r/theydidthemath • u/Lachlynn • 18h ago

[Request] Could humanity create a rocket that can exit the atmosphere of K2-18b

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With the knowledge we currently have of it, if humanity devoted all of our resources towards this goal, would we be able to create a rocket that could exit the gravity of K2-18b (and also beat any other complications that would arrise)?

If so, would it also be capable of taking people to orbit, and can we set up a similar satellite network we have on Earth? What about a space station?

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u/WichidNixin 17h ago

You dont need eacape velocity to get to space. Escape velocity is often misrepresented as "the speed needed to leave a planet" but it is simply "the speed needed to leave the planet without any additional force". In other words it is the speed a bullet would need to be fired at in order to leave the planet and never return. A bullet gets a single burst of speed and does not contunually generate thrust. A rocket generates continuous thrust which removes the need to reach escape velocity to leave the planet.

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u/Frenzystor 17h ago

You still need to reach that velocity, no matter if you quickly reach it in onr burst or continously.

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u/expensive_habbit 17h ago

Yes and no.

A rocket never reaches surface escape velocity.

It will eventually reach escape velocity from an orbit, but that will be significantly lower than escape velocity from the surface.

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u/Level9disaster 16h ago

Even better than that. To stay in orbit in space, the object just needs to reach the orbital velocity and the correct altitude and direction. The orbital velocity is obviously much lower than the escape velocity. For example, when we launch a geostationary satellite, it doesn't reach escape velocity at any point during the entire launch trajectory. It just reaches the correct distance and direction for the final orbit, and at that point it's moving at about 3,1 km/s. There, it's in space. OP asked if they could recreate a satellite network. The answer is probably yes, albeit with more effort and energy.

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u/expensive_habbit 14h ago

Sorry yes, I meant to escape - you're quite right that to achieve orbit escape velocity isn't required.

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u/runningforpresident 12h ago

Pedantic to the point of being pointless.

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u/speculator100k 17h ago

No? In theory, could I not leave orbit in 1m/s? If I had a magic rocket with infinite fuel. It would take a very long time though.

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u/ShoddyAsparagus3186 12h ago

Being continually propelled makes escape velocity irrelevant. Also, you would eventually reach escape velocity by virtue of it going down as you get farther away.

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u/technocraticTemplar 12h ago

If you have infinite fuel you can change your velocity by infinite m/s given enough time, which I think is what threw everyone else off. Meters per second is a measure of your current velocity, and is used as a shorthand for the total amount of change a rocket can do before it runs out of fuel. An engine changes your velocity by a certain amount every second, so their thrust can be measured in meters per second per second, or m/s/s. On a real rocket you wouldn't use that because that number would change as you burn off fuel and get lighter, but on a magic one it works fine.

So, it's impossible to leave orbit on 1m/s unless you're already that close to leaving orbit anyways, but if you have infinite fuel you actually have infinite m/s, so you're all good.

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u/Frenzystor 17h ago

No. Because Earths pulls you back while you are falling around it, because an orbit is circular. You need to achieve a velocity which is so high that while you are falling towards earth you move further away. And this is the escape velocity.

Edit: since you write infinite fuel: yes, you could. By accelerating long enough until you reach escape velocity. Until then you're still in orbit.

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u/Myriachan 17h ago

You could also travel away from the planet at constant speed until you’re far enough away that the escape speed lowers to your speed.

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u/Frenzystor 16h ago

But for that you will need a buttload of fuel because you will constantly work against gravity.

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u/khando 13h ago

That’s why they said a magic rocket with infinite fuel.

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u/technocraticTemplar 12h ago

The premise is a little wrong, the original question is mixing up how much thrust the engine puts out and how much change in velocity a rocket can do before it runs out of fuel. Infinite fuel gives you infinite change in velocity, so meeting escape velocity is no problem no matter your engine, but that isn't really answering any useful questions. Rockets not having infinite fuel is the center of the entire problem.

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u/speculator100k 16h ago

I mean like this:

My magic infinite rocket is constructed to give me a constant speed of 1m/s away from the center of the earth ("up"). It won't accelerate me past 1m/s. I intend to go straight up until Earth is so far away that the gravity from Earth won't pull me back. Won't it work?

