Will Starship exceed 28000 km/h (=LEO orbital speed) at an altitude of 200km or higher during the "orbital" flight test?
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resolved Apr 20
Resolved
N/A

The upcoming Starship–Superheavy test flight is planned to be slightly suborbital.

There are two ways to achieve a suborbital trajectory, you can either be going too slow, or you can be pointed in the wrong direction.

Which one will apply to Starship's trajectory?

Orbital speed for a circular orbit at an altitude of 200 km is approximately 28000 km/h. If Starship exceeds this speed at this altitude or higher, then it will have been travelling at orbital speed.

This market resolves YES on credible reports of Starship's trajectory being one that exceeded a speed of 28000 km/h at an altitude of at least 200km. Ideally this data will come from the live stream, but I'll get it from any credible source if needed. If the exact data isn't available, but can be calculated from other available data to within reasonable uncertainty, then I will resolve the market on that basis. If actual data isn't available but credible reports strongly imply "orbital speeds" were or were not reached, I'll resolve on that basis. If it's not possible to figure out with reasonable confidence, I'll resolve N/A.

Scrubbed launches result in the market remaining open.

A launch with Starship reaching an apogee lower than 200km resolves NO.

Destruction of the vehicle on the pad or before Starship reaches an altitude of 200km resolves N/A

If it turns out that these specific criteria were foolishly chosen, I reserve the right to resolve based on the spirit of the question, which is supposed to be about whether the suborbital trajectory was achieved by keeping speed below orbital speed (resolves NO), or by high eccentricity so as to intersect the atmosphere (resolves YES). In this case I will close the market and people can discuss - if there are strong objections to how I think it should resolve and reasonable consensus can't be reached, I'll resolve N/A.

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Followup market here, with more generous threshold of only having to reach orbital energy for a 160km altitude circular orbit:

Exploded before reaching 200km, resolves N/A

Here's how I'm thinking I'll proceed. Assuming it looks the same as on other SpaceX launches and there is no good evidence to the contrary, I'm going to treat the speed shown on the livestream as if it is in the rotating frame of the launch site.

That means one should add 1,501 km/h to the eastward component of Starship's velocity to calculate its velocity in the non-rotating frame of the Earth.

But we won't have an accurate full velocity vector, we'll just have speed. So what I'll first do is add 1,501 km/h to the displayed speed to get an upper bound for the speed in the nonrotating frame. If that's 28,000 km/h or lower, resolves NO.

If it's higher, I'll estimate the vertical component of velocity from the altitude display, and get a closer upper bound for the speed in the nonrotating frame by adding 1,501 km/h only to the horizontal component of the velocity. Then if that's less than 28,000 km/h to good confidence (the calculation will have some uncertainty), resolves NO.

Otherwise, I'll estimate the heading of starship from maps of its planned trajectory and whatever other info I can find, and do the correct calculation of adding 1,501 km/h only to the eastward component of Starship's velocity in the rotating frame. Then the question resolves on whether that's higher or lower than 28,000 km/h to good confidence. If neither, question stays open pending better data or official statements, eventually resolving N/A if none is forthcoming.

If telemetry cuts out after MECO but before reaching an altitude of 200km, I will calculate what speed would be required at the altitude at which telemetry was last available, in order to be on a trajectory with the same energy as one with speed 28,000 km/h at 200km altitude, and resolve based on whether that speed was reached at that time.

By the way, if launch happens tonight then it may be a late night for me here in Australia, so if brainpower is required to figure out how this resolves, I'm unlikely to get to it today. So don't expect quick resolution unless the result is obvious or someone else convincingly calculates whatever is needed.

The FCC filing says Starship is intended to reach an apogee of 250km. Even for very high perigees (100km) that puts the speed at 200km altitude below 28000km/h.

@Mqrius For what it's worth, if you consider the lowest possible orbit to be 160km, then a 250km×70km "orbit" would have the same energy as a 160km circular orbit. So if apogee is 250km, perigee would need to be 70km or greater in order for someone to argue that the test flight exceeded orbital energy.

Regardless, I'll resolve based on exceeding the energy requirement for a 200km circular orbit, which is sounding unlikely.

predicted NO

Yeah, doesn't look like they'll go for that, even if the difference is quite small practically.

@Mqrius I heard from someone who heard from someone that at engine shutdown the target orbit is one with perigee -20km.

