I have been teaching a space exploration unit to a secondary school science class. I want them to have some appreciation of the amount of fuel required to lift a payload into orbit or further out to space.
Specifically, we are considering the logistics of a manned mission to Mars. I want them to consider the emmense problem of transporting / caching sufficient fuel on Mars for the return trip. That is, if it takes 180kg of fuel to get a 1kg payload to Mars and 30 kg to get it back then the fuel required is 180×31kg.
I have put together a rough sort of a spreadsheet which kind of shows this. You enter the payload and amount of fuel and it tells you how far above the earth’s surface it reached before falling. If it escaped earth’s gravitational well then it tells you the velocity attained and gives an estimate of the time for the journey.
Lesson done and I think the message got across, but I can’t help wondering if I could have improved my formulas a bit and gotten more realistic results. I had to make a number of assumptions –
[li]I calculated height reached using the principle of energy conversion: chemical to gravitational. I assumed that the fuel, on average was lifted for half the height attained.[/li][li]Having calculated the energy to get free of the gravitational well I converted any remaining energy to kinetic energy and solved for speed.[/li][li]I used an unreasonably large deltaH for my chemical reaction to provide the energy so that my results fit in a reasonable range.[/li][/ul]
This gives me an ugly looking piecewise function that, while indicative, gives some odd values for some parameters. It also omits the fuel requirement for braking once the planet is reached. I would welcome a refinement of my assumptions or a better rule-of thumb formula that will illustrate what I need.
That’s how I understood it, not that it makes much sense. But the 180 kg of fuel per trip isn’t just to get 1kg of payload to Mars, it’s to get the empty ship plus the 180kg of fuel (which diminishes as the flight progresses) plus the 1kg of payload to Mars.
Which means that if you took two kg of payload in one trip, you’re not increasing the weight by 100%, you’re increasing it by way less than 1%. So you wouldn’t need 360+kg of fuel to do it, you’d need (pulling a number out of my ass) like 180.2kg. For 3kg of payload, you’d need like 180.3kg of fuel. And so on.
Sorry if I lacked clarity.
I’m looking at how much fuel is required to get 1kg of – whatever to Mars. For the purpose of this discussion the mass of the rocket counts as “payload”.
If it takes 30 kg of fuel for each kg I bring back from Mars, then it I obviously need to send that 30kg of fuel to Mars – either in the same trip or cache it somehow.
If each kg I send from earth to Mars requires 180kg of fuel to get it there then it folows that I need to use 180×30 kg to get the necessary fuel to the red planet.
The point being made is that a return manned trip is logistically a lot more difficult and expensive than a one-way trip.
Note that I am not wanting to put together something that will withstand the scritiny of a design team - I’m not a rocket scientist. What I do want is something that serves well as an illustration and gives ballpark reasonable results when you play with a few basic input parameters. I know we have a few rocket scientists on the board and could think of no better group to ask.
The Saturn 5 used for the Apollo program provides a good real-world sanity check; you might use those mission parameters in your spreadsheet, and see how your output compares with the real numbers.
For the lunar missions, the Saturn could carry a payload of 45,000 kg, and the total vehicle mass was 3.04M kg; that’s a vehicle-to-payload ratio of 68:1. That allowed for a return flight from the moon. ISTR that the Saturn vehicle was doing just a few thousand MPH by the time they were in a position to start orbiting the moon. Mars has more gravity than the moon (0.376g, versus 0.165g for the moon), so you need a bigger rocket to get off the surface for the return trip. Plus it’s a longer distance, so you want a higher interplanetary speed so you can reach Mars in a reasonable amount of time.
Just a SWAG, but if the Saturn was running a 68:1 vehicle-to-payload ratio in order to reach the moon, 180:1 (vehicle-to-payload, not fuel-to-payload) seems not too outlandish for a Mars mission, and a 30:1 ratio for the return flight seems reasonable, too.
This much, I can tell you, is no good. Rockets are a horribly inefficient way of converting chemical energy to gravitational potential energy.
On the other hand:
If you can get a payload to the Moon, you can get it almost anywhere, since any realistic mission to Mars or other points beyond will use lunar slingshots to get up to speed. The reason a mission to Mars is so much harder is just because the mission must take a longer time, and thus must generally have a larger payload (especially if it’s manned, in which case you need a lot more life support).
Thanks for that. I had avoided using the rocket equation directly since I didn’t have specific impulse values to hand. This looks good. I think I will redesign the spreadsheet for next time I teach the class.
I am aware of that. But I thought that it would at least be approximately proportional. I wasn’t sure of the height that fuel was typically lifted – except to say that the quicker the burn and the faster you could jettison un-needed hardware the better.
Again, I suspected as much. The kids do have a research project and while I have hinted that these kind of options are available I have left these kinds of solutions for them to discover for themselves. My spreadsheet was merely to present the enormity of the task and show them that creative solutions are needed.
The really expensive parts of the trip are the beginnings & endings – going up or down in the gravity wells of Earth & Mars. So if that could be avoided, like for example, shuttling the needed fuel & supplies up to the space station, and then launching the rocket from near there – how would that affect the design?
It might be an interesting discussion to ask your kids to brainstorm about this, and maybe do some rough calculations about how the size & fuel load of the rocket would change under this plan. Also note that the overall fuel load would not decrease; in fact would probably increase (significantly?) because of the inefficiencies of many shuttle trips to bring supplies up to the space station vicinity. This might bring up an interesting class session about the trade-offs needed in any such design.
