Need some help figuring out how much propulsion would be necessary
September 16, 2022 7:39 PM Subscribe
imagine you have an object that is 18,000,000 metric tons. I want to know how much propulsion would be necessary to get it to escape earth's gravitational pull. I'm not sure how difficult this is to calculate, but my guess is that for someone who knows, there are probably simplified equations?
assume it's just one gigantic thruster at the bottom firing at a constant acceleration...though honestly, you can assume whatever you want to simplify things! ultimately, I'm curious how much thrust would be necessary, and then secondarily would be, how much of the earth would a rocket firing that thrust incinerate? because I'm assuming that even with generous assumptions, the amount of thrust would be absolutely mind boggling, and it would completely ravage the earth. but how much?!
let's ignore how big such a massive object would be, and how much drag it would have. BUT if you are a nerd (I love nerds!) who wants to factor that in, extra credit for assuming it is the size of say...a 1 km^3 cube.
also, I don't know how hard it is to factor in, but I'm fine with assuming that there is some sort of fancy sci-fi energy source that has no weight...but again, for extra credit, it'd be very interesting to know how much fuel would be necessary (can assume some sort of dense, near-perfectly-efficient fuel source), and how much weight it would add, since I believe this is called the tyrrany of the rocket equation!
I wonder if there would be a subreddit of people excited to calculate these sorts of things...
for context, the object in question is....the SDF-1 macross lol.
assume it's just one gigantic thruster at the bottom firing at a constant acceleration...though honestly, you can assume whatever you want to simplify things! ultimately, I'm curious how much thrust would be necessary, and then secondarily would be, how much of the earth would a rocket firing that thrust incinerate? because I'm assuming that even with generous assumptions, the amount of thrust would be absolutely mind boggling, and it would completely ravage the earth. but how much?!
let's ignore how big such a massive object would be, and how much drag it would have. BUT if you are a nerd (I love nerds!) who wants to factor that in, extra credit for assuming it is the size of say...a 1 km^3 cube.
also, I don't know how hard it is to factor in, but I'm fine with assuming that there is some sort of fancy sci-fi energy source that has no weight...but again, for extra credit, it'd be very interesting to know how much fuel would be necessary (can assume some sort of dense, near-perfectly-efficient fuel source), and how much weight it would add, since I believe this is called the tyrrany of the rocket equation!
I wonder if there would be a subreddit of people excited to calculate these sorts of things...
for context, the object in question is....the SDF-1 macross lol.
Have you heard of Randall Munroe of "xkcd" fame? He has a few books that might be helpful -- or you could even appeal directly for an answer!
posted by wenestvedt at 8:17 PM on September 16, 2022 [1 favorite]
posted by wenestvedt at 8:17 PM on September 16, 2022 [1 favorite]
Project Orion had penciled out designs for an 8,000,000 ton rocket. It took a mere 1080 nukes to get to low earth orbit.
posted by rockindata at 8:23 PM on September 16, 2022 [4 favorites]
posted by rockindata at 8:23 PM on September 16, 2022 [4 favorites]
Another way to solve it is to take the payload capability of existing rockets. However the key is to look up the payload for Earth-escape orbits, not the easier LEO or GEO orbits. I have referred to this excellent table on Wikipedia many times for this purpose. What you want to do is sort that table by the "other" payload mass, and then look for the "TMI" numbers. Those are the payload capabilities for "trans Mars injection", so the capability of each rocket to send mass all the way to Mars.
Picking the two biggest rockets right now, the Falcon Heavy can send 16,800 kg (16.8 tons) on TMI, and the Delta IV Heavy can send 8,000 kg (8 tons) on TMI.
18 million tons therefore would take about a million Falcon Heavy rockets, or over 2 million DIVHs. It's a ludicrous amount of mass, honestly.
Now, if you're talking about mass that is already in Earth orbit, that's a lot easier. Still a monstrous number of rockets though.
* As soon as SLS and Starship launch, they will be at the new kings of that chart.
posted by intermod at 8:30 PM on September 16, 2022
Picking the two biggest rockets right now, the Falcon Heavy can send 16,800 kg (16.8 tons) on TMI, and the Delta IV Heavy can send 8,000 kg (8 tons) on TMI.
