Comment by tonmoy | Hacker News Reader

Comment by tonmoy | Hacker News Reader

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1. It is much easier to land on/get into the orbit of a planet. For example, transfer from low earth orbit to Mars requires a delta V of about 4.3 km/s. Now to get into Mars' orbit, this needs to be reduced to about 3 km/s (rough napkin-top calculation, take with a grain of salt). However if you want to "stop" at a massless orbital platform/asteroid, you would have to reduce all 4.3 km/s (without any aerobrake assist).

2. If we assume an orbital platform is like a ring (similar to Halo), and if the diameter of this thing is 2km, then it would need to spin very fast (70m/s). This is okay, until you realize, that so much speed means a lot of difference in gravity from the floor to ceiling. For example, a man who is 2m tall, would feel 1g on his feet and about 0.95g = 9.3m/s^2 on his head. The difference in g would give rise to different health issues. You could make the station bigger, but then it would be probably better to stick to a natural body that is similar size instead

I believe you're mistaken about the gravity. Position is as Re^(iv/rt), so |acceleration| is as Rv^2/r^2 -- right? For a fixed station of floor radius r, acceleration is directly proportional to distance from the center, R, so with r=1000m, a 2m astronaut will feel 99.8% of normal gravity at their head.

Assuming 5% variation from head to toe is tolerable, a 40m radius habitat would suffice, which could be rotated at 20m/s.

There has actually been research done on how humans tolerate rotational gravity (although not long term, AFAIK.) A 30% gradient at 0.5g is reportedly "disturbing". But perhaps long term one could become acclimated, just as one does at sea.

http://www.artificial-gravity.com/Dissertation/2_2.htm

I thought delta-v from LEO to Mars was over 6km/s? There are near-Earth asteroids at much lower delta-v. https://www.planetaryresources.com/2015/08/how-we-choose-our... gets into some of this.

At the 1km radius you give, a 2m human would have head at 0.2% lesser acceleration than feet. Am I miscalculating? Note that 1g is just one place in the design space, though an attractive one for a big enough station.

I am probably wrong with the 4.3 number (compiled the first result from google without double checking). And it won't be 0.2%, the head's speed would be 0.2% of the feet, but since g=v^2/r, the gravity difference would be about 3%

No, g also equals omega^2 r. The angular velocity omega is the same for head and feet, leaving only the change to r. https://en.wikipedia.org/wiki/Circular_motion#Formulas

(So what's wrong with g = v^2/r? Nothing -- both v and r change in that change to g. To first order, they cancel out in the change to v/r, leaving only a linear factor.)

You are right, I was had made a very ugly mistake I guess

No worries! It's a natural mistake.

You could spin the station at half that speed and still have half a g and mitigate those other issues.

At half the speed, it would be quarter the gravity. And 2km diameter is huge, it would be easier to make a smaller base camp on a planet where there already is gravity present

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