Assuming some devastating event will destroy every habitable planet in the system and we have to evacuate. We build a fleet of spaceships 200 miles long and 50 miles wide. Everything from earth we don't take with us will be vaporized so we need to simulate all the interconnected layers of the oceans as well as surface life here on earth. Assuming the cylinders are hollow in the middle, with a hydrogen ramjet principle for thrust. At less than 1/10 g acceleration, would the aforementioned 50 mile wide cylinders have the required 1g simulated gravity for 3 miles of depth for ocean life to survive on this ship?
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$\begingroup$ Also, are you requiring that the cylinders (assuming you meant that the spaceships were cylindrical) have constant simulated gravity for some region of depth? Because that's impossible. $\endgroup$– probably_someoneMay 26, 2017 at 23:37
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$\begingroup$ It is not necessary for the cylinders to be propelled. They will keep rotating at constant speed without any thrust. $\endgroup$– sammy gerbilMay 28, 2017 at 8:33
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The effective "gravity" goes by $$g_{eff}(r) = \omega^2 r \;,$$ where $\omega$ is the angular frequency with which the ship turns and $r$ is the radius at which you measure the effective acceleration.
Obviously it will only be exactly $1\,\mathrm{g}$ at one radius, and it varies linearly as you move away from that point. It is up to you do determine how much slop you are happy with, and from there figure out the necessary size of the craft.
answered May 27, 2017 at 17:38
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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$\begingroup$ Sorry, let me be clearer. Yes, it is a cylindrical spaceship. Make one too small and the difference from 1g would be noticeable between your head and your feet. How wide would a cylindrical spaceship have to be to go from 1.03 g at the outermost deck to .97 g on the innermost deck in the habitation decks if the total distance between them is three miles vertical? Assume acceleration forward is not an issue. I wish I could do the math myself. But a grade of 206 out of 600 in Calculus proved I'm not able to. Lets not mention my physics grade. $\endgroup$– EricMay 28, 2017 at 13:08
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$\begingroup$ You don't need to do calculus to answer that question with the result that I gave you. I've reduced it to a algebra problem. I'm not going to work the figures for you because that would leave you believing that this is a hard problem when it simply isn't. Centrifugal pseudo-gravity varies linearly with radius. If you want a 3% change from the current value go 3% further from (or close to) the axis. $\endgroup$ May 28, 2017 at 17:42
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$\begingroup$ So to conclude, the thickness can range from $50 \times 1.03=51.5$ miles ,(where you would feel $1.03g$) to $50 \times 0.97=48.5$ (where you would feel $0.97g$). So total thickness of livable spaces 3 miles $\endgroup$– MrBrushyMay 30, 2017 at 13:16