@kimholder I looked and the scaling is all jumbled up in the text and not clear to me. But since the surface gravity only increase by 14.5%, that earth is made of swiss cheese. In my question I ask that the average density be the same. So if the radius is 1.1 times earth radius, then the mass is (1.1)^3 times the earth mass.

Can we see "the main thing" as an equation, not a slogan? If the answer is that there is no limit, then let's just get it out there in math rather than paragraphs.

I understand that in free space, the final velocity (delta-v) is limitless. See this answer. But is it really true when accelerating in the gravity field of a planet? No matter how strong the gravity, a chemical rocket can produce enough thrust to counter it all the way to orbit? If this is known to be true, then it must have been calculated and proven. This is rocket science, not rocket folklore.

@kimholder I know, I am sure there are other ways to get off the planet. Maybe it's like Apollo plus CERN plus the Human Genome Project just to get a cubesat suborbital, but it would happen. This question however is about conventional, chemical rockets, and it's really a math problem with that constraint. There's a number, probably between 1 and 2, and Don Petit suggests it's around 1.5 but with unclear constraints. I'm just after the correct answer with the math or a reproducible calculation - like a Kerbal YouTube or something.

@kimholder If so, I need to see the math then. Not just a classic equation, but someone needs to work the problem and if what you are saying is true, make a plot (for example) of total rocket mass to put say 10kg in some kind of LEO as a function of planet_radius / earth_radius. An abstract, simplified detla-v is not sufficient for this question. If launching in air, and turning in gravity, and attaining orbital velocity is actually possible for arbitrarily large, same-density earths, this should be demonstrable with math.

@kimholder wow! OK that's great, thank you very much for your help and interest!

@Lex oh! I see what you mean. Indeed, those should be altitudes. Okay I'll fix that. Thanks! edit: how that it look now?

@Lex knowing that someone has actually read one of my questions through to the end makes my day!

@RoryAlsop That question's OP has already voted for that question to be marked as duplicate of this one. It has three close votes already in the last few hours. space.stackexchange.com/q/5320/#comment78283_5320 That question's OP also added a bounty to this question a while ago to help make sure this also received an answer. Beyond that, there is no answer to this question there.

I wasn't sure you knew the other question was closer to closing. Just trying to help.

@uhoh , just a nit, but postulating a constant mass density as radius increases means you're changing the planet's composition! We typically regard liquids and solids as "incompressible", but at the huge pressures in Earth's interior even iron's density increases from 7874 kg/m^3 to ~13000 kg/m^3. Scientists modeling planetary interiors have to consider "compressed density". If you increase a planet's radius without changing its composition, interior pressures increase and the average density increases. Growing while maintaining density requires higher fractions of the lighter constituents.

@TomSpilker I specified same density profile and same average density as Earth to keep the problem simple enough that someone might actually take the time to answer. Asking people to find the compressibility for each strata at some assumed temperature profile, and then solving for the density profile would result in zero answers.

@uhoh , Second sentence after Question: is "I'm looking for a number with backup, based on scaling the earth radius and maintaining the same average density."

@TomSpilker the same density profile (say 1.5 at the surface to 15 in the center) would result in the same average density. It doesn't mean the density would be constant.

@uhoh , yep, it's beyond what a lot of people would be willing to tackle. That's why I say it's a "nit". But I find that phenomenon very interesting in the way it must be included when doing interior models.

@uhoh , "the same density profile (say 1.5 at the surface to 15 in the center)," Yes, that's what I assumed you meant, and it's a perfectly good postulate. It's just not what actually happens. If we want to continue this we should probably move to chat. I've never done that before, have you?

If you see a message inviting you, you can activate it. It will create a new chat room for us and include copies of our recent activity. Sure, go for it!

@TomSpilker good evening!

Do I just type something into the box here?

It seems so! ;-)

I used the (at) sign to generate a flag and a beep for you, once we start talking it's not necessary

OK, great! I've never done anything like this before so there might be a learning curve!

How do I activate the audio comm channel on my end?

I don't know if there is such a thing here. There is an annoying sound if you get a new message in chat while you are in a different window. Apart from that I don't know of any audio.

So if you hadn't opened the chat tab in your browser, you would have gotten a sound letting you know I was here. That's what I meant.

Oh, OK. Can you pull up things from the web while we chat? If so, take a look at en.wikipedia.org/wiki/Structure_of_the_Earth#/media/…

That's a beautiful curve!

