massive gravity effects

[spirochete]

In-vitro may be better... certainly cheaper to send up genetic material first. Mother after.

Passenger space aboard the maternity launchers would be low availability and high demand. There would certainly be a waiting list. How would prospective mothers be selected? Lottery? Highest bidder? Selection Bureau sounds good, its even more distopian - and creepy.

 

[mr31337]

Call me dirty minded, but wouldn't procreation be a bit of a strain in 3G let alone smoking a ciggie afterwards? Maybe I'm just getting old.

 

[F33D]

It should be within the TL of 2300 to have rail-gun type launchers placed on really high altitude mountains to help get the processed material into orbit more cheaply that via chem rocket..

Just a thought for those who want to run with the extreme grav planet as is...

 

[spirochete]

It should be within the TL of 2300 to have rail-gun type launchers placed on really high altitude mountains to help get the processed material into orbit more cheaply that via chem rocket..

Mountain height scales in direct proportion to surface gravity, so the "broken" King's mountains are generally 1/3 of the height of Earth's mountains.

 

[spirochete]

Mountain height scales in direct proportion to surface gravity, so the "broken"
King's mountains are generally 1/3 of the height of Earth's mountains.

Silly me. The atmosphere is also compressed closer to the surface by roughly the same proportion.

 

[Lemnoc]

It should be within the TL of 2300 to have rail-gun type launchers placed on really high altitude mountains to help get the processed material into orbit more cheaply that via chem rocket.

Yes; in fact I was thinking it might be more expedient to shoot lots of smaller "bullets" of ore into space, where they can be captured, rather than fewer really large (and heavy) loads of ore. The shockwave out of the end of that rail gun, however, many times Mach, would probably level anything near it.

And, yes, like someone else wrote, ore would be refined and processed below to ensure the best grades got shipped. In fact, I think most jobs on King would revolve around ore processing with robots doing most of the mining.

King seems like a place that could REALLY use a space elevator, although the physics of that might be beyond 2300 tech. The thing would have to be several times the length of Gateway, carry several times the weight, including its own materials.

A skyhook might indeed be the best option.

---

EDIT: Perhaps instead of ore, I should say "product."

 

[Lemnoc]

Mountain height scales in direct proportion to surface gravity, so the "broken" King's mountains are generally 1/3 of the height of Earth's mountains.

I'd expect that, too. But I wonder if the hideous internal heat of King might also push up some really incredible volcanoes.

The whole place is likely to be radioactive as hell.

 

[Wil Mireu]

I'd expect that, too. But I wonder if the hideous internal heat of King might also push up some really incredible volcanoes.

The whole place is likely to be radioactive as hell.

It would have more radioisotopes making more internal heat, but the environment wouldn't really be "radioactive" any more than Earth's is. As for volcanoes, I don't know... volcanism on larger planets may be quite different to that on Earth (higher pressure gradients with depth might affect how easily volcanoes form, higher pressures would be required in magma chambers, higher pressure could affect partial melting etc). That's not really been investigated a lot in the literature.

 

[Lemnoc]

It would have more radioisotopes making more internal heat, but the environment wouldn't really be "radioactive" any more than Earth's is.

Yes; bad term. I'd expect radiative sources to be more pronounced—a hellish Van Allen Belt, for example, an oppressive Heaviside layer. So maybe not "radioactive" in the sense of plutonium being more common than chalk, but perhaps more intensive E-M radiations.

I suppose it doesn't have to be that way, e.g., Venus.

 

[F33D]

Yes; bad term. I'd expect radiative sources to be more pronounced—a hellish Van Allen Belt, for example,

That belt is a result of captured solar radiation not, from the Earth itself..

 

[Wil Mireu]

That belt is a result of captured solar radiation not, from the Earth itself..

True, but if the planet had a more powerful magnetic field, its radiation belts would be more dangerous too. That would partly depend on its rotation period though, and given it's a big planet that should be tidally locked to the star, it may actually not have a particularly strong field.

 

[F33D]

That belt is a result of captured solar radiation not, from the Earth itself..

True, but if the planet had a more powerful magnetic field, its radiation belts would be more dangerous too. That would partly depend on its rotation period though, and given it's a big planet that should be tidally locked to the star, it may actually not have a particularly strong field.

Agree. Too may unknowns. Both Mars (smaller) & Venus (almost same size as Earth) have negligible magnetic fields...

 

[Wil Mireu]

Agree. Too may unknowns. Both Mars (smaller) & Venus (almost same size as Earth) have negligible magnetic fields...

Mars doesn't have a magnetic field because its internal dynamo seized up from lack of heat (no molten conductive material in its interior in which to generate the field) - it just has remnant fields above small areas of its surface that retain their ancient magnetism.

Venus has no field probably because its rotation period is so slow (243 days) that the required convection currents aren't strong enough in its almost certainly molten core. The hot surface temperature also messes up the planet's geothermal gradient, which may prevent convective currents forming in its core anyway.

King, being bigger, would presumably have more radioisotopes and it'll lose its heat more slowly than Earth anyway given its size. The unknowns come from the effect of its size on its internal structure, mineral phase changes in its deep interior, whether it would have a solid inner core or not, and so on. Even if all the other factors were favorable to the formation of a magnetic field though, the fact that it should be tidelocked to its star means that it has a slow rotation period, which would significantly limit its strength (as described in the text, King does have a short rotation period, but there's no way that can be possible).