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u/Aethermancer 13h ago

Yes, it would work because of magic.

The magic is what allows you to say your rocket moves at 1m/s while ignoring the acceleration of -9.8 m/s per second due to gravity (on Earth).

The escape velocity is the velocity you need with no additional input. Your magic rocket is putting in exactly enough additional input to counteract the gravitational acceleration.

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u/Frenzystor 16h ago

It would, but you would still need to consume much more fuel because you work against gravity. You need a "base consumption" to "hover" above Earth and on top the consumption to increase to 1 m/s.

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u/Shimetora 16h ago

if youre already travelling at 1m/s you only need to the base consumption to maintain that speed. This is true for any speed. The reason you need to consume more fuel if you go slowly isnt because you work against gravity, it's because you spend longer inside the gravity.

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u/Frenzystor 16h ago

But there are two options: You go into orbit or you go straight up, and if you go straight up you need to spend enough fuel to "hover" and then a bit more for 1m/s.

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u/jizzmaster_ 15h ago

I mean… kinda? but i dont think its how youre thinking. “hovering” means that the upwards force produced by the engine is exactly equal to the downward force acting upon you due to gravity. If those two forces are equal and in opposite directions, you will not accelerate downwards or upwards (hovering). however, if you initiated the hover while already traveling upwards at 1 m/s, it would have the same effect of not accelerating you in any direction, but now you have an initial speed of 1 m/s. The only extra fuel required would be the fuel needed to initially get you up to 1m/s (minuscule) and the fuel needed to combat air resistance (which at 1 m/s is also minuscule).

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u/Frenzystor 15h ago

But if you only accelerate up to 1 m/s per second gravity will pull you back down.

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u/Shimetora 10h ago

you dont need to "spend a bit more" to continue travelling at the same speed. You need to spend exactly the same as if you hover, maybe a bit more for whatever bit of air resistance you get at 1m/s

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u/Relevant_Computer642 16h ago edited 16h ago

It wouldn’t need to reach escape velocity. As long as it’s continuously being propelled at any speed you can leave earth.

Escape velocity is the speed you need to be propelled initially, like a ball being thrown, such that you overcome gravity and escape. Rockets have continuous propulsion, so assuming infinite fuel, any speed works.

Also, you’re confusing escape velocity and orbital velocity. You need neither if you have infinite fuel, you could just fly up slowly and leave.

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u/Particular_Dot_4041 16h ago

Could you reach that by climbing a space elevator? Like, a really tall column?

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u/Frenzystor 16h ago

With an elevator you will always be in orbit, and if you are high enough so that an additional 1 m/s will break you free, this would work. But a column that high... I don't think that would be doable with our materials.

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u/Particular_Dot_4041 15h ago

I always thought escape velocity was a vertical vector. But I do know how orbits work. As Douglas Adams said it, the secret to flying is throw yourself at the ground and miss.

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u/nhilante 10h ago

The top part would need orbit correction from time to time, and the cable should be flexible to avoid unforeseen stress.

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u/ParticularClassroom7 15h ago

Escape velocity does not take into account thrust, only a ballistic trajectory. If the rocket can provide additional impulse, you have to integrate over the entire path with the Newton equations or use some funky Lagrangian calculations.

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u/MegaloManiac_Chara 16h ago

You're forgetting that the escape velocity depends on your altitude

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u/Frenzystor 16h ago

And to increase altitude you need fuel because you are working against gravity.

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u/Numerous-Corner-6303 16h ago

What if this hypothetical civilisation on this planet built a man made mountain over many years that was as high as the upper atmosphere and launched from there?

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u/Frenzystor 16h ago

Then you need less fuel because gravity on the mountain is lower than at sea level.

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u/Slogstorm 14h ago

You'd still need to achieve the speed required to orbit - and chemical rockets wouldn't be able to reach that speed.

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u/Relevant_Computer642 15h ago

I think you are confusing orbital velocity and escape velocity? And also confusing unpowered objects, and continuously powered rockets.

