This is a straightforward space exploration question, belonging better to the related community. space.stackexchange.com/search?q=launch+from+moon $\endgroup$
    – L.Dutch
    Oct 1 '21 at 6:37
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    $\begingroup$ This is more than one question: 1) Launching from Moon to Mars; 2) Launching from Earth's L2 to Mars; 3) Minimizing travel time by extra fuel burn. $\endgroup$
    – Alexander
    Oct 1 '21 at 7:40
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    $\begingroup$ Note that 1g continuous thrust is INSANE (and likely unfeasible no matter our tech progress)-- it would allow to cross the whole milky way in 25 years (ship-time) => en.wikipedia.org/wiki/Space_travel_using_constant_acceleration $\endgroup$
    – radioflash
    Oct 1 '21 at 15:54
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    $\begingroup$ You would save a fraction of a fraction of a fraction of a percent of your fuel. a 1g continuous burn uses obscene amounts of fuel. Billion of trillions of times more than you actually need. Launching from orbit, or from the Moon, will save you a few tons, out of the zillions of tons you are going t need in any case. $\endgroup$
    – PcMan
    Oct 1 '21 at 18:48
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    $\begingroup$ physics or space-exploration are more appropriate $\endgroup$ Oct 7 '21 at 3:34
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There are slightly too many unknowns here for a straightfoward answer, but it can be shown that a continuous thrust transit (also known as a brachistochrone) between Earth's orbit and Mars uses up vastly more fuel than lifting from Earth's surface would, so the savings as a proportion of the fuel used are low.

Earth's gravity well to Mars, keeping 1 G of thrust the entire time (accelerate halfway and decelerate halfway).

Time: arithmetic says it'd take about 35 days.

You haven't thought very carefully about this! After 17 days of thrust at 1G, you'd be travelling at over 14000km/s and have travelled over 7 billion kilometres. The maximum distance between Earth and Mars is about 400 million kilometres. Your flight plan gets you to the Kuiper Belt, not Mars!

To travel the average Earth-Mars distance of ~225 million kilometres with a continuous thrust of 1G and a flipover in the middle takes a little over 3 days, given $t = 2\sqrt{\frac{d}{a}}$ where $d$ is the distance and $a$ is the acceleration.

However, if the ship were to launch from outside of the gravity well -- say, the L2 Earth-Moon-LaGrange Point -- how much fuel would be saved? Or, if using the same amount of fuel, how much time would be saved?

An important figure in rocketry is delta-V, or change in velocity. Earth's escape velocity, for example, is a bit over 11km/s. Ignoring the effects of atmospheric and gravity drag for the moment, a rocket which had a delta-V of 11-and-a-bit km/s could escape from Earth's gravity well and fly into interplanetary space. The moon's gravity is much lower, so its escape velocity is a bit over 2km/s. At the Earth-Moon L2 point it is lower still... well under 1km/s.

Now lets consider your 1G continuous burn trajectory. If we run the engine for ~1.75 days, we reach a maximum velocity of nearly 1500km/s. We then need to slow back down to a relative stop. That requires a total delta-V budget of nearly 3000km/s... slightly more than 270 times the minimum delta-V required to escape from Earth's surface!

Clearly, if you have rocketry powerful enough to sustain that much thrust for that long, getting out of a deep gravity well is a) child's play, b) cheap and c) fast. The difference will be negligible, if you ignore environmental issues.

You might consider asking a separate question about operating a rocket which could have a delta-V of the best part of 3000km/s inside Earth's atmosphere. Spoiler alert: it'll probably involve an awful lot of antimatter and be a bit like a continuous nuclear explosion that runs for a few minutes and probably ends with high-altitude EMP causing widespread issues across the hemisphere the rocket launched from. Rockets of this power level are exceptionally hazardous.

