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So an "EARTH" gravity is used in the math in the game even though a Kerbin surface gravity is less then an earth gravity.
Which leads to confusion because the word *GRAVITY* is used BOTH as a denotation of the *WEIGHT* of an object and as *SHORTHAND* for the *FORMULA* (9.8 M / s^2) used in acceleration formulas.
The "SIZE" of parts is a compromise between 2 different measurement sizes metric and imperial.
KSP size, Imperial, Metric. (Note values are approximations from the conversion)
TINY, 2 Feet, 0.65 Meters
SMALL, 4 Feet, 1.25 Meters
LARGE, 8 Feet, 2.5 Meters
EXTRA LARGE, 12 Feet, 3.5 Meters
HUGE, 32 Feet, 10 Meters
Kerbals stand roughly 0.75 meters tall (2 Foot 5½Inches).
It's easy to look up:
Kerbin has a density of about 484 kg / m^3, and Earth has a density of about 2,700 kg / m^3. But acceleration due to gravity is more a function of mass than density. Albeit, an argument can be made that density is a function os mass anyway. That's not really the point though.
I don't understand why the dev's would use the same acceleration due do gravity as Earth but use a different mass than Earth. What was the point of making Kerbin then? Why can't this just be on Earth with all the real Earth constants? And if the acceleration due to gravity is really 9.8 m / s^2 on Kerbin, how are they getting into space so fast (like in seconds)? Because they are smaller and lighter like @Ogre said? Because their engines are more powerful?
The radius of Kerbin is about 1/10 that of Earth, so that r^2 factor differs by about a factor of 100 between the two.
As for density, plugging in the measured values of a, r, and G and then solving for M is how they find the mass of a planet. From there, you get that density is mass / volume. Doing that naively will give that Kerbin is about 10 times the density of Earth.
I have no idea where you got the 484 kg/m^3 figure. That's much less dense than water, which would lead water to try to sink into the planet rather than sitting on the surface. Regardless, density doesn't have to be uniform. We can't dig 1000 miles into the Earth to directly measure its density at that depth, or any other depth other than very near the surface. Getting the total mass from the measured force of gravity is the only way that we have to do it. For what it's worth, the average density of Earth is about 5514 kg/m^3, which isn't remotely close to your quoted figure.
My theory is that the universe of KSP has about 10 times as high of a universal gravitational constant (G) as ours. If you assume that they have exactly the same value of G, then you'd get that Kerbin has an average density of over 50000 kg/m^3. The only known substances with a density of even half of that are things crushed into such a small space by incredibly strong pressures, such as neutron stars. It's much cleaner if the game just uses a different value of G.
The 2,700 kg / m^3 I quoted was an error. That's just for Earth's crust. The density I quoted for Kerbin was a typo. According to the Wiki it is 58,484 kg / m^3 (not 484). Again, density has nothing to really do with my question.
Your last paragraph delivered a reasonable and interesting hypothesis though. I thought Kerbin was in our universe following our laws of physics (just a different planet in our universe). I'll put my calculator away, I guess I won't be needing it now.
Thanks for your thoughtful answer.
We expect things to fall back quite fast on earth and that's most likely the starting point kerbin should get gameplaywise as well, because it feels more intuitive and makes other planets or satelites with less gravity feel more like what we see in videos from moon landings.
It would be strange to see, if we get that feeling already on kerbin and it wouldn't feel that great, if we fly 2 minutes through space, just because we accidentally jump on the mun, whil it doesn't have enough gravity to pull us back fast enough.
Neutron stars are Trillions of times more dense than even Kerbin, Kerbin is only about as half as dense as the suns core
Gravity is a function of mass and distance. Kerbin[wiki.kerbalspaceprogram.com]'s mass is 5.29e22 kg and its radius is 600 km. When you do the math, you get to 9.81 m.s^-2
So as not to confuse newbies with a wildly different gravity on the beginner planet which is where you'll be doing most of your experimenting and learning at the start.
Kerbal rockets tend to be more generous on the amount of thrust provided wrt mass, and we don't have to worry much about aerodynamic forces breaking the rocket apart, so we generally enjoy higher accelerations. We also have much reduced constraints on deltaV, so it's entirely feasible to take a faster (vertical) but more wasteful route, which is the only way you'll get to space "in seconds" even in KSP.
Mind you even human rated rockets don't typically take much longer than a couple minutes to get to 70km in real life.
Even the density of the top of the earth's crust is only a rough approximation, which is probably why it was quoted with only two significant figures. We can measure the density of the very top of the earth's crust just by picking up some dirt and measuring it in a bunch of places in the world. But we can't reach the bottom of the earth's crust to measure it, too, and it might not be the same as what is further up. The only way to tell the difference between the bottom of the earth's crust and the earth's mantle or core from gravity alone is to get below the earth's crust and then measure, which we can't do.
The value of G is not measurable from inside of the game. Furthermore, it's unlikely that it's explicitly defined in the game's code, as it would be much cleaner to just define GM for each celestial body rather than defining G and M separately and having to multiply them every time.
It's easy to measure the force of gravity that a very large object (such as a planet) exerts on a very small object, and from that, we can get the mass of the small object. But that doesn't give you the value of G, but only GM for the large object.
In order to compute G, you have to measure the force of gravity of one small object on another, as you need to know the mass of both objects as well as the force of gravity. That is, if you have F = GMm/r^2 and you know everything except for G, you can solve for G.
But you can't do that measurement in game, as celestial bodies are the only things that exert a force of gravity. If you try to compute the force of gravity of one rocket on another, you'll always get F = 0 and then compute G = 0. That's a simplifying assumption in order to make the game work, but it means that you can't measure G.
It's quite difficult to measure G in real life, too. That's why more than a century passed between when Newton's Law of Gravity was stated and the time that anyone actually tried to measure G, as Cavendish eventually did. Even then, the measurement was pretty rough, as the force of gravity of one small weight upon another is very small.
The density of the Sun's core isn't really known, and measuring it directly is pretty hopeless. The average density of the Sun isn't very high. There are models that predict that the Sun's core should have very high density, but we know that the models are wrong, and it's only a question of whether they're only slightly wrong or wildly wrong. Regardless, even if the Sun's core does have very high density (as is likely), that's very much a case of needing incredibly high pressures in order to crush something into having high density. That's not the case with Kerbin.
BTW, the two sig figs with the density of the crust (2,700 kg / m^3) were my approximation. I wasn't thinking about sig figs.
You are mistaken. If the value of G were multiplied by some x > 0 and the value of M for all celestial bodies were divided by x, there is no experiment that can distinguish that from the original values, at least apart from extremely large or small values of x that overflow or underflow a floating-point data type. You can only ever compute the product GM, not the individual values of G and M. The mass that you're citing is computed from the assumption that G uses the same value in KSP as in our universe. Attempting to use that to compute G is circular reasoning.
https://courses.lumenlearning.com/suny-earthscience/chapter/earths-interior/
Seismographs tell us that there are particular depths at which something changes. They don't tell you how homogeneous each layer is, as it could have several very different sub-layers that all treat earthquake waves the same way. And they don't tell you anything about the density of the earth at any given layer, other than the standard effect that relatively denser objects tend to sink to the bottom while less dense ones float to the top.
As far as gravity goes, any rotationally symmetric object looks identical to one in which all mass below your altitude is exactly at the center of the object, while all mass above your altitude does not exist. That prevents gravity measurements from giving us any information about the composition of a planet beyond its total mass.