Previously, we looked at single stage to orbit, SSTO. We found that SSTO isn’t generally practical for launching useful payloads from Earth, although it will definitely have applicability launching from other bodies, like the Moon or Mars. We looked at the numbers and verified Musk’s statement that a Starship with no payload could (theoretically) reach orbit on its own.
In this article, we’ll examine multistage rockets. First, though, let’s return briefly to SSTO, and ask a similar question, “Can the Super Heavy booster make orbit on its own?” Obviously, it would need some type of streamlined fairing, but without an upper stage or cargo, is it possible?
In the SSTO article, we discussed a concept called mass ratio, MR, which is the initial mass of the vehicle (fueled, with payload, on the pad) divided by the final mass (vehicle mass after fuel is expended, but before payload deployment—so, including the vehicle’s dry mass plus payload). For now, we’ll ignore the mass of any special fairing, and the definition of this thought experiment says there’s no payload.
Super Heavy’s dry mass is about 230 tons. With a propellant mass of about 3300 tons, that gives an initial mass of 3530 tons. Remember we’re not using significant figures in these calculations; rather, we’re taking literally some rather casual comments by Musk. We’re doing that so it’s easier to trace the provenance of the numbers, so we can refine them later.
These figures yield a mass ratio of:
MR = mo/mf = 3530/230 = 15.3
Or a mass fraction:
mass fraction = mf / mo = 230/3530 = 0.065 which also equals the deadweight fraction, s.
Our propellant fraction, z is:
z = mp / mo = 3300 / 3530 = 0.935
From our earlier work, the DeltaV needed to make low earth orbit is 8471 m/s. For Starship, we concluded the average exhaust velocity for our Raptor engines is 3433 m/s, but that was based on a mix of sea level and vacuum nozzle engines, which also means a mix of Isp. Recall that Isp = c/g, or c = Isp * g. With all sea level Raptors:
c = 330s * 9.81m/s2 = 3237 m/sec
That means the Mass Ratio needed is:
e(exp(DeltaVdesign/c)) = mo/mf = MR
2.71828(exp(8471/3237)) = 13.69 = MR (needed)
In other words, a Super Heavy first stage could easily reach orbit on its own, if it wasn’t pushing a second stage. Of course, that raises the question, “To what end?” But it also raises a different question: how is Musk, ahead of all others, suddenly able to send SSTO vehicles to orbit without any problem?
The answer lies in comments Musk has made in previous International Astronautical Congress sessions: all other things being equal, the larger the vehicle, the more efficient its design. Getting a vehicle to orbit in one stage is all about minimizing the structural fraction. Since volume scales more rapidly than area, as vehicle size increases, the amount of propellant you can carry increases faster than the structure required to hold it. Starship and Super Heavy are enormous vehicles, and that’s the trick. This also explains why the original Interplanetary Transport System/Mars Colonial Transporter was so enormous. If you can build to that scale, geometry is on your side. Important caveat: that’s a big “if.”
We could run the numbers and see how much payload a Super Heavy could put into orbit on its own, but since SpaceX doesn’t seem interested in that at the moment, let’s move on to multiple stages.
For determining the capabilities of the first stage, the initial mass of the vehicle again includes everything: fueled, with payload. For this calculation, the payload is now the entire second stage, including its fuel and payload. The final mass of the first stage calculation is the mass of the vehicle just before stage separation—so, empty first stage plus full second stage (structure, fuel, and payload).
The second stage calculation (assuming just two stages) is essentially just the SSTO calculations from before.
So, for s Super Heavy/Starship combination, let’s do the math. For the standard Super Heavy/Starship combination, a fully loaded Starship second stage is part of both the initial and final first stage mass calculation:
Starship dry mass + Starship propellant + Starship payload = Starship total mass
120 t + 1200 t + 150 t = 1470 tons
To this, we add the Super Heavy final mass, 230 tons, to arrive at the Super Heavy/Starship final mass just prior to stage separation = 1700 tons. Super Heavy/Starship’s initial mass is 1700 tons + 3300 tons propellant = 5000 tons. Then, for Super Heavy, we calculate:
DeltaVdesignfirstStage = c * ln(mo/mf) = 3237 m/sec * ln(5000 t / 1700 t) = 3492 m/sec
And for the Starship second stage:
DeltaVdesignstarship = c * ln(mo/mf) = 3433 * ln(1,470/210) = 6680 m/sec.
The vehicle is capable of a total DeltaV of 10,172 m/sec. That’s well above the 8471 m/sec needed to reach LEO, which makes intuitive sense—SpaceX has spoken extensively about the capability of Super Heavy/Starship. It’s important to note, though, that Starship needs to keep some of its propellant in order to get home. SpaceX has previously said the propellant penalty for a Falcon 9 first stage recovery is about 10-15%. Using those figures for a first cut estimate, 10% of 1200 tons propellant = 120 tons. Starship’s skydiver reentry won’t have to do a boost back burn, so 10% seems like a reasonable starting figure.
