Calculating the relevant variables for a rotating habitat

AI Thread Summary
The discussion focuses on calculating the rotational period and surface velocity of a Bishop Ring with a radius of 80 km and a surