Starship deorbit process

Question

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?

Answer 1

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

• Dragon takes 12 minutes

• Shuttle took three-to-four minutes

• Soyuz takes 4 minutes and 21 seconds (I think that's too precise, but yell at the press office, not me)

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.

Answer 2

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)