 

[Lemnoc]

Even if all the other factors were favorable to the formation of a magnetic field though, the fact that it should be tidelocked to its star means that it has a slow rotation period, which would significantly limit its strength (as described in the text, King does have a short rotation period, but there's no way that can be possible).

While I agree with you about all probabilities King should be tidelocked, it's not 100% certain. Could be a captured world, could have been shifted in its orbit,* could have been born with a hellishly huge rotation that has taken aeons to spin down. Etc.

* http://www.universetoday.com/88374/how-did-jupiter-shape-our-solar-system/

 

[Jeraa]

Even if all the other factors were favorable to the formation of a magnetic field though, the fact that it should be tidelocked to its star means that it has a slow rotation period, which would significantly limit its strength (as described in the text, King does have a short rotation period, but there's no way that can be possible).

While I agree with you about all probabilities King should be tidelocked, it's not 100% certain. Could be a captured world, could have been shifted in its orbit,* could have been born with a hellishly huge rotation that has taken aeons to spin down. Etc.

According to the 2320AD book from QLI, King isn't tidal locked at all (it doesn't have a notation about it, like the other tidal locked planets do.) It has a day length of 19.25 hours. The 2300 Colonial Atlas from GDW says 19.65 hours. (It also says it has a density of 1.3 standard, giving it a mass 17x that of Earths.)

 

[Wil Mireu]

While I agree with you about all probabilities King should be tidelocked, it's not 100% certain. Could be a captured world, could have been shifted in its orbit,* could have been born with a hellishly huge rotation that has taken aeons to spin down. Etc.

* http://www.universetoday.com/88374/how-did-jupiter-shape-our-solar-system/

A solid world that big, that close to the star? I'd say it's pretty certain. It would become tidelocked VERY quickly: the calculations show that Io, the innermost major satellites of Jupiter, was tidelocked in only a few million years, and that timescale decreases with increasing size of the satellites. Even if you scale the mass up to stars and planets and to King's distance, it happens on the order of a few 10s to 100s of million years, which is very rapid in geological terms).

Even if it had its own independent rotation period, King would be tidelocked so quickly that the chance of us finding it in a non-tidelocked state is very tiny. And if it wasn't tidelocked, it'd be busy turning itself inside out due to the ridiculous amount of tidal heating it'd be suffering as the tides slow down its rotation.

 

[Wil Mireu]

According to the 2320AD book from QLI, King isn't tidal locked at all (it doesn't have a notation about it, like the other tidal locked planets do.) It has a day length of 19.25 hours. The 2300 Colonial Atlas from GDW says 19.65 hours. (It also says it has a density of 1.3 standard, giving it a mass 17x that of Earths.)

It isn't tidelocked in the books. But for a world of that size and mass, there's no way it can't be. The only way to make that aspect of it work is to have it orbiting a G star - then it'll be further from the star and the tidal despinning would be much slower.

 

[F33D]

Venus has no field probably because its rotation period is so slow (243 days) that the required convection currents aren't strong enough in its almost certainly molten core. The hot surface temperature also messes up the planet's geothermal gradient, which may prevent convective currents forming in its core anyway.

That's total guesswork based on guesswork. (not yours I know) The fact is, we don't really know why Venus has no field.

Even if all the other factors were favorable to the formation of a magnetic field though, the fact that it should be tidelocked to its star means that it has a slow rotation period, which would significantly limit its strength (as described in the text, King does have a short rotation period, but there's no way that can be possible).

You got something there for sure.

 

[Wil Mireu]

That's total guesswork based on guesswork. (not yours I know) The fact is, we don't really know why Venus has no field.

I would rather provide some likely reasons based on what we know about physics and chemistry and geology than just shrug and say "we don't really know, it's all guesswork" (which helps nobody at all). You obviously don't know much about the subject, but please don't make the mistake of using your ignorance to be authoritative about what is actually known.

 

[GamingGlen]

Is there an equation for this? The only one I found is linear, so my spreadsheet keeps the surface gravity seemingly low for super Earths (which does keep such planets more viable for adventuring on them).

The radius-density relationship is better than scaling exponents. You can derive density from radius and vice-versa. Once you have these two values, you can solve for everything else. Where radius r and density d are expressed in earths (radius and density = 1), the radius-density relationship appears to be:

Ln(2d) = Ln(2) × r and 2d = 2^r

Therefore
d = Exp(r × Ln(2))/2 = 2^r/2 and
r = Ln(2d)/Ln(2)

Watery/icy bodies (icy moons, failed cores, oceanic superterrestrials) have 1/3 to 1/6 of this value (Ceres 0.67, Pluto 0.651, Triton 0.649, Titania 0.574, Charon 0.562, Oberon 0.546, Ganymede 0.531, Titan 0.517, Callisto 0.513, Dione 0.508, Umbriel 0.477, Rhea 0.414, Mimas 0.409, Iapetus 0.365, Enceladus 0.343, Tethys 0.339).

With mass m expressed in earth masses,
m = 0.5 Exp( r × Ln(2) ) × r³ = r³ × 2^r / 2

You can derive radius from mass by working this in reverse, but you need a Lambert power log function.
r = 3 LambertW( Ln(2) / 3×CubeRt(2m) ) / Ln(2)

Hey, professor, is this all going to be on the test? :mrgreen:

Thanks for the information. I was using the right equation after all for surface gravity, so it was not as linear as I had thought when I posted.