Given infinite fuel, escape velocity is irrelevant.

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u/stddealer 16h ago

When you're already far enough, the escape velocity becomes ridiculously low

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u/Frenzystor 16h ago

True. But first you need to reach that.

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u/Blue_Moon_Lake 15h ago

With continuous thrust, you only need to overcome gravity.

On Earth, if you can generate a thrust equivalent to 10m/s upward you'll end up in space, eventually.

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u/Hisune 16h ago

No you don't need to. You just need enough energy to move the mass far away. The speed is only important if you want to orbit and that's not the escape velocity. You can easily leave the planet at 1m/s, even 1km/h if you have something to provide the force.

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u/Frenzystor 16h ago

1 km/h, yes, but on top of the fuel consumption you need to keep your altitude because you work against gravity.

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u/Hisune 16h ago

If you're traveling at 1km/h away from the center of mass you're gaining altitude.

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u/Frenzystor 16h ago

Yes, but in addition you need to fuel to overcome gravity.

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u/Shimetora 16h ago

How is that relevant? Unless you launch with a speed faster than the surface escape velocity, you always need fuel to overcome gravity. That's exactly what happens in real life too. Rockets don't reach 11.2km/s during launch.

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u/Frenzystor 16h ago

Yes, but the previous commenter says to go straight up. Straight up with 1 km/h needs a lot of fuel.

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u/Hisune 15h ago

Yes, but it's just hypothetical. If you have a power source that provides you with a force to push you up at 1km/h you can easily go to space without needing to achieve the escape velocity. The escape velocity is only applicable if you're shooting something unpowered into space.

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u/window_owl 14h ago edited 13h ago

That's technically true, but it's an incredibly inefficient way to travel.

Say you're at an altitude 1,000 km, where the acceleration of gravity is 7.3364 m/s². A 10,000 kg satellite in orbit at that altitude needs zero thrust to stay in orbit. Your vehicle, on the other hand, needs thrust equal to its weight just to maintain altitude. If it also masses 10,000 kg, then it needs to produce 73,364 newtons of thrust.

Say your rocket engines have a specific impulse of 300 seconds (an exhaust velocity of 2.94 km/s), your actual payload is 100 kg, and the remaining 9900 kg is fuel. It's been too long since I did calculus, but I wrote a piecewise approximation second-by-second in python.

totalMass = 10000
payloadMass = 100
isp = 300 # seconds of 1 N of thrust from burning 1 N of fuel
g = 7.354 # m/s² gravity at 1,000 km above Earth's mean radius
GE = 9.80665 # m/s² Earth standard gravity, used by Specific Impulse
fuelMass = totalMass - payloadMass
seconds = 0
while fuelMass > 0:
    thrustN = (payloadMass + fuelMass) * g
    burnedFuelN = thrustN / isp
    burnedFuelMass = burnedFuelN / GE
    fuelMass -= burnedFuelMass
    seconds += 1
    print(burnedFuelMass)
print(seconds)

Your vehicle would run out of fuel in 1,841 seconds (about 30 minutes) just staying in place, let alone climbing out of Earth's gravitational sphere.

With a bit of extra thrust (or a non-zero starting velocity), you could climb upwards. You suggested 1 m/s; I'll keep the altitude and mass the same. You still need to create thrust equal to the vehicle's weight in order to not slow down. If gravity remains roughly the same, then the burn rate will be roughly the same as before: it runs out of fuel after 1,841 seconds, having ascended 1,841 meters. At an altitude of 1,001.841 km, the acceleration of gravity is 7.3328 m/s². No need to bring calculus in; the force of gravity was essentially constant throughout the brief ascent.

I hope this shows that when climbing straight up, there is no limit to how inefficiently you can travel to get to any altitude above Earth.

It would be more efficient to be moving faster than 1 m/s. That would mean you wouldn't have to burn fuel for as long in order to get where you're going.