18 million tons therefore would take about a million Falcon Heavy rockets, or over 2 million DIVHs. It's a ludicrous amount of mass, honestly.
Now, if you're talking about mass that is already in Earth orbit, that's a lot easier. Still a monstrous number of rockets though.
* As soon as SLS and Starship launch, they will be at the new kings of that chart.
posted by intermod at 8:30 PM on September 16, 2022
Project Orion had penciled out designs for an 8,000,000 ton rocket. It took a mere 1080 nukes to get to low earth orbit.
To be clear, that's the mass of the rocket, very little of which makes it to orbit (it's mostly the propellant that gets burned on the way up). The mass of payload that reaches that orbit is three orders of magnitude less -- a thousand times less at 8,000 tons. Modern rockets already beat that, or at least certainly SLS and Starship will.
posted by intermod at 8:34 PM on September 16, 2022 [1 favorite]
To be clear, that's the mass of the rocket, very little of which makes it to orbit (it's mostly the propellant that gets burned on the way up). The mass of payload that reaches that orbit is three orders of magnitude less -- a thousand times less at 8,000 tons. Modern rockets already beat that, or at least certainly SLS and Starship will.
posted by intermod at 8:34 PM on September 16, 2022 [1 favorite]
Response by poster: ooh some great answers. I particularly like intermod's! something I wondering then is how one would quantify the destructive capacity of a 1 km^2 (or smaller of course) thruster putting out the thrust of a million falcon heavies.
god I would love for "what if?" to get into this...
posted by damedeshou at 8:58 PM on September 16, 2022
god I would love for "what if?" to get into this...
posted by damedeshou at 8:58 PM on September 16, 2022
Falcon Heavy is about 45m2 in cross section, so you'd be looking at about 45 times as much thrust per area for your 1km2 cube. Definitely more thrust per area, but I don't see why that would necessarily be particularly destructive.
If your surface couldn't sustain the launch for thermal reasons, you'd have to launch slower, which means you have to launch more fuel in total, which means you'd need more thrust...
posted by ssg at 9:54 PM on September 16, 2022 [1 favorite]
If your surface couldn't sustain the launch for thermal reasons, you'd have to launch slower, which means you have to launch more fuel in total, which means you'd need more thrust...
posted by ssg at 9:54 PM on September 16, 2022 [1 favorite]
https://what-if.xkcd.com/24/
This is pretty relevant to some of the physics concepts behind rocketry without actually answering the question
posted by Jacen at 10:56 PM on September 16, 2022 [1 favorite]
This is pretty relevant to some of the physics concepts behind rocketry without actually answering the question
posted by Jacen at 10:56 PM on September 16, 2022 [1 favorite]
seems like the kind of thing you could simulate in kerbal space program
maybe you can find a subreddit dedicated to that game and ask them?
posted by onya at 4:56 AM on September 17, 2022 [1 favorite]
maybe you can find a subreddit dedicated to that game and ask them?
posted by onya at 4:56 AM on September 17, 2022 [1 favorite]
You asked about volume.
There are about 8 metric tons of steel in a cubic meter. So 18,000,000 metric tons comes to 18,000,000/8 = 2,250,000 cubic meters, or a cube about 131 meters on a side.
As for propulsion, the Project Orion approach would probably be best due to the energy density of nuclear fission. The A-bombs of the original version were small compared to the city-destroying H-bombs of the cold war era. Best to launch in a remote area, though.
posted by SemiSalt at 7:04 AM on September 17, 2022
There are about 8 metric tons of steel in a cubic meter. So 18,000,000 metric tons comes to 18,000,000/8 = 2,250,000 cubic meters, or a cube about 131 meters on a side.
