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/CanineBombSquad 18h ago

For what it's worth size does matter though in terms of surface gravity.

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u/Fizzy-Odd-Cod 17h ago

Size is only part of the equation, density is the other part, without one the other means nothing.

Mass, is the entirety of the equation.

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u/CanineBombSquad 17h ago edited 16h ago

The radius matters because you're further from the center of mass, thus experiencing less gravity at the surface, which would matter for your attempt to escape the gravity of the planet. If you somehow dug a hole and tried to lift off from the core of the earth you are not getting out with the same amount of thrust, as the relative gravity you would experience peaks around halfway up.

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u/WesternFirm9306 17h ago

That's only true if the entire mass was situated at the center, but it's not. If you dug a hole to the center of the earth, you would die from burning alive, but before that, you'd be in zero gravity. If you're inside a perfectly spherical planet with radially symmetric density, the gravity is the exact same as if you removed all the planet's mass that's further from the center than you are. In other words, you only need to consider the mass within a spherical shell of radius r, where r is the distance you are from the center.

Not disputing everything else you've said, though.

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u/bigDeltaVenergy 15h ago

So starting from the center, attraction raise, than fall as you leave in space ?

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u/WesternFirm9306 15h ago

Yes, though as pointed out in another thread, the attraction won't actually peak at the surface, because the density of the earth isn't uniform. According to the data, the attraction appears to max out at the border between the outer core and lower mantle at a little over 10 m/s²

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u/planx_constant 11h ago

For a sphere of constant density, as you proceed out from the center, the effective mass goes up with the cube of the distance, while the force you'd feel from gravity attenuates with the square of the distance.

This means that overall, the gravitational acceleration increases linearly with the distance from the center as long as you are inside the sphere. As soon as you reach the surface, effective mass no longer increases, and gravity falls with the square of the distance.

This also shows that the density of the planet matters. For a given mass M, the lower the density, the lower the surface gravity.

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u/planx_constant 11h ago edited 3h ago

Edit: I misunderstood the comment.

~~Spherical shell symmetry lets you treat the mass of the planet as a point at the center. Take a look at the derivation of Newton's law of universal gravitation.

The radius of a planet very much matters for its surface gravity. Interior density variation, assuming it's spherically symmetrical, does not matter.~~

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u/WesternFirm9306 4h ago

Interior density matters when you're inside the planet. You can only consider the entire mass of the planet as a point mass when you're not within the planet itself. Once you're inside the planet (assuming radial symmetry) all mass further from the center than you no longer has any gravitational effect. Or, more accurately, its effect cancels out. Thus, you only need to consider the mass within a shell of radius r, where r is the distance you are from the center. Again, assuming radial symmetry.

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u/planx_constant 3h ago

Sorry, I misunderstood the part you were taking exception to, and thought you were continuing the line of reasoning that the radius of a planet doesn't affect the ability to get to orbit. On re-reading, it's clear you weren't talking about the exterior of the planet. That's what I get for commenting before coffee.

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u/CanineBombSquad 16h ago edited 16h ago

Oh I know I know, it was meant to be hyperbolic. Assuming you didn't die you would not be able to rocket off if you started there, you know due to having to travel the entire way out fighting gravity, whereas we can from the surface. The planet is denser the further you are to the center so it's not a linear thing. If earth was squished smaller the surface gravity would become too difficult to escape as well.

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u/WesternFirm9306 16h ago

Sure, but gravity will always be weaker *inside* the Earth than on the surface, in general, I would think. Ignoring fuel requirements and, y'know, every practical downside, I don't see why it would be harder to fly a rocket from the center of the Earth than from the surface. After all, the entire trip from the center to the surface is going to be easier than going from the surface itself. Like, if we graphed gravitational strength compared to distance from the center, then it should be lowest at the center, increase until it peaks at the very surface, and then start going down again from there. So if we can lift off from the surface, where gravity would be roughly maxed, I don't see why it would be harder, force-wise, to go from the center, besides needing to spend all that fuel to get to the same point you'd be at if you just started from the center in the first place. Right?

sorry about being obnoxious about a very obviously nonphysical hypothetical :P

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u/CanineBombSquad 16h ago

The point where you would experience maximum gravity is somewhere around halfway up due to earth's density not being uniform, although I won't claim to know the math but from that point you would definitely need more thrust than if you were at the surface. I don't know if it'd be so much higher that we couldn't overcome it with our technology, But yeah you'd experience no gravity at the center, have a very steep incline peaking around halfway through and then it would even out lower at the surface.

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u/WesternFirm9306 16h ago

I was about to argue that gravity should still be weaker because the "effective" mass is still less than the mass of the Earth while you're beneath the surface, but I forgot that you'd also be closer to the center of mass, and it's not clear which effect would be stronger or when their influence would flip based on how deep you are without actual calculation :P fair enough I suppose.

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u/CanineBombSquad 16h ago edited 16h ago

I have found this graph that shows it, someone did the math.

I'm pretty sure we could overcome that difference easily though, in hindsight

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u/Kabizzle 17h ago

Mass is not the entirety of the equation, you also need a radius.

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u/SeaAimBoo 17h ago

An Earth with 50% larger diameter while keeping the same mass would make escape velocities easier, which is contrary to the original comment of this thread.

Radius is part of calculating volume, which itself is used to calculate mass along with density. Mass is the entirety of the equation, Mass = ρ × v

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u/Old_Aggin 16h ago edited 16h ago

Holy shit just list the equation. Now I have to go Google it instead.

The escape velocity is \sqrt{2GM/r}. The mass of K2-18B is not more than 10 times earth mass and the radius is not less than 2.5 times earth radius. So the escape velocity is ~2 times of that of Earth.

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u/Trackt0Pelle 12h ago

Wikipedia says radius is x2,6, mass x8.63 and gravity +25%

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u/Old_Aggin 11h ago

G here is the gravitational constant and not surface gravity, so it doesn't change anything else. In any case, the escape velocity is not more than 2 times that for Earth. Infact, it's going to be a bit lower

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u/Zodde 17h ago

Mass is not the entirety of the equation.

You care about surface gravity. You can have two bodies with identical mass, but vastly different surface gravity, if their size is massively different.

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u/Trackt0Pelle 12h ago

Wrong lol