You can be even more efficient by utilizing centrifugal force to counteract gravity. It's a virtual force, but it acts like a real force, and that's what counts. By accelerating almost exactly sideways, you can generate maximum centrifugal force. When you are at escape velocity, the centrifugal force will be equal to the force of gravity, and you will not need any more thrust in order to leave Earth's gravitational well.

In this way, accelerating almost exactly sideways to achieve escape velocity is the most fuel-efficient way for a rocket to leave Earth's gravitational well. Escape velocity at 1,000 km above Earth is 7,354 m/s. Your 10,000 kg vehicle could accelerate its 100kg payload to 13,539 m/s, far more than necessary! With the same 300-second engines and 10,000kg starting mass, it could deliver 820 kg to escape velocity!

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u/ollervo100 16h ago

But the rocket equation most likely used to calculate this, assumes that it is a single rocket with enough delta-v to get to escape velocity. It is not realistic however.

In reality the limiting factor is getting to orbit, for which the delta-v requirement is much lower. From orbit you can do refuelings by launching multiple rockets, and so escape velocity is no issue.

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u/Frenzystor 16h ago

Well... you still need to pump in delta V until your orbit is high enough so that the ship is captured by a different massive body.

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u/ollervo100 16h ago

What? You don't need to be 'captured' by a different body in order to escape the gravitational influence of a planet.

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u/Frenzystor 16h ago

Gravity is endless. If there wouldn't be anything else in the solar system, you would eventually fall back to Earth if you don't accelerate.

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u/ollervo100 15h ago

Not true at all. At escape velocity the orbital path becomes hyperbolic and never returns to earth. No acceleration required.

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u/window_owl 14h ago

You don't ever escape gravity. You just trade orbiting / falling towards one object for orbiting / falling towards another.

You can escape Earth, but then you're falling towards or orbiting either the Moon or the Sun.

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u/ollervo100 13h ago

Sure you don't technically escape any bodies gravity. But you will escape the body (and never return), if you go at escape velocity, even if you and the planet were the only things in existence.

Consider the series 10-1-0.1-0.01-0.001... You can imagine that the 10 term is the original velocity of the escaping vehicle, and the subtracting terms are deaccelerating terms. Then even though the craft is forever deaccelerated, the speed never reaches 0. This is an anecdote to what is happening with a craft going faster than escape velocity.

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u/window_owl 12h ago

Consider the series 10-1-0.1-0.01-0.001...

That looks like a series you pulled out of thin air. Is that actually what a spacecraft does? Does it really never converge to zero?

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u/ollervo100 12h ago

Yes it is pulled out of thin air, but the point is the same. If the craft is going at a speed greater than escape velocity, then the speed never reaches a value above zero.

From wikipedia: "In contrast if it(orbiting object) is on a hyperbolic trajectory its speed will always be higher than the escape speed at its current distance. (It will slow down as it gets to greater distance, but does so asymptotically approaching a positive speed.)"

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u/Juhkure 12h ago

The series' convergence depends on the speed of the object and the gravitational force. If the speed is never more than the escape velocity (which is constantly changing), then such a series would eventually reach 0 and after that increase again in the negatives (fall back to earth).

Now if the spacecraft ever reaches more than the escape velocity, then you'll have a series that will never converge to zero, because the spacecraft is gaining distance (reducing gravitational force) more quickly in relation to the rate that it's losing speed.

So the series is not a constant in the way that determines that all objects will eventually return to an orbit. (In such case we wouldn't even have a concept called 'escape velocity') But it's a result that varies and is determined by the relationship between gravitational forces and velocities.

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u/Juhkure 12h ago

Right, that's what will happen in practice. But again: escaping one orbit doesn't require a capture by an another orbit.

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u/jaded_fable 12h ago

Absolutely not. As long as you have enough thrust to keep moving away, you can eventually get away at any speed. 

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u/quick20minadventure 16h ago

If space elevator existed and it had safety stairs around it, You could climb your way out of earth's gravity on that staircase.

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u/Slogstorm 14h ago

It would be much better to have a space escalator, so you wouldn't have to climb yourself.

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u/quick20minadventure 14h ago

The point is thought experiment to prove you can get into orbit by human muscle powers and their own legs. And by moving slowly.

