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COle Bubbles, and Gravity questions

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Sun, 2015-03-01 01:21

jasonbrisbane jasonbrisbane's picture

COle Bubbles, and Gravity questions

Hi, Im looking at putting the players into a cole bubble, in an asteroid in the Main belt. Im having trouble trying to figure out what size the interior of the bubble would be (inside a hollowed out asteroid). Firstly, does anyone have a formula or two that tells what gravity is generated from a size asteroid and spin rate? If the whole asteroid is spinning "chaotically" through space, would the interior inhabitants get various g forces generated? Or would it simply be constant? (I imagine constant..) What is a "safe rate of spin" for an asteroid to stop inhabitants from getting sick? If the cole bubble has one quarter of the space for habitable land and 3/4 for "generated sky", and it is spinning end over end constantly through space, then wouldnt the inhabitants believe it is close to normal gravity (earth?) I am wanting a cole bubble not too large but big enough for at least 5000 people (or a 100km square parkland and a few scattered buildings...(that may or may not be inhabited by a few people). What would be the minimum asteroid size that would generate a reasonable (0.3-0.6 G) gravity force? Thanks

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Regards, Jason Brisbane

Sun, 2015-03-01 06:24

TheGrue

Calculator

The formula to solve for centripetal acceleration is f = (mv)^2 /r Where "m" is the mass of the object experiencing centripetal acceleration, "v" is its rotational velocity, "r" is the radius of the circle it travels as it rotates, and "f" is the force acting on the object expressed in Newtons. Since we're dealing with the interior of a spinning asteroid, we can assume "r" is the radius of a given point on the interior hollow surface. We can also alter the equation; a "g" is a unit of gravitational force, equivalent to one Newton/kilogram (g = f/m). With a little bit of high school algebra, we can eliminate "m" from the above equation: r(g) = v^2 Plug in the radius of your asteroid and how much apparent gravity you want inhabitants to experience at the rim, and you'll get your velocity. The radius of your asteroid will depend on how much "vertical" space you want there to be inside. As for the precise dimensions, you already have an idea how many people you want on it. Figure out how much surface area you need for 5,000 people to live, work and occasionally have some time to themselves, and then abstract the interior of your asteroid as a cylinder. The formula for calculating the surface area of a cylinder is 2(pi)rh+2(pi)r However, the end caps of your cylinder are not usable area so we don't care about them, which leaves us with: A = 2(pi)rh You'll have an idea what A is, but you'll have to pull either r or h out of your hat; you can calculate r given h, or h given r, but you're going to have to make up one number or the other. The good news is asteroids come in all shapes and sizes so you can get away with it.

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Thermonuclear Banana Split - A not-really-weekly Eclipse Phase campaign journal.

Mon, 2015-03-02 02:33

jasonbrisbane jasonbrisbane's picture

ok, so an asteroid of 5km by

ok, so an asteroid of 5km by 1km high gives 31.4 km2 of area. SHorten it down if you want an entry point into the interior. Spin the asteroid at 1.58 rpm and you get a Gravity of .5 that of earth? Or is gravity measured in different units? Is 1G equivalent to earth?

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Regards, Jason Brisbane

Mon, 2015-03-02 02:51

TheGrue

Yeah; 1g is equivalent to the gravitational force experienced on Earth at sea level, or 9.8 N/kg. I think the standard for most habitats is Mars gravity, which is 0.37g.

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Thermonuclear Banana Split - A not-really-weekly Eclipse Phase campaign journal.

Mon, 2015-03-02 07:22

ORCACommander ORCACommander's picture

http://en.wikipedia.org/wiki

http://en.wikipedia.org/wiki/Gravity_of_Earth fun fact gravity is not a universal constant on this planet :P although i doubt the variance is easily noticed by us physically.

Mon, 2015-03-02 19:19

TheGrue

ORCACommander wrote:http://en

ORCACommander wrote:
http://en.wikipedia.org/wiki/Gravity_of_Earth fun fact gravity is not a universal constant on this planet :P although i doubt the variance is easily noticed by us physically.

Wrong. Gravity is absolutely a universal constant, on this planet and everywhere else in the observable universe. What is not constant is the force of acceleration due to gravity at any randomly-chosen point on the "surface" of the Earth. This is because the Earth is not perfectly spherical, and even if it were is not perfectly homogeneous in density. If it were, however, the force of acceleration due to gravity would be precisely the same at any given point on its surface. Even in that case, force of acceleration due to gravity varies with the distance between two objects. Be very careful with your terminology. ;)

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Thermonuclear Banana Split - A not-really-weekly Eclipse Phase campaign journal.

Mon, 2015-03-02 21:00

ORCACommander ORCACommander's picture

I stand corrected but its not

I stand corrected but its not me being lax with terms more like only half understanding them. I opted out of hs physics because trig or precal was a prerequisite and i would not want another course with that teacher.

Mon, 2015-03-16 20:42

Chernoborg Chernoborg's picture

http://www.artificial-gravity

http://www.artificial-gravity.com/sw/SpinCalc/SpinCalc.htm Courtesy Atomic Rockets http://www.projectrho.com/public_html/rocket/ This is a really useful page for getting a feel for what radius, RPM, and desired G forces you'd get for any particular values you assign. It even has warning signs if the results are uncomfortable for humans. There's some question as to the validity of those thresholds, and basic biomods would probably counter most of that.

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