Kerbol's Habitable Zone

[stupid_chris]

Magratheans.

It's even more than 10 characters long, so there's your proof.

What if Deep Thought just decided to emulate KSP after it concluded the answer was 42???

[Spatzimaus]

The gravitational constant is different.

No, the gravitational constant is exactly the same. Seriously, that's WHY Kerbin's so dense; it's got Earth's surface gravity (9.8m/s^2), and the only way to do that while having a much smaller radius (600km) is to have its mass scale as radius squared instead of cubed. The physical constants are the same in KSP, it's what they DID with those constants that's different.

[Brotoro]

What if Deep Thought just decided to emulate KSP after it concluded the answer was 42???

How many Jebediahs does the average KSP user kill?

No...too obvious.

[stupid_chris]

How many Jebediahs does the average KSP user kill?

No...too obvious.

Then it said "create earth, give them this game, they'll find the answer themselves"

[Merinsan]

No, the gravitational constant is exactly the same. Seriously, that's WHY Kerbin's so dense; it's got Earth's surface gravity (9.8m/s^2), and the only way to do that while having a much smaller radius (600km) is to have its mass scale as radius squared instead of cubed. The physical constants are the same in KSP, it's what they DID with those constants that's different.

The gravitational constant (G) is not the same as the surface gravity of Kerbin (g). Increasing the gravitational constant would allow us to have a planet like Kerbin exert more gravity if it were composed of similar material as Earth.

Force exerted between 2 bodies is calculated like so:

F = G *m1*m2/r^2

Clearly, increasing the gravitational constant, will increase the force, and thus the acceleration. So, starting with a smaller earthlike planet, you have a lower surface gravity, then increasing the gravitational constant sufficiently, you can get back to 9.8m/s^2.

Of course, simply increasing the gravitational constant affects other things, and there is a fine balance between all the natural forces, but the point is, you could possibly alter them to achieve a universe like the Kerbin system.

[Brotoro]

I've been trying to calculate Kerbol's Habitable Zone to see if it'd actually be in one like Earth is, or whether it should be a dead block of ice. Haven't had much luck. I'm no math hero, so all the formulas out there aren't help me find the answer anytime soon. Would anyone else be able to calculate this?

I went in the other direction (in the "What Star is Most Like Kerbol" thread). If we assume that the devs want Kerbin to have an environment similar to Earth's (which makes sense since they chose to give it the same surface gravity, and it appears to be covered with water oceans and has ice caps), this tells you the temperature of Kerbol. The values I derived for the temperature and luminosity are what you'd expect for a typical red dwarf star (and the radius listed in the Kerbal Celestials Wiki also fits a red dwarf).

The only thing that does NOT fit is the mass of Kerbol, which is too low to be a red dwarf. (The mass is determined from the orbit sizes and periods of its planets.) So Kerbol is not your run-of-the-mill red dwarf. Either somebody has fiddled with it's innards using some high density material, or somebody has been fiddling with the fundamental constants of the universe.

[stupid_chris]

The gravitational constant (G) is not the same as the surface gravity of Kerbin (g). Increasing the gravitational constant would allow us to have a planet like Kerbin exert more gravity if it were composed of similar material as Earth.

Force exerted between 2 bodies is calculated like so:

F = G *m1*m2/r^2

Clearly, increasing the gravitational constant, will increase the force, and thus the acceleration. So, starting with a smaller earthlike planet, you have a lower surface gravity, then increasing the gravitational constant sufficiently, you can get back to 9.8m/s^2.

Of course, simply increasing the gravitational constant affects other things, and there is a fine balance between all the natural forces, but the point is, you could possibly alter them to achieve a universe like the Kerbin system.

Alright we're going to test this.

First we'll try to see what force is exerced between the Earth and someone weighting 60kg. So F is what we're looking for, G is the constant, 6,67384x10^-11Nm^2/kg^2, m1 is the mass of the Earth, 5,9736x10^24kg, m2 is the mass of the person, 70kg and r is the diameter of the Earth, 6371000m.

So: F=(6,67384x10^-11*5,9736x10^24*70/6371000^2=686.5349165N

Now, let's see what the gravitational constant would have to be on Kerbin to have the same force between a 70kg Kerbal and Kerbin, where F is the force exerced, 686.5349165N, m1 is the mass of Kerbin, 5.2915793x10^22kg, m2 the mass of the Kerbal, 70kg and r is the diameter of Kerbin 600000m:

G=686.73*600000^2/5.29x10^22*70=6,682114461x10^-11

Equations show it, the gravitational constant is the same in KSP.