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    $\begingroup$ " if you have rocketry powerful enough... getting out of a deep gravity well is ... b) cheap" Ummm... I wouldn't call "cheap" a propellant budget that goes exponential with its mass. Have any idea what exp(150) translates into? If you could get with 1kg propellant for the minimal delta-V, to get 150*delta-V you would need 1.4e+65kg (Earth mass = 6.4e+24kg for comparison) $\endgroup$ Oct 1 '21 at 9:39
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    $\begingroup$ @AdrianColomitchi the OP presupposes the existence of a rocket powerful enough to do a 35 day brachistochrone transit at 1G. Spending an extra 20 minutes operating the engine to develop an initial 11km/s is not going to tax it unduly. $\endgroup$ Oct 1 '21 at 9:42
  • $\begingroup$ Got it, thanks. So that I'd need to direct my observation to OP showing that a rocket powerful enough to do a 35 day brachistochrone transit at 1G to Mars would involve Earth masses of propellant or ultrarelativistic ejection energies. Unless you want to do it as additional considerations in you answer. $\endgroup$ Oct 1 '21 at 9:48
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    $\begingroup$ @AdrianColomitchi looks like a proper beam-core antimatter rocket could do it with a mass ratio of a mere 1.14, following Frisbee's paper (with exhaust velocity of a non-relativistic .33c). Leaving aside the implausibility of such a rocket, something with an wet mass of 1140kg would need a mere 70kg of pure antimatter to get to Mars, and its engines would develop a power of about half a terawatt on takeoff. Cooling and radiation shielding left as an exercise for the reader. $\endgroup$ Oct 1 '21 at 10:14
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    $\begingroup$ @JohnDvorak nah, a mere 70kg of antimatter isn't going to make that much of a dent in the Earth... that's only a 3 gigatonne yield if it all goes up at once... maybe a 5km crater. Bad news for the witnesses and neighbours, but you'd barely notice if you were a country or two away. Of course, if it were a ship more the size of the space shuttle than a light aircraft it'd be a bit more problematic... $\endgroup$ Oct 1 '21 at 12:13
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Fuel: I read somewhere that this would take an astronomical amount, which is why, as of how much bang for the buck we currently get, the best thing to do is accelerate of to some sort of cruising speed, coast for a while (many months), and decelerate.

If you insist on going 1G then you would need that astronomical amount of fuel. But then there won't be much difference between L2 and the ground. in the amount of fuel you need to take with you that is. But when you coast for a while to save fuel then you don't save any meaningful time.

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Traveling to the Moon to refuel (re-tank) is not a good way to get to Mars. At least not at the moment. The details depend on what propellants you plan to use, what infrastructure is in place on the Moon and where you are going to land on the Moon.

If you plan to re-tank with LOX and LH2 (and re-tanking with both these propellants is most efficient) then you are limited to a polar location where the ice is. But propellants still have to be expended to land on the surface which must come from Earth, and a lot of propellant must be loaded on the Moon to escape the Moon and fly on to Mars. Producing large volumes of propellants on the Moon isn't going to happen any time soon.

re-tanking with just LOX is very inefficient because even more energy must be expended to land the fuel that is needed to take off again. But LOX could be produced almost anywhere from the Lunar regolith.

Note it would not be possible to accelerate/decelerate at 1g all the way to Mars. Chemical engines or even nuclear engines would not be able to provide sustained thrust for a long enough period. A thrust of 1g could only be sustained for minutes before the propellants ran out (regardless of propellants).

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    $\begingroup$ Surely if you have some kind of lunar fuel refinery, it makes sense to launch the products into (Lunar) orbit and transfer there? Taking the whole starship down the moon and back up again can't possibly be the efficient choice. $\endgroup$
    – Cadence
    Oct 1 '21 at 9:58
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    $\begingroup$ Yes true. There are all sorts of possibilities and permutations. The main issues are the minimal gain and the vast outlay in infrastructure to generate such a small gain. Although propellant produced on the Moon does not have to travel from Earth that does not make it cheap. Because the vast solar array or nuclear power plant required, all of the storage tanks and material processing facilities and repair needs would have to come from Earth at great cost. Maybe revisit the issue a hundred years from now and the answer might be different. $\endgroup$
    – Slarty
    Oct 1 '21 at 10:06
  • $\begingroup$ The other problem with lunar orbits is the changing alignment with Mars due to the Moons orbit around the Earth. So delays might be costly $\endgroup$
    – Slarty
    Oct 1 '21 at 10:08
  • $\begingroup$ "Because the vast solar array or nuclear power plant required, all of the storage tanks and material processing facilities and repair needs would have to come from Earth at great cost." Then send one factory that build solar arrays and storage tanks on the Moon first. Otherwise it's like saying "It doesn't make sense to colonize America, because we'll need to send them axes and hoes at great cost" Granted, it's not going to finish this stage tomorrow, but neither a trips to Mars scheduled the day after. $\endgroup$ Oct 1 '21 at 10:50
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    $\begingroup$ Yes that works, but at the cost of even more infrastructure $\endgroup$
    – Slarty
    Oct 3 '21 at 18:12
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If we built spaceships on the Moon, would we have more fuel to burn for trip to Mars