We’ll change Starship’s final orbital mass to reflect retaining 180 tons of propellant needed to get home: mf = 210 tons + 120 tons = 330 tons.
Then,
DeltaVdesignStarship = c * ln(mo/mf) = 3433 * ln(1,470/330) = 5128 m/sec.
That gives a vehicle total Delta V of 8620 m/sec. Above the needed 8471 by a small, but comfortable, margin.
Just for fun, let’s return to the idea of putting a Super Heavy into orbit, specifically asking the question, “Why would we want to do that?” Starship is designed to reenter from orbital velocity. Super Heavy is not; it’s designed to do a flip, deceleration burn, reentry burn, and landing burn, all at a fraction of orbital velocity. A Super Heavy in orbit would be trapped, never able to return. So, what’s the point?
Two words: “orbital refilling.”
Let’s look at DeltaVdesign again, but just for a Super Heavy with no second stage or payload:
DeltaVdesign = c * ln(mo/mf) = 3237 m/sec * ln(3530 t / 230 t) = 8840 m/sec.
The first observation is that this confirms our Mass Ratio calculation—8840 m/sec is certainly more than the 8471 m/sec needed to reach orbit. But put another way, it also means a Super Heavy in orbit, after refilling, is capable of generating a whopping Delta V of 8840 m/sec. Alternatively, it could provide an extra 3492 m/sec to a Starship leaving on an interplanetary mission.
But what of Musk’s comment that a Super Heavy could operate with as few as 24 engines? Why would we care? (When this was written, Super Heavy was planned with 37 engines; that's now down to 28 as Raptor proves itself capable of higher chamber pressure and thrust.)
A Raptor engine tips the scales at 1500 kg. If we use 24 engines instead of 37, that’s a savings of 13 * 1.5 = 19.5 tons. Call it 20. We probably can’t load another 20 tons of propellant, but just reducing the structure by 20 tons gives us:
DeltaVdesign = c * ln(mo/mf) = 3237 m/sec * ln(3510 t / 210 t) = 9116 m/sec.
That’s 275 m/sec Delta V gain at a cost of…not putting 13 extra engines on the rocket. Pretty good ROI. Or, more likely, those 20 tons could offset the solar panels we’d need to add to keep our Super Heavy alive during refueling and return. We also won’t want to throw away the Super Heavy after just one use, so we’ll need to retain some propellant.
Exploring this a little further, DeltaVneeded from LEO to Earth escape velocity is 3220 m/sec. That figure just gets us to Earth escape with zero additional velocity, so we’ll need additional DeltaV to go anywhere. For Super Heavy, though, 3220 m/sec is the maximum DeltaV we want from the vehicle, because we want it to go back to Earth. That’s a little less than our Super Heavy can theoretically give us (we can get 3492 m/sec pushing a Starship), but we won’t burn to exhaustion because we need to retain propellant for the return. Super Heavy will find itself on a very elliptical orbit, with perigee back at LEO. We’ll be hauling the mail at perigee, so we may want to just do a retro burn vs. skimming the atmosphere on the first pass. Regardless, a couple retro burns followed by a little aerobraking, and our Super Heavy will be back in LEO ready for the next trip.
The DeltaVneeded for trans-Mars injection is 3600 m/sec for a Hohmann transfer, so a Super Heavy push wouldn’t be needed for Starship going to Mars. On the other hand, the DeltaVneeded for a Jupiter-bound Hohmann transfer is 6300 m/sec. Looking at our calculations above, Starship is capable of this DeltaV, but regardless of how it gets into Jupiter orbit, Starship wouldn’t have enough propellant to get home. On the other hand, with a Super Heavy push, we have a maximum DeltaV of 10,172. That leaves 3872 m/sec to play with—which still isn’t nearly enough. We’ll need on the order of 5600 m/sec to get into a Hohmann transfer trajectory from Jupiter to Earth, and that’s also a really long trip—a Hohmann transfer takes 2.75 years to get from Earth to Jupiter, and the same to get back.
We’re well into the realm of speculation here, but if the video of Starship landing on Europa is to ever take place, at minimum, we’ll need another stage—something about the size of Starship, docked between Starship and Super Heavy. This stage will pretty much need to be expendable—it will be well past Earth escape velocity at burnout. But it can also be very simple—notionally, it might just carry the 3 gimballed core Raptors and a whole lot of propellant.
Of course, we’ll also need quite a bit of orbital infrastructure to support assembling and filling such an enormous vehicle—exactly the sort of infrastructure that moves from science fiction to an engineering problem with the Super Heavy/Starship combination.