This simulation doesn't get above 28000km/h at an altitude above 200km, though it does get that speed at lower altitudes.

@Mqrius Yep, that sim falls short of orbital speed for a 200km orbit, going faster at lower altitude isn't sufficient. The sim does have faster than orbital speed at its current altitude for all altitudes up to 172km, though.

Apparently the lowest possible orbit is about 160km, so the sim does have enough energy to put something in orbit, just not at 200km.

This is kind of a weird way of going about it, but I guess it's a valid question :D

I'm expecting they'll go for an orbit of 50x350km, more or less. If that's the case, then at 200km altitude we can calculate the velocity with the vis-viva equation, v^2 = GM(2/r - 1/a), with a being the semi-major axis.

import math

# Define constants

G = 6.6743e-11 # Gravitational constant

M = 5.9722e24 # Mass of Earth

R = 6371e3 # Radius of Earth

# Define orbit parameters

periapsis = 50e3 # Periapsis altitude

apoapsis = 350e3 # Apoapsis altitude
a = (periapsis + apoapsis + 2*R) / 2 # Semi-major axis

# Calculate speed at 200km altitude

r = 200e3 + R # Convert altitude to distance from center of Earth

v = math.sqrt(G*M*(2/r - 1/a)) # Speed

print(f"Speed at {(r-R)/1000} km altitude: {v*3.6:.2f} km/h")

> Speed at 200.0 km altitude: 28038.63 km/h

So in that case it would resolve Yes.

We can also calculate what the apoapsis should be, given a certain periapsis and a speed at 200km altitude.

import math

# Define constants

G = 6.6743e-11 # Gravitational constant

M = 5.9722e24 # Mass of Earth in kg

R = 6371e3 # Radius of Earth in m

periapsis = 50e3 # Periapsis altitude

vkmh = 28000 # Speed at 200km altitude, in km/h

v = vkmh/3.6 # Speed in m/s

# Calculate distance from center of Earth at 200km altitude

r = R + 200e3

# Solve for apoapsis distance using vis-viva equation

# v**2 = G*M*(2/r - 1/axis)

# v**2/(G*M) = (2/r) - (1/axis)

# 1/axis = (2/r) - v**2/(G*M)

axis = 1/((2/r) - v**2/(G*M))

apoapsis_altitude = 2*axis - 2*R - periapsis

print(f"Apoapsis altitude assuming {periapsis/1000:.0f} km periapsis: {apoapsis_altitude/1000:.0f} km")

> Apoapsis altitude assuming 50 km periapsis: 314 km
> Apoapsis altitude assuming 20 km periapsis: 344 km

As for the rotation of Earth, if they launch from the equator, so that would add 460km/h to the orbital speed at most. It will be a bit less since they are not launching from the equator. We can figure that bit out later if necessary, it gets a bit messy. Ideally they just tell us the orbit it was in, with apogee and perigee. Or maybe Jonathan McDowell will post it.

predicted YES

Okay the spoiler tags didn't collapse, now it's just annoying 😂 my bad

predicted YES

Also equator ground speed is 460 m/s, not km/h. In km/h it's 1656

@Mqrius ha, your 50x350km orbit has a semimajor axis of exactly 200km, so a circular orbit with the same energy would have an altitude of 200km. If you are correct about this being the target "orbit" then it'll resolve YES only because I rounded the speed down by 39 km/h 😬

It occurs to me that I have no idea what the "speed" display on SpaceX livestreams (e.g. of Falcon 9 missions) actually shows. Is it groundspeed? It clearly isn't speed with respect to the nonrotating frame of the Earth, because if it were it would be over 1000 km/h whilst still on the launchpad.

The figure 28000 km/h refers to the speed in the nonrotating frame of the Earth, so any speed figure displayed on the livestream may not be sufficient to resolve.

It seems like the speed indicator on the Falcon 9 livestreams is either a) speed in the rotating frame of the launch site, or perhaps something a bit ad hoc like b) speed in the nonrotating frame of the earth, minus the earth's rotation speed at the launch site (i.e. scalar subtraction of speeds, not normally a meaningful thing to calculate).

The Earth's rotation speed at the launch site is 1500km/h, so in case b) the true speed will be 1500 km/h faster than what is displayed on the live stream, and in case a) it's a bit more complex but can be figured out if we know the vehicle's vertical velocity (which we will if they show an altitude indicator) and approximate heading.