Will the first Starship–Super Heavy launch to reach SECO be travelling at orbital speeds?
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Jul 1
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chance

A followup to this market.

The recent Starship–Super Heavy full stack flight test was intended to be slightly suborbital.

There are two ways a trajectory can be suborbital: either the vehicle can be going too slow, or it can be pointed in the wrong direction to complete a full orbit.

Which will apply to Starship, on the next flight test of a Starship–Super Heavy full stack that makes it as far as 2nd stage engine cutoff (SECO)?

This market will resolve YES if, at any altitude, Starship has a total energy (kinetic plus potential) greater than that of a 160km altitude circular orbit (close to the lowest possible circular orbit).

That is, this market resolves YES if, for any altitude h, Starship's speed v satisfies:

where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.

For example:

h = 150 km, v > 7824 m/s

h = 175 km, v > 7794 m/s

h = 200 km, v > 7765 m/s

h = 225 km, v > 7735 m/s

Note that the speed indicator on SpaceX's live stream is in a frame of reference rotating with the Earth, but the above speeds refer to the speed of Starship in the equivalent non-rotating frame.

Thus depending on how close Starship is to the above thresholds, resolution may require estimating Starship's vertical velocity from the altitude indicator, and perhaps its heading from maps of the planned trajectory as well, in order to add Earth's rotation speed only to the eastward component of Starship's velocity in the rotating frame, to convert to the nonrotating frame. If, due to uncertainty, this calculation is too close to give a clear result, and barring clarifying official statements or other data that could help resolve it, the market will resolve N/A.

This question is equivalent to asking whether Starship will at any point be on a trajectory with a semimajor axis greater than 160km (plus Earth's radius). So official numbers on the apogee and perigee would also be sufficient to resolve it (unofficial estimates would have to come from the velocity data anyway).

Market stays open until a Starship–Super Heavy launch reaches SECO, or resolves N/A if the development of Starship/Super Heavy as we know it ceases.

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Not that anyone doubted it, but confirmation from SpaceX that IFT-2 did not reach planned SECO:

https://www.spacex.com/updates

A leak in the aft section of the spacecraft that developed when the liquid oxygen vent was initiated resulted in a combustion event and subsequent fires that led to a loss of communication between the spacecraft’s flight computers. This resulted in a commanded shut down of all six engines prior to completion of the ascent burn, followed by the Autonomous Flight Safety System detecting a mission rule violation and activating the flight termination system, leading to vehicle breakup.

bought Ṁ300 of YES

Edit ignore this comment lol, I just saw Chris' initial comment about the possibility of an apogee of 250 km and perigee of 50 km. I didn't think about the possibility of such low flight, it makes all the fancy math I did below totally pointless.

For the last few days I've had this nagging thought in the back of my head: Is it even possible to go more than halfway around the Earth without reaching the total energy required for orbit? Intuitively it seems to me the lowest energy trajectory that would take you halfway around the Earth would be a circular one, which would be orbital.

I did some back-of-the-envelope math this afternoon that I think confirms this, but I may well have made errors, so don't take this as the word of god. If you have an ellipse with the x axis as the semi-major axis and the center of the Earth at one of the foci, then in order to just barely make it halfway around the Earth, the points on the ellipse directly above and below the focus must be the radius of the Earth (or the radius plus 160 km to be out of the atmosphere) away from that focus. The points where the ellipse would intersect the Earth would be { (a^2-b^2)^1/2, R }, if you plug those into the formula for an ellipse and solve for the semi-major axis, you get a=b^2/R. All elliptical orbits of the same semi-major axis have the same energy, so minimizing the semi-major axis subject to the constraint above will get us to the minimum energy trajectory. Since I'm assuming the larger axis is the x axis, a>=b, so the lowest a can be while satisfying a=b^2/R is b, making a circular orbit the lowest energy trajectory to get halfway around. I think this means that if you go halfway around the Earth or more, you need to achieve the total energy that is required for orbit.

bought Ṁ100 of YES

According to this Reddit comment summarizing yesterday's talk, they plan to test a deorbit burn from the header tank, I think that makes it quite likely that they will be at orbital speeds before the deorbit burn: https://www.reddit.com/r/spacex/comments/19251pv/starship_development_thread_53/khlal8m/

predicts YES

@TimDuffy You can be suborbital and still do a flip maneuver and try to reignite.