Yep! You see it flattens out as you get close to the center of Earth. That's because the gravitational acceleration is going to zero as you approach the center, so the pressure isn't increasing much any more.

I see. Pressure is maximum but gravity is zero.

If for example compressibility were constant, would the density profile be something simple like parabolic?

(compressibility, temperature, composition were all constant)

If you increase the radius of the planet with the same materials and the same relative profile (core radius, mantle thickness, etc. are all at the same ration as surface radius) the surface acceleration increases, so the pressure profile steepens. So at the same depth, the density is higher than in the original case. The de4nsity at the center is higher also.

Yes I am sure that's correct.

In stars, astronomers use polytrope models as simple, analytical starting points.

I wondered if there is a simple analytical solution to the problem for a homogeneous compressible material.

The net result is that, with a density profile that is everywhere higher than in the original case, the average density increases. To keep average density constant as size increases, you have to have a higher ratio of lighter constituents.

Hmm, there probably is an analytical solution, I'm not sure how simple it would be. I'm pretty sure there'll be an exponential in there somewhere.

I wonder how much stellar physicists have had to modify their models on the basis of the data we've gotten from helioseismology. That has identified stratification in the sun's interior, notably levels at which the primary heat transfer mechanism (apparently) goes from convective to radiative and then back again.

I only mentioned polytropes as an analog, a different example of a problem where an analytical form for a density profile is used by "space people" to obtain a scaling law.

I'm thinking about the analytical solution problem and how to set it up. Specifically, how to model the density as a function of pressure. You might have to model that via interatomic spacing, assuming that spacing goes as something like 1/p^n : the higher the pressure, the smaller the spacing, with zero spacing being forbidden.

oh i see!

So a constant compressibility would give rho(P) = rho_0 + const.*P

P = pressure

That already prevents zero distance.

Hmm. Tonight this'll probably displace some work I'm supposed to do! But my wife (Project Scientist of the Cassini mission) is in Paris, so...freedom!

I had a suspicion you might be a bit bored

Yes, the rho(P) = rho_0 + CpP might work, good idea.

I figured out the connection between you two yesterday. You guys have been on the Space show a few times.

Not actually bored; I'm in the middle of writing up my sections of reports for a couple of studies I've been involved in, one about aerial platforms at Venus, one about small (~30 kg) atmospheric entry probes at Uranus or Neptune. I love doing calculations! I loathe writing reports!

I think that rho(P) = rho_0 + CpP makes the integral fairly easy. That'll be fun!

I can't do math any more. I'll guess the shape of the density is a Gaussian or Lorentzian just because those are the only two I know

I'll have to look up the equations of state for things like iron/nickel mixtures, and silicate minerals, and see how closely rho(P) = rho_0 + CpP models that.

I mean, I can't solve differential equations any more.

You'll have to include temperature

and therefore phase

You'll have to build a planet!

Yeah, I don't do it nearly as much as I used to. But they're still more fun than management tasks, which JPL kept trying to get me to do. Eventually I put a sign on my office door: "I solve problems with units like km/s or W/(cm^2 - sr); if your problem involves units like $ or full-time-equivalent-employees, I can recommend someone else to help with it."

Yep!

Oh yes, phase transitions would be important. Fortunately at terrestrial planets the pressures don't get high enough that quantum-mechanical transitions, like molecular hydrogen to metallic hydrogen, don't occur.

So as the mass of planet Spilker increases, at some point will the liquid metal disappear and it will be all solid iron?

I know people who get paid to solve these kinds of modeling problems!

Those are the lucky ones.

Unlucky ones are told "we're going to start paying you for management, because we know you can't stop doing the fun stuff"

Solid iron: it depends on the temperature profile. For any profile halfway physically reasonable, there'll probably be a (possibly very thin!) layer of liquid iron. For odd profiles, you might get multiple transitions from solid to liquid and back.

Eventually I told JPL management that "I never want to earn my living in a job with the word 'manager' in its title."

good for you!

(parenthetical: there's a question about Cassini that could use another answer space.stackexchange.com/q/16839/12102)

OK, I'll take a look. I should exit now, before I do tackle that question. It's well after my usual dinner time here. My stomach is going after my liver, and I don't even like liver! But it's been a pleasure, uhoh, thanks for setting up the chat!

Go for it! I don't normally chat much but that's because most people don't chat about science. Happy anytime, and thanks for the planetary science tutorial!

Buona notte!