Yes, you will have save fuel. Gravity is a conservative field - the energy one spends (or gains) moving between two points is the same no matter the trajectory.

More or less, the extra energy available to you if you start the journey to Mars from L2 instead of the Earth surface is equal to the energy you need to reach L2 from Earth.

Some caveats:

  • when you fall into a gravity field, the energy that you gain will not fill back your rocket fuel tank
  • a good amount of energy when taking off from inside Earth atmosphere is lost to aerodynamic friction.
  • while the energy you save is constant, the percentage of the saved fuel to the total necessary fuel for your speed/time regime will depend on that regime.
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    $\begingroup$ Energy isn't usually a useful way to look at things in orbital dynamics. The amount of propellant required is related to the change in velocity, not the change in energy, and in fact the same burn can make different changes to a craft's orbital energy depending on what direction it's in and what the craft's orbital velocity is at the time of the burn. Falling into a gravity well does not put propellant in your tank, but does mean you can get a bigger trajectory change out of it. $\endgroup$ Oct 1 '21 at 15:16
  • $\begingroup$ @ChristopherJamesHuff First, the amount of propellant is already derived from energy considerations, so you'll use energy implicitly anyway. Second, energy is important when you get to compute trajectories that don't neglect gravitational interaction with other bodies and start thinking of grav assist (slingshot or braking) maneuvers on it. delta-V is for beginners. $\endgroup$ Oct 1 '21 at 20:42
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As others point out, such a continuous-thrust trajectory implies capabilities that make it pretty much irrelevant where you start from. You could pretty much just point your ship where you want to go, without any care put into trajectory design and minimal thought toward propellant budgets.

Real spacecraft do high-thrust burns that tend to last tens of minutes at most, or use continuous acceleration with thrusters that are propellant-efficient, but so power-hungry that the vehicle is limited to milli-gee accelerations. And realistically, those use periods of coasting to get into position too. There's parts of the trajectory where it's more effective to expend propellant, you can get more out of what you carry if you're more careful where you expend it. The ideal case is actually to have the burns be instantaneous, real vehicles of course being limited to finite accelerations.

The low-thrust craft are good candidates for starting from the vicinity of the moon, because it saves months of slowly spiraling out through Earth's radiation belts. The moon is not a good source of the best propellants for these systems, though, so the propellant would need to be imported...most likely from Earth.

As for high-thrust vehicles, yes, they could get to Mars from lunar orbit with less propellant than needed to do so from Earth orbit. However:

  • You need to get back, too. Out in lunar orbit, that means burning propellant to brake into orbit. If you instead return to Earth, you can use the atmosphere to brake, considerably reducing your propellant requirements (allowing you to carry useful payload instead of propellant).
  • There isn't really anything the moon has that Mars doesn't. You need to get passengers/cargo from Earth, and it takes nearly as much propellant to get them from LEO to lunar orbit as it does to go straight to Mars. Stopping there actually makes the trip more expensive.
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  • $\begingroup$ The Moon, with an escape velocity of 2.4 km/s, could place the fuel tanks into orbit using non-rocket based means. Granted, not on short term, but then neither is the original question restricted to "how would you do it today". $\endgroup$ Oct 2 '21 at 0:14

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