[Brotoro]

Alright we're going to test this.

First we'll try to see what force is exerced between the Earth and someone weighting 60kg. So F is what we're looking for, G is the constant, 6,67384x10^-11Nm^2/kg^2, m1 is the mass of the Earth, 5,9736x10^24kg, m2 is the mass of the person, 70kg and r is the diameter of the Earth, 6371000m.

So: F=(6,67384x10^-11*5,9736x10^24*70/6371000^2=686.5349165N

Now, let's see what the gravitational constant would have to be on Kerbin to have the same force between a 70kg Kerbal and Kerbin, where F is the force exerced, 686.5349165N, m1 is the mass of Kerbin, 5.2915793x10^22kg, m2 the mass of the Kerbal, 70kg and r is the diameter of Kerbin 600000m:

G=686.73*600000^2/5.29x10^22*70=6,682114461x10^-11

Equations show it, the gravitational constant is the same in KSP.

Yes, but "what Gus is trying to say" is that you are using the mass for Kerbin given in the wiki, which somebody calculated ASSUMING that the gravitational constant G was the same in the kerbal universe as it is in ours. But if you DON'T make that assumption, and you make Kerbin out of normal density materials and figure out what its mass would be, then you can calculate what the value would be for G in a kerbal universe where we assume normal matter densities.

[stupid_chris]

Changing G affects more than simply the gravitational pull of things. G would need to be at least 100 to 1000 times more powerful to create such a gravitational pull from Kerbin, and this would have MAJOR consequences. For example, Kerbol would never ever ever ever logically be so big, it would be way smaller. It's density is ridiculously low and a stonger G would contract it down about five times.

[stupid_chris]

Well, I found an official way to confirm the gravitational parameter is the same in KSP:

The new knowledge system is telling us the mass of Kerbin is the same as the one stated on the wiki, meaning the developers want it to be actually super dense. Thus I guess we can safely assume the laws of physic are the same in KSP.

[Fel]

Why can't we just agree that Kerbol isn't actually a star?

I mean, the light spectrum (hmm, can you get an accurate spectra off a computer screen?) claims it has a greater mass than is possible. You experience no extreme amounts of heat while orbiting only a few km away from it... and disappear when you get too close.

Perhaps it is a hole to... somewhere... and... idk XD.

[AceMgy]

Why can't we just agree that Kerbol isn't actually a star? . . . Perhaps it is a hole to... somewhere... and... idk XD.

You mean a white hole? A theoretical "other end" to a black hole which is its inverse, meaning one can never enter it, but constantly spews energy. They even apparently have mass. I like the idea, but I don't know if you can make them as small as Kerbol.

You experience no extreme amounts of heat while orbiting only a few km away from it....

I imagine when re-entry heat is added to the game this will no longer be possible to do.

[Fel]

I imagine when re-entry heat is added to the game this will no longer be possible to do.

Well perhaps, but I was basing that off of the thermometer which I had attached to the ship, reentry heat probably would be a factor as you got closer to the corona (note that I have no clue what that means, and just... oh right, highly charged surrounding the star?), but I forgot what I was saying.

A whitehole actually sounds interesting, and if it is spewing out ungodly amounts of energy from who /cares/ where, it could account for how impractical it is to be that size

[stupid_chris]

Everything that is temperature, atmospheric pressure, aerodynamics, buoyancy or whatever concerns weather is a placeholder for. It's not correctly implemented and won't be until the developers redo it correctly. If you go on rerbin or any other planet, the temperature at the poles is the same as at the equator. The temperature model is really basic, and will be reworked someday, I bet at the same time as the reentry heating. So maybe things don't match reality for now but those things will be re done eventually.

[AceMgy]

Yeah the temperature probes are placeholders at the moment, but you can estimate that the average temperature/pressure of Kerbin is the same as Earth's because water exists in all three phases. That gives you a fairly accurate stepping point to estimate the effective temperature of Kerbol, neglecting albedo.

They say in astronomy that if you're within an order of magnitude (multiple of ten), you're close enough. Simply because the accuracy we can measure things with is so abysmal.