It does not seem to me that SpaceX will use thrusters similar to Draco to deorbit Starship, and I think firing a Raptor would be too powerful. All I can imagine is that it will use ullage gases through the existing vents to slow the spacecraft's orbital velocity. Am I correct?

  • $\begingroup$ FYI I have nearly reversed my previous answer. I'm also one of the downvoters, but have to wait a bit to retract that. I no longer think there's an obvious answer. $\endgroup$
    – Erin Anne
    Feb 26 at 23:31

2 Answers 2


NOTE: after doing the below sanity check I have softened my position a lot. I don't think it's completely obvious from what's publicly available how the deorbit will be conducted. A lot of the prior version of this answer may not have been sound reasoning.

With the caveats that Starship CONOPS probably aren't entirely defined, and the publicly-available information on Starship is basically limited to the animations released by SpaceX and data released haphazardly through tweets and off-the-cuff remarks: you might be correct. It's hard to say.

For a sanity check, here's a back-of-the-envelope calculation based on the numbers currently available from Wikipedia on the Raptor engine and on Starship. For simplicity, let's say we're running one Raptor 2 at minimum thrust (20% of 2.3 MN), and that the Starship is at its dry mass (~100,000 kg, it should be heavier, but that's the worst case for shortening the burn) and the burn mass change is negligible (it isn't, but including this effect would extend the burn rather than shorten it). How long of a burn is a 90m/s deorbit-from-LEO?

$\frac{2.3 MN * 20\%}{100,000 kg} = 4.6 m/s^2 \approx 0.5 G$

$\frac{90 m/s}{4.6 m/s^2} \approx 20 s$

There are a LOT of assumptions we could refine from above. For example, if it was landing with 10% propellant (which is probably not a realistic figure for landing propellant, I just chose 10%), we'd add another 340,000 kg of mass, extending the burn to more like 86 seconds. For burns longer than a few seconds it's also much more responsible to use the Tsiolkovsky equation to obtain the delta-V (and then compute duration from mass flow rate, which are more reasonably assumed constant) rather than naively assuming that the mass and acceleration are constant. I've also neglected to think about any rotation consequences from the torque of a single-engine firing. I just wanted a reasonable-ish guess for the

So what's a reasonable length of deorbit burn? I assume now that your guess that the Raptor is too powerful might come from prior discussions of the Dracos being too powerful for deorbiting Dragon; if you'd already run some numbers to come to that conclusion, I wish you had shared them! A quick review indicates that

These are all roughly an order of magnitude off from the burn duration we've SWAG'd for Starship. A Raptor deorbit would be a lot faster burn than other vehicles'. Maybe that means the ullage vents would be used instead.

That said, if you'll excuse my opinion as an aerospace engineer who never worked on propulsion systems professionally, Raptors are expected to be precisely-controllable enough to vertical-land Starship and SuperHeavy. When your worst-case-too-quick burn is tens of seconds, and you'd certainly expect to be able to control the engine to within half-a-second, surely the sensible thing is to burn most of that delta-V efficiently with the Raptors and then correct the residuals with the ullage vents. The deorbit burn isn't the last time the engines and fuel are going to be needed.

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    $\begingroup$ Note: there will be no RCS thrusters in the traditional sense. The old sub-orbital prototypes had cold-gas thrusters (possibly just Falcon 9 ones) as an interim solution and SpaceX talked about development of hot-gas thrusters. But the hot-gas thrusters have been deleted from the design, both booster and ship will use ullage thrusters for attitude / reaction control. You generally have to vent (some of) the ullage overboard anyway, so why not make use of it? See youtu.be/t705r8ICkRw?t=2591 $\endgroup$ Feb 25 at 21:09
  • $\begingroup$ @JörgWMittag oh right that's the sequence where Everyday Astronaut prompts Elon (who I think also said he was on painkillers during that trip) to think about why Starship doesn't do the same thing as Superheavy. Did that really result in a design change? I assume there's an Elon tweet out there about it somewhere. $\endgroup$
    – Erin Anne
    Feb 25 at 21:17
  • $\begingroup$ There was one single mockup of a hot-gas thruster spotted at Starbase, presumably for fit-checks. That was even before that video, IIRC. Since then, nothing that even remotely relates to hot-gas thrusters has been spotted at Starbase, and the orifices in the hull that had been speculated to be for mounting the thrusters have vanished from later ships. I don't think S24 has any thrusters, neither hot-gas nor cold-gas, but it has the same "cowbells" that B7 has. photos.cosmicperspective.com/Starship/i-z4qps9L/A $\endgroup$ Feb 25 at 21:38
  • $\begingroup$ @JörgWMittag ok, thanks very much for the correction $\endgroup$
    – Erin Anne
    Feb 25 at 21:39
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    $\begingroup$ Shuttle deorbit burn was ~90 m/s. Just a perigee lowering orbit adjustment really. $\endgroup$ Feb 26 at 13:08

StarShip will need to drastically reduce vehicle speed from ~ mach 25, in order to re enter the atmosphere, this will require firing of one or more Raptor engines. (RCS, Ullage & Draco engines aren't strong enough)

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    $\begingroup$ You do not need to drastically reduce speed to re-enter, for spacecraft that can take an atmospheric re-entry, all that's needed its to reduce speed enough to drop the periapsis of the orbit to the desired aerobraking altitude. For instance, from the ISS' orbit, it takes less than 100 m/s of delta-v to re-enter for landing. $\endgroup$
    – notovny
    Mar 23 at 12:09
  • $\begingroup$ Thanks, but is it not Also true that by reducing speed before re entry the heating will be reduced and time required for that re entry, thanks $\endgroup$ Mar 23 at 12:42
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    $\begingroup$ The heating comes from the atmosphere doing your deceleration work for you, so you don't have to spend large fractions of your mass budget on deceleration fuel, and can use that for payload instead. It's much more efficient to use the atmosphere and withstand a few minutes of re-entry heating, which is why every spacecraft that returns to Earth does it that way. $\endgroup$
    – notovny
    Mar 25 at 16:26

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