As for propulsion, the Project Orion approach would probably be best due to the energy density of nuclear fission. The A-bombs of the original version were small compared to the city-destroying H-bombs of the cold war era. Best to launch in a remote area, though.
posted by SemiSalt at 7:04 AM on September 17, 2022
To get a sense of the destructive power of thatwhichfalls’ estimate, you can read about Tsar Bomba:
All buildings in the village of Severny, both wooden and brick, located 55 km (34 mi) from ground zero within the Sukhoy Nos test range, were destroyed. In districts hundreds of kilometres from ground zero, wooden houses were destroyed, stone ones lost their roofs, windows, and doors, and radio communications were interrupted for almost one hour. One participant in the test saw a bright flash through dark goggles and felt the effects of a thermal pulse even at a distance of 270 km (170 mi). The heat from the explosion could have caused third-degree burns 100 km (62 mi) away from ground zero. A shock wave was observed in the air at Dikson settlement 700 km (430 mi) away; windowpanes were partially broken for distances up to 900 kilometres (560 mi).[54] Atmospheric focusing caused blast damage at even greater distances, breaking windows in Norway and Finland. Despite being detonated 4.2 km (3 mi) above ground, its seismic body wave magnitude was estimated at 5.0–5.25.posted by Winnie the Proust at 8:43 AM on September 17, 2022
The ideal rocket equation can be derived fairly straightforwardly by conservation of momentum. It's all on wikipedia if anyone wants to see the full derivation. It's given as
(1) Δv = ve ln( m0 / m1 )
where Δv is the change in velocity of the rocket, ve is the exhaust velocity of the gas, ln is the natrual logarithm, m0 is the initial mass and m1 is the final mass. I'll use the substitutions m0 = mp + mf (the mass of the payload plus the fuel)and m0 = mp .
There are a lot of assumptions that go into making that equation so neat, (no gravity or air resistance) but it's pretty much valid for the case of a rocket floating in empty space. Let's solve it anyway, rearranging the above equation gives:
(2) mf = mp ( exp(Δv / ve) -1)
And after plugging in some values which I just googled we end up with:
mf = mp (exp(104 / 5x103) -1)
mf = 6 mp
So in this idealised version, for a given mass you need about 6x the mass of fuel to get it going fast enogh, this is the tyranny of the rocket equation you mentioned. Of the mass that I called mp probably 90% plus is taken up with the rocket structure, (dropping some of this halfway up makes a big difference, hence multistage rockets), so a more realistic estimate might be 60x the mass of the actual payload.
To make the equations a little bit more accurate we can take gravity account. Again, we conserve momentum but this time there's am external force acting so we end up with
(3) Δv = ve ln( m0 / m1 ) - gΔt
which can be rearranged to give:
(4) mf = mp ( exp((Δv + gΔt) / ve) -1)
where g is the gravitational constant and Δt is the time taken to burn the fuel. Annoyingly g changes quite a lot when you're on a rocket going into space, which makes the maths tricky. But, we've also completely ignored air resistance too, so lets take the largest value of g and hope it cancels out. Taking values of 10m/s2 and 100s I get a fuel to rocket ratio of 1: 8, which actually isn't that far away from what Nasa manages.
This ratio would still be true for your massive payload, though the structure holding it together might start to get unfeasibly heavy. So a payload of about 2x107 kg, will need a rocket that weighs about ten times that, and fuel that weighs another ten times that, so we're getting close to 1010kg of fuel. Wolfram alpha gives some... interesting comparisons. The rocket would weigh about the same as (and probably resemble) one of the great pyramids, but you could instead use 1/10th of the dry biomass of all living humans if you were that way inclined.
For your extra credit question, a fancy fuel with no weight would be useless in a rocket, as it works on conservation of momentum: if you have no mass to push away from you you can't go anywhere. What the equations are really sensitive to is the the size of the push (or ve), and how long it has to hang around in gravity, or Δt, so that's where I'd expect a sci-fi energy source to do better.