Were I doing their mission planning, I would not attempt full orbit for a while. I would test that I can get to orbital velocity, I would test reigniting engines, but I would assume that at some point I’m going to lose comms or something and not be able to detonate/deorbit, and you really don’t want something that large up there for a long time.

sold Ṁ16 of YES

@ChuckLauerVose Yeah I think you're right that Starship's trajectory will be suborbital even before the deorbit burn after reading more about it last night an this morning. This snippet from a document submitted to the FCC suggests that they plan to touch down in the Indian Ocean, west of the planned touchdown site for IFT-2. So I think that before the deorbit burn, Starship might be on a suborbital trajectory heading towards the Pacific like they had planned for IFT-2. The document also suggests a lower apogee than I anticipated, so I now think the probability of reaching orbital speeds is quite low, maybe a bit over 10%.

predicts YES

@TimDuffy That’s helpful. thank you!

bought Ṁ500 YES

If they do a prop transfer, they're very likely going for full orbit. (This will be an internal prop transfer from the header tanks to the main tanks)

This also matches the FCC filing that says they'll do a powered reentry into the Indian Ocean for IFT3

predicts NO

@Mqrius Do you think there's a chance this might end up being IFT-4, and they'll do another one with the same profile as IFT-1 and 2 first? I forget where I read it or who, but someone Spacex affiliated (I think) said some words implying a similar trajectory on the next flight (something like "next time, all the way to Hawaii!"). Basically, what tells us that these FCC filings pertain to a third flight specifically, as opposed to a fourth?

predicts YES

@chrisjbillington I guess the fact that it's the only FCC filing they have at the moment? Not entirely sure though

predicts NO

@Mqrius I'm not sure they need new FCC filings for every flight, if things are the same.

Plenty of people are talking about this as IFT-3 though, so I would guess it is.

predicts NO

@Mqrius If this is the propellant transfer demonstration for NASA, they have to transfer 10 tons of LOX (https://spaceflightnow.com/2020/10/16/nasa-selects-companies-to-demonstrate-in-space-refueling-and-propellant-depot-tech/), which is probably the size of the header tank. This means that if they want to test propellant transfer into the header tank, they have to launch with an initially empty header tank.

But they will likely need full header tanks for the deorbit burn, and they don't want to rely on the untested propellant transfer being successful for a safe deorbit. So it makes sense that they do this propellant transfer demonstration on a suborbital flight that doesn't need a deorbit burn.

In fact I assume that this propellant transfer test was already planned for IFT-2, SpaceX just didn't tell anybody. This would also explain why no flip maneuver and powered water landing was planned at the end of IFT-2.

predicts YES

@dp9000 they could launch with full header tanks and then transfer from the header tank to the main tanks. Not sure how big the header tanks are, but if that doesn't leave enough they could transfer some back.

For prop transfer to work they need to do ullage for propellant settling. If they can do that, then they can also do ullage for the deorbit burn. So either the transfer doesn't work and they can use the full header tanks for deorbit, or it does work and they can refill the header tanks or use ullage.

In theory anyway. Your concerns are somewhat legit, but I still think they're going for full orbit.

bought Ṁ50 of NO

FWIW Jonathan McDowell estimates IFT-2 Starship's trajectory before telemetry was lost had a perigee of -1740km, and an apogee 150km. It didn't reach SECO so not relevant to this market.

sold Ṁ121 of NO

Currently filed FCC plan for IFT-3 is planning to go for orbital speeds and powered reentry. It's still early though and they might change their plans based on the outcome of today's IFT-2, but this market has a chance of Yes now!

bought Ṁ95 YES
predicts NO

@Mqrius Ha, wow. Not what I was going for with this market, but the resolution criteria are unambiguous about this, so it is what it is! I'm going to bet a little NO based on the possibility they won't want to shoot for orbit without a bit more success on the slightly-suborbital flights..

bought Ṁ50 NO from 52% to 44%
predicts YES

@chrisjbillington I could see them going for orbit next tbh! They've always taken 2 steps ahead where you'd expect them to take only 1. Remember the first successful Falcon landing? It was on land after they failed to land in the ocean. It was surprising then too, since you want to keep your land safe, in theory.

predicts YES

@chrisjbillington Probably time to extend the deadline btw, unless we get a SpaceX christmas present.

predicts NO

Did this flight reach SECO?

predicts YES

Musk tweeted back in August

I think we have ~50% probability of reaching orbital velocity, however even getting to stage separation would be a win.

https://twitter.com/elonmusk/status/1687617123647111168

predicts NO

@jack Orbital velocity as defined by Musk isn't the same as the one defined by this market though.

bought Ṁ10 of YES

@Mqrius I know, just giving another data point.I haven't been able to find the details of the planned trajectory and what the planned velocity is, anyone know?

predicts NO

@jack We know it will reenter passively near Hawaii, so that limits the possible trajectories a bit. I don't know details for the upcoming flight but the previous one was aiming for a 250km apogee and 50km perigee according to wikipedia [source needed tho], which wouldn't be enough for this market to resolve Yes.

bought Ṁ20 of YES

Great question. I couldn't find a market on whether it reaches SECO, so made one:

predicts NO

I've written some code for analysing the livestream telemetry to extract orbital parameters:

https://github.com/chrisjbillington/starship_telemetry

This is the kind of thing I'll be doing to resolve the market in the absence of official statements that would be sufficient to resolve.

It is still an unproven assumption that the telemetry speed data is in the rotating frame of the earth, but we will be able to see whether telemetry is consistent with this, for example by verifying that under this assumption the computed orbital parameters do not change after SECO (if they do we've done something wrong!).