I have to take minor issue with the underlined statement; although it would seem to be beneficial to use the Earth’s overly large Moon to get a little bit of extra free impulse, the reality is that it is almost impossible to do this. The primary reason is that the plane of the Moon’s orbit about the Sun is ~5° off the solar ecliptic, while both Earth and Mars are about 1.5° to 1.7° degrees off the ecliptic. This means that it is relatively easy to plot synodic transfer trajectories between Earth and Mars given a mission window of a few years to work with, but plotting a course that uses the Moon to significantly shape the trajectory of a Mars trajectory requires a particular family of conjunctions that is much rarer, since you now have to select a trajectory that only only nulls out the in-plane velocity as you approach the intercept but also have a dramatic out of plane component that has to match up. (For the same reason, using the Moon as a staging base for any other interplanetary missions is also problematic.)
However, Chronos is correct that if you have the capacity to achieve translunar injection, you also have the capacity to achieve interplanetary injection. Getting from geosynchronous orbit to any place outside of Earth’s orbit is a trivial exercise, in energy terms, provided that you have nothing but time to spend. You can use the outer planets to alter your trajectory (and the typical purpose in performing planetary flybys is to effect a radical change in direction, not pick up speed) but it isn’t really necessary, so long as you aren’t really in a hurry.
The Artemis Project website cited early has the Tsiolkovsky rocket equation, which is the governing principle of reactive thrust based propulsion. The equation as presented, however, only considers a single stage vehicle; because the propellant mass fraction (the fraction of mass that is propellant in proportion to the overall vehicle) of a single unitary vehicle is so low, it is almost always necessary with chemically-powered rockets to use multiple stages. This reduces the inert (dead) weight that has to be carried to orbit (i.e. tankage and support structure) and also allows for optimizing thrust performance of the nozzles in upper atmosphere and vacuum where a small but significant percentage of thrust can be gained from expansion of the plume in the extended diverging portion of the nozzle. (In order to account for this you have to use the rated I[SUB]sp[/SUB] at the correct altitude, as I[SUB]sp[/SUB] at ground level will be significantly less than at vacuum; for a rough estimate you can interpolate the I[SUB]sp[/SUB] linearly with pressure if you have both numbers.) You can just chain the equation together (i.e. m[SUB]0[/SUB], m[SUB]1[/SUB], m[SUB]2[/SUB], et cetera) to get the delta-v.
Note that while delta-v times the resulting mass (the inert weight, including payload, and for multiple stages the upper vehicle) gives you the impulse (thrust times time) the velocity change will be significantly less than the delta-v due to losses from gravity and atmospheric drag. To calculate this requires a much more sophisticated model, but if you set your example to launching from an airless body like the Moon (and assume that the action time of the rocket is about the same regardless) then the gravity losses cancel out of a comparison, and instead you can directly compare the payload capacity for different mass fractions and stage configurations. You can even see the result of having a “pure” rocket that is all propellant and no inert mass other than the payload. (This is not as absurd as it sounds; some large solid motors using lightweight composite cases can approach a propellant mass fraction of 0.95, though more typical mass fractions are around 0.91-0.92.) Even without any dead weight there is a limitation based upon the propulsive efficiency of the propellants that limits the carrying capacity, and as noted by the o.p., every unit of additional propellant you carry to orbit is many units of propellant you need to carry it.
Although in theory you can scale up any particular design linearly to carry any amount of load (i.e. you could make a Saturn V that can carry 450 tonnes to translunar injection just by giving it ten times the propellant capacity and adding 20 F-1 motors to the first stage, et cetera) the reality is that there are some limiting factors to just how big you can make a rocket structure and how much thrust it is feasible to develop in a single stage. With the Soviet N-1 rocket (intended to be their launch vehicle for their Moon shot program) they attempted to gang a bunch of smaller engines together to achieve the necessary liftoff thrust. However, they failed to account for resonances between the motors and in the propellant feed system that caused the vehicle to be uncontrollable or exceeded structural margins. Making a stage too long, or with inadequate stiffness can cause it to be uncontrollable; making it too large can require a disproportionate amount of inert mass in the structure, et cetera. So for a given vehicle configuration there is, in practice, a maximum amount of payload it can feasibly be scaled to carry. For solid motors, one of the biggest limitations is the size of the solid propellant grain that can feasibly be poured and handled in transportation, and the pressure that the case material can be designed to withstand. For liquids, one large limitation is the mass flow rate you can conceivably dump into a combustion chamber and the combustion efficiency of the chamber. (Most large liquid engines use multiple combustion chambers for this reason, and development of the gigantic F-1 engine was a major risk item for the Saturn V.)
As for the specifics of the o.p., a booster to payload ratio of 180:1 seems excessive; for the Delta IV (a typical modern vehicle), the propellant mass fractions for the individual liquid stages (Common Booster Core and Upper Stage) is about 0.88, and the resulting m[SUB]0[/SUB]:m[SUB]1[/SUB] to GTO is around 60:1 (depending on configuration), so the actual propellant to payload ratio is closer to 53:1. It is still pretty onerous, although from a cost standpoint it isn’t the propellant that is the major expense; it is the vehicle itself and all of the launch facilities and support necessary to utilize the propellant.
Thanks Stranger. I was hoping that you would turn up and I appreciate your knowledge of these things and willingness to share that information.
I am set for this year, but I think I now have enough info to put together a much better simulation for next year’s class.