Finally a bit about what it would look like to burn all of that fuel. As the energy will be dissipated across the path of the rocket I don't think it would be as destructive as some might expect. Perhaps the easiest way to consider it is that the thrust force Ft will be a little bit greater than the force due to gravity. Ft = ve x ṁ where ṁ is the mass flow of fuel, so I make that about 105 kg of fuel per second. Plugging in the energy density of rocket fuel you end up with 1013 Joules per second. Apparently that's 1/5th of the energy of Little Boy nuclear bomb, but going off each second, so you definitely wouldn't want to be standing near it. But I don't think it's anything like a crazy planet destroying amount of energy, in total it looks like the rocket would release as much energy as a hurricane, so I imagine it's like a nuclear bomb if you're nearby, with something like a hurricane started on top due to the sheer amount of hot gas released into the atmosphere.
posted by Ned G at 9:53 AM on September 20, 2022 [1 favorite]
(1)
where Δv is the change in velocity of the rocket, ve is the exhaust velocity of the gas, ln is the natrual logarithm, m0 is the initial mass and m1 is the final mass. I'll use the substitutions m0 = mp + mf (the mass of the payload plus the fuel)and m0 = mp .
There are a lot of assumptions that go into making that equation so neat, (no gravity or air resistance) but it's pretty much valid for the case of a rocket floating in empty space. Let's solve it anyway, rearranging the above equation gives:
(2)
And after plugging in some values which I just googled we end up with:
mf = 6 mp
So in this idealised version, for a given mass you need about 6x the mass of fuel to get it going fast enogh, this is the tyranny of the rocket equation you mentioned. Of the mass that I called mp probably 90% plus is taken up with the rocket structure, (dropping some of this halfway up makes a big difference, hence multistage rockets), so a more realistic estimate might be 60x the mass of the actual payload.
To make the equations a little bit more accurate we can take gravity account. Again, we conserve momentum but this time there's am external force acting so we end up with
(3)
which can be rearranged to give:
(4)
where g is the gravitational constant and Δt is the time taken to burn the fuel. Annoyingly g changes quite a lot when you're on a rocket going into space, which makes the maths tricky. But, we've also completely ignored air resistance too, so lets take the largest value of g and hope it cancels out. Taking values of 10m/s2 and 100s I get a fuel to rocket ratio of 1: 8, which actually isn't that far away from what Nasa manages.
This ratio would still be true for your massive payload, though the structure holding it together might start to get unfeasibly heavy. So a payload of about 2x107 kg, will need a rocket that weighs about ten times that, and fuel that weighs another ten times that, so we're getting close to 1010kg of fuel. Wolfram alpha gives some... interesting comparisons. The rocket would weigh about the same as (and probably resemble) one of the great pyramids, but you could instead use 1/10th of the dry biomass of all living humans if you were that way inclined.
For your extra credit question, a fancy fuel with no weight would be useless in a rocket, as it works on conservation of momentum: if you have no mass to push away from you you can't go anywhere. What the equations are really sensitive to is the the size of the push (or ve), and how long it has to hang around in gravity, or Δt, so that's where I'd expect a sci-fi energy source to do better.
Finally a bit about what it would look like to burn all of that fuel. As the energy will be dissipated across the path of the rocket I don't think it would be as destructive as some might expect. Perhaps the easiest way to consider it is that the thrust force Ft will be a little bit greater than the force due to gravity. Ft = ve x ṁ where ṁ is the mass flow of fuel, so I make that about 105 kg of fuel per second. Plugging in the energy density of rocket fuel you end up with 1013 Joules per second. Apparently that's 1/5th of the energy of Little Boy nuclear bomb, but going off each second, so you definitely wouldn't want to be standing near it. But I don't think it's anything like a crazy planet destroying amount of energy, in total it looks like the rocket would release as much energy as a hurricane, so I imagine it's like a nuclear bomb if you're nearby, with something like a hurricane started on top due to the sheer amount of hot gas released into the atmosphere.
posted by Ned G at 9:53 AM on September 20, 2022 [1 favorite]
This thread is closed to new comments.
Earth's escape velocity is 11.2 km/s = 11,200 m/s
The kinetic energy an object of 18,000,000 tonnes (18,000,000,000 kg) would need to reach that velocity is:
0.5 x 18,000,000,000 x (11,200)^2
= 1.12896E+18 Joules
That's about one exajoule. About five of the largest nuclear weapons ever exploded. Or ten kilos of antimatter.
It would make a mess.
posted by thatwhichfalls at 8:03 PM on September 16, 2022 [1 favorite]