The booster was falling at 4500 Km/h 30 seconds before the catch with 2-3% fuel left. How is that amount of fuel remotely enough to stop the downward momentum?
First off, the booster was going about 1250 km/h when it started its landing burn, it relied purely on drag to get it slowed down to that speed.
Going by the telemetry of the seconds before the landing burn and noting the speed vs time, it seems drag was around 40 m/s^2 when it was going at around 3000 km/h. Since drag depends on velocity squared though, it had reduced to just above 10 m/s^2 just before the engines lit at 1250 km/h, and so would quickly become negligible once the engines lit.
Going by Wikipedia, the Super Heavy[1] has 3400000 kg of fuel at launch, so 3% of that is about 102000 kg. For the landing burn, it used 13 Raptor v3 engines[2] to scrub speed. Each Raptor flows about 650 kg/s max, so 3% fuel is enough for about 12 seconds for the 13 engines.
The empty mass of the Super Heavy is about 275000 kg, so about 377000 kg before the landing burn with 3% fuel.
Using the sea-level vs vacuum performance of the Raptor v2 engines, one can estimate that each Raptor v3 produces about 2.45 NM of force at sea-level. So 13 of them would produce about 31.85 MN of force.
Using Newton's second law, F=ma, this gives an initial deceleration of about 84 m/s^2 and about 104 m/s^2 when empty. If we do a rough spreadsheet integration, we get that a burn of roughly 4 seconds is needed to scrub the speed assuming no other forces.
Now, comparing this with reality, the full 13 engines were lit for a little over 5 seconds.
In my simplified calculations I was assuming full throttle the whole way, which obviously isn't realistic, and I also assumed 3% fuel. So over all I think that's a pretty decent estimation.
Methane has 28% more energy per kg than kerosene and also produces slightly less CO2 (2.75kg CO2/kg burned vs 3.00 for kerosene) when burned [1]. SpaceX uses a 78:22 LOX to CH4 ratio, so for 34M kg of fuel burned, 20.57M kg of CO2 are produced (34×0.22×2.75).
Somewhat tangential, but as far as rocket fuels go energy per volume is also an important metric to consider. It's one of the (several) reasons hydrogen isn't quite as good for rockets in practice as it is in theory - while hydrogen has tons of energy per unit mass (120-142 MJ/kg for hydrogen vs. 50-55 MJ/kg for methane and 43.1-46.2 MJ/kg), it has a far lower density (70.85 kg/m3 vs. 422.8 kg/m3 for liquid methane and 820 kg/m3 for kerosene). As a result, you need quite a bit more tankage for a given amount of energy from hydrogen compared to what methane/kerosene requires.
IIRC there's a tradeoff between efficiency and thrust as well. Heavier fuels aren't quite as energy-efficient, but it's easier for them to develop a lot of thrust, which is important for the initial stages of launch. If I'm remembering events described in Ignition! correctly this led to "thrust density" being something that was optimized for - to the point that there were experiments with mixing mercury into the fuel!
That's excellent work, only nit is I think these were not the v3 raptors. I think they're just now starting production of those, so they still have a bunch of earlier ones they need to use up in these early test flights. This actually makes your calculation more accurate, as earlier versions will need a slightly longer burn time.
Ah, good catch, I must have misread the Wikipedia page. I misread that they were included in the rocket in August. Reading the Raptor page more closely I also see that the 2.75 MN of the Raptor V3 are supposed to be at sea-level.
Anyway, plugging in the Raptor V2 thrust numbers the approximation increases to 4.25 seconds. This is in line with the thrust I used for the V3 being ~8% higher than the V2 thrust figures.
Just a programmer, though been interested in physics since I was a teen and did take a bachelor degree in simulation (mainly physics).
Long ago though so rusty, $dayjob doesn't involve any advanced math at all.
edit: To expand, the "rough spreadsheet integration" was just the Euler method[1] assuming a constant acceleration. So
v(t+dt) = v(t) + a * dt
The acceleration comes from F=ma as mentioned, where F is the force of the engines (Newtons), m is the mass of the rocket (kg) and a is the acceleration (m/s^2). Solving for a we get a = F/m and we get
v(t+dt) = v(t) + F/m(t) * dt
To make things easy I assumed the weight of the rocket was constant at each timestep, but if we take dt to be small enough it's a decent enough approximation. For each timestep I also updated the mass using the estimated mass flow:
m(t+dt) = m(t) - 650 * dt
I started with m(0) = 377000 kg, v(0) = 1250 km/h = 347 m/s, and a constant -31850000 N force from the engines.
Using dt = 0.1 seconds, I got almost exactly 4 seconds until the velocity reached zero.
Newton's laws of mechanics are high-school physics IIRC; my son studied them at 8th grade or so. They are really simple; an evening with Wikipedia or 3blue1brown or whatever floats your boat will let you get sufficient understanding, provided you're also comfortable with high-school math.
The atmosphere does its bit to slow down the booster as well. It’d be interesting to see a plot of the power output over time on reentry but I’ve always assumed the motors aren’t doing a lot of work other than keeping the booster stable, until the very end.
Yes, for most of the booster return it’s ‘gliding’ with the rocket engines completely shut down.
They ignite a subset of engines just a few seconds before landing for the final slowdown and maneuvering.
Edit: here is a video from further away that shows the rocket gliding in under control of the grid fins before the engines light and execute the final landing maneuver:
> Yes, for most of the booster return it’s ‘gliding’ with the rocket engines completely shut down.
Watching the video, it looked like the bottom of the rocket was glowing hot, but the engines were cool. I imagine that means they were probably running some amount of methane through the engine bells to cool them.
Super heavy probably makes certain aspects (unrelated to the tower catching the booster) of landing much easier, by virtue of its greater mass. Timing with the Falcon 9 was always essential, because the minimum thrust (with one motor, throttled all the way back) of one of the engines was enough that if you left the motor running, the booster would start to rise. Time the beginning of the final burn incorrectly and you had a real problem, with the booster either crashing or rising at the end and... then crashing, but with Super Heavy, it might simply be a matter of having to compute a slightly different sequence for shutting off the engines.
Super Heavy is easier to control on landing because it's using 3 engines so has better directional control, and it can throttle those engines down into a sustained hover, which is what it does before being "caught" by the arms.
Surprising to see this work first time though - I don't recall them doing any hover and lateral movement tests, but I assume they must have done.
What's also wild is that the booster isn't being caught/supported by those giant grid fins, but rather by small lifting pins just below them, and seems to only have two of these (one on either side), so it also has to get it's rotational position right so those pins engage with (are supported by) the arms.
Yeah, but even without fuel the booster weighs 200 tons. It's pretty wild to have 200 tons of steel incoming at supersonic speed, then nailing a gentle pinpoint landing like that !
Yeah if the engines didn’t light and it hit the launch facility that whole place would be a crater. The whole thing was shocking to me but how fast it was coming down was extra shocking haha. I wonder what deceleration forces were at play because when the engines lit the booster slowed down very fast.
It weighs about 10% of what it did at liftoff, but half of the engines fire to slow it down.
Also I don't think the telemetry on the feed is that accurate, so with all of the atmospheric braking, it was probably going a bit slower than the 1200km/h at engine reignition.
Note that the energy of 3% of the propellant (~100 GJ) could theoretically get the empty booster (100,000 kg) to a little over 5000 km/h if properly applied.