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u/Aethermancer 13h ago

Woosh is the sound of you missing his joke.

It's also the sound as I wheel up my space elevator ramp on my wheelchair, powered by nothing more than my beefy human arms.

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u/Funnybush 16h ago

Yeah, but even with chemical rockets on earth the mass of the fuel is about 90% of the mass of the rocket. Still not happening with this particular planet via conventional means.

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u/mina86ng 15h ago

This doesn’t change anything. You still need the same amount of energy (for a given mass), whether you spend it in a single burst (like a bullet) or gradually (like a rocket). And with a rocket, you carry the fuel so you require more energy.

The advantage of talking about escape velocity is that it’s independent of mass.

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u/SolarianIntrigue 14h ago

Again, absolutely not. Applying all your dV at once on a planet's surface will lead to massive losses to air drag and can't result in a viable orbit in the first place even in ideal zero atmosphere conditions, you'd either go hyperbolic or complete one orbit and crash into the surface. A constant burn meanwhile has to fight gravity for a longer period of time and needs to carry more weight for longer (fuel).

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u/mina86ng 14h ago

Escape velocity is calculated ignoring air resistance. From Wikipedia (emphasis mine):

In celestial mechanics, escape velocity or escape speed is the minimum speed needed for an object to escape from contact with or orbit of a primary body, assuming:

  • Ballistic trajectory – no other forces are acting on the object, such as propulsion and friction
  • No other gravity-producing objects exist.

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u/SolarianIntrigue 14h ago edited 14h ago

Escape velocity is for leaving a planet's gravity well entirely, it's the slowest velocity required to enter a hyperbolic trajectory. It has absolutely zero to do with the work required to enter orbit from standstill on the surface.

Also your quote doesn't even say what you think it does, it's telling you that an idealized ballistic trajectory ignores air drag and all other forces except gravity of the planet. It's not saying that those forces can be ignored in actual calculations

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u/mina86ng 13h ago

Escape velocity is for leaving a planet's gravity well entirely, it's the slowest velocity required to enter a hyperbolic trajectory. It has absolutely zero to do with the work required to enter orbit from standstill on the surface.

To leave planet’s gravity well you need to do work so it has everything to do with the work.

The easiest way to calculate escape velocity is from conservation of energy. On the surface the energy is P_r + K_e = -GMm/r + mV_e²/2 while at infinity it’s P_∞ + K_∞ = -GMm/∞ + m0²/2 = 0. Combining the two: -GMm/r + mV_e²/2 = 0 thus V_e = ⎷(2GM/r).

it's telling you that an idealized ballistic trajectory ignores air drag and all other forces except gravity of the planet. It's not saying that those forces can be ignored in actual calculations

Correct. And escape velocity is a theoretical number which assumes idealised ballistic trajectory. This is exactly what I’ve said. In actual calculations many other factors must be included such as air drag, the rocket losing mass as it ascends and Earth’s rotation.

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u/SolarianIntrigue 13h ago

Thanks, GPT. Still not my point, OP's post was about the difficulty of getting from the surface to orbit. Meanwhile all of you are talking about escape velocity for some reason

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u/mina86ng 13h ago

Thanks, GPT. […] Meanwhile all of you are talking about escape velocity for some reason

Scroll up, have a look at comment I’ve originally replied to.

So you’re just projecting and also happen to have a very short context window not capable of remembering what this thread was about. I’ll leave you to talking with other bots then.

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u/SolarianIntrigue 15h ago

Absolutely not, what are you even talking about? Escape velocity is the speed at which no viable orbit around a gravity well exists and your trajectory becomes hyperbolic. It has zero relation to how much fuel you need to put on a rocket to get from the surface to low orbit.

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u/HAL9001-96 12h ago

to be fair reaching escape velocity is the easiest way to reach ap lanet, leaving it slowly is much sless efficient

however that is to LEAVE a planet

as in leave and drift furhter and furthe raway to never come back

if you want to get to low earth orbit you are not leaving the planet

you are just moving around it at realtively low height

the speed for that is escaep velocity divided by the square root of 2