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Calculating the relevant variables for a rotating habitat

  • Thread starter Bizmuth
  • Start date
  • #1
31
9

1. Homework Statement


I want to calculate the rotational period and surface velocity of a Bishop Ring with a radius of 80 km and a surface 'gravity' of 0.86 G

I should note that this isn't for homework. It's for personal interest. I'm looking for a sanity check on this. The homework forum just seemed the best place for this type of question.


Homework Equations



From https://en.wikipedia.org/wiki/Artificial_gravity the formula to calculate the rotational period for a given radius and desired gravity is

t = 2 * pi * (r/a)^.5

where t is time, r is radius of the habitat, and a is desired acceleration (gravity)

The Attempt at a Solution


[/B]
0.86 G = 8.428 m/s^2

2 * pi * (80,000/8.428)^.5 = 612 seconds

circumference = 2 * pi * 80 km = 502.65 km

linear velocity = 502.65 / 612 = 0.82 km/s = 2957 km/hr


So the habitat will rotate once every 612 seconds, and the velocity of the rim will be 2957 km/hr

Did I brain-fart anywhere in here?
 

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
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Homework Statement



I want to calculate the rotational period and surface velocity of a Bishop Ring with a radius of 80 km and a surface 'gravity' of 0.86 G

I should note that this isn't for homework. It's for personal interest. I'm looking for a sanity check on this. The homework forum just seemed the best place for this type of question.


Homework Equations



From https://en.wikipedia.org/wiki/Artificial_gravity the formula to calculate the rotational period for a given radius and desired gravity is

t = 2 * pi * (r/a)^.5

where t is time, r is radius of the habitat, and a is desired acceleration (gravity)

The Attempt at a Solution


[/B]
0.86 G = 8.428 m/s^2

2 * pi * (80,000/8.428)^.5 = 612 seconds

circumference = 2 * pi * 80 km = 502.65 km

linear velocity = 502.65 / 612 = 0.82 km/s = 2957 km/hr


So the habitat will rotate once every 612 seconds, and the velocity of the rim will be 2957 km/hr

Did I brain-fart anywhere in here?
It looks good to me. :smile:

There is the possibility of a very minor rounding error in your work. But given your choice of g = 9.8 m/s2, which has only two significant figures, any such rounding errors would be pretty meaningless anyway.

Oh, and one last thing, the period -- the amount of time for a single rotation -- is typically denoted as capital T rather than lowercase t.

Btw, nice work!
 
Last edited:
  • #3
31
9
Thank you!
 

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