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EDIT: Here are some BF numbers from the video:

total vehicle mass  :  4,400 ton  (2nd stage propellant fill fraction unclear)
booster thrust:            52,700 kN
max ascent payload:           150 ton  (to LEO)

2nd stage dry mass:            85 ton
max propellant mass:        1,100 ton (240/860 CH4/LOX)

Merlin SL engine ISP  (x2):   330 sec (SL)
                              356 sec (vac)
Merlin vac engine ISP (x4):   375 sec

On-orbit refueling of the SpaceX BFR upper stage is necessary for trips to the Moon and Mars. In the presentation at the "International Astronautical Congress (IAC) in Adelaide, Australia, SpaceX CEO and Lead Designer Elon Musk (provided) an update to his 2016 presentation regarding the long-term technical challenges that need to be solved to support the creation of a permanent, self-sustaining human presence on Mars."

In the YouTube video after 27:17 the value of orbital refueling is discussed, along with a calculation of payload mass versus delta-v for what looks like different numbers of tankers and refillings.

At a later time 32:06 the task of bringing cargo to the Moon for a base station is described, and the refilling(s) is(are) shown to happen in elliptical Earth orbit.

If five refillings are used as shown earlier, how much cargo can be brought to the moon, roughly?

Could five refillings be used to get to Mars? Would the last one or two actually have to take place beyond Earth orbit to be useful? note: unlike the proposed Moon mission, the proposed Mars mission includes a refueling at Mars with propellent synthesized from Martial water and carbon dioxide.

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First of all, let's look at what the delta V is for a fully loaded BFR upper stage in orbit. Given the Delta V tables shown, and assuming 150 tons, a delta V of 6.2 km/s is shown. 150 tons is assumed because that is the mass to LEO provided. As I understand it, these numbers come from a tanker orbiting Earth approaching the payload, and refueling it.

It seems to me that the Moon mission would involve at least some refilling in LEO, and some in an elliptical orbit. I'm not sure of the number of refillings, but I would suspect there is more than the standard (5?) that is required for a deep space mission, say to Mars. The Delta V to land on the Moon from LEO is 5.9 km/s or so. Thus, a BFR could get from LEO to the Moon's surface, but couldn't return without refueling. I can't find the exact delta V required to get to Earth from the Moon's surface, but it is around 2.74 km/s. Therefore, the elliptical orbit must use around 2.5 km/s of delta V, and be fully loaded, in order to get to the Moon without refueling. That is entirely possible, but it will be a significantly elliptical orbit, something on the order of to the GEO belt, so far as I can tell.

I assume that what will happen is that the tankers will fuel the lunar bound BFR in LEO, and then top off the rocket when it is in the correct transfer orbit. That at least is how I understood it. I suspect it will take about as many missions to refuel the BFR in the elliptical orbit as it will take to refuel it in LEO, although there will be less fuel per load, the rest is lost to match the orbit of the payload rocket.

As for Mars, the Delta V required to get to Mars is about 4.3 km/s. I'm not sure what it takes to land the BFR on Mars, but it seems perfectly reasonable that a fully loaded BFR can take 150 tons to Mars, from LEO.

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  • $\begingroup$ "Would the last one or two actually have to take place beyond Earth orbit to be useful?" $\endgroup$ – uhoh Oct 6 '17 at 12:39
  • $\begingroup$ Made much clearer. The answer was quite poor as it was written, it is much better now. $\endgroup$ – PearsonArtPhoto Oct 6 '17 at 12:49
  • $\begingroup$ If I watch the 150 ton point, for $n_{refill}$ of 0, 1, 2, 5 I see a $\Delta v$ of about 0, 2, 3, 6 km/s. Each one adds less presumably because the tankers must burns more and more of their cargo to match the higher and higher orbits of the target ship. I'm having trouble following your answer though, have you demonstrated that all tankers can remain in Elliptical orbit for a moon landing and return, and for a one-way to Mars, or must some tankers leave Earth orbit? Can you show the arithmetic/logic more explicitly, it's hard for me to follow a prose-only answer. $\endgroup$ – uhoh Oct 6 '17 at 13:04
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    $\begingroup$ When the target ship uses it's refilled fuel, it burns it. Then it is empty, and moving in an orbit with a faster velocity at periapsis, and a higher altitude at apoapsis. Each tanker would have to match each increasingly elevated orbit in order to dock and transfer fuel. So each tanker would probably burn increasing amounts of it's cargo in order to match the higher and higher orbits of the target ship. Is this not so? $\endgroup$ – uhoh Oct 6 '17 at 13:45
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    $\begingroup$ Oh, I understand what you are saying now. So do you think that the animation showing the full/empty → emtpy/full transfer is not so accurate, or perhaps the full tanker shown has already accumulated fuel from multiple, maybe up to five previous tankers perhaps? I think I'm going to have to wait for an answer with a more quantitative analysis; this is not an easy question to answer. $\endgroup$ – uhoh Oct 6 '17 at 14:52
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Of course everything about planned SpaceX Starship configurations and missions is still tentative and subject to change. However, there are a few points worth establishing:

  • Despite the animation, one tanker on a refuelling mission can't come close to fully refuelling one starship. The Starship burns about 1100 tons of fuel to put 150 tons of payload (plus the ship) into LEO.
  • Counterintuitively, a moon surface mission and return is actually harder in terms of delta-V than a Mars mission, because you can't aerobrake and you can't hope (in the near term anyway) to manufacture methane/LOX fuel on the Moon (not enough carbon or hydrogen accessible). It also doesn't make much sense to land your whole rocket, with all the fuel needed to return to Earth, on the Moon, rather than leaving most of it in orbit.

I don't have figures for the new designs, but the basic capabilities of the BFS are described on wikipedia. The dry mass of the BFS is 85 tons. The propellant mass for a full load is 1100 tons and the payload to LEO of the BFR is quoted on that page as 100 tons plus. I think I've heard 140 elsewhere, but I don't recall where and that might just be for the tanker variant.

So you can deliver your BFS to LEO with 100 tons of whatever you want on board. That could be unused fuel (in which case you could use the tanker variant, which is a bit lighter in itself (no life support, windows, ...) so can carry more) but it's more likely to be your mission payload. So you probably arrive in LEO With no fuel. Each refuelling flight to LEO carries 100 to 150 tons of propellant, so it takes somewhere between seven and eleven of them to fill your tanks completely.

The presentation shows a fully refuelled (although this isn't clear, it might be as refuelled as it can get in 5 flights) BFS to have a delta-V of between about 6 and 9 km/s depending on payload. A moon surface and return mission is about 9 km/s, a Mars surface mission (planning to refuel there before return) is about 4.8 km/s (more if you want to shorten the transfer). So fully refuelling in LEO is enough to get you to Mars, but not enough to get any useful payload to the Moon and back.

To manage the lunar mission, the concept is to add a couple of refuelling transfers in a high elliptical Earth orbit (GTO or even a bit higher). This will deliver less fuel per tanker launch (since more of it will be burned getting there) but reduces the delta-V from last refuelling to end of mission by about 2.5 km/s.

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  • $\begingroup$ Are you able to be more quantitative about the remaining fuel going to LEO at least? Half? 10%? Thanks! $\endgroup$ – uhoh Jan 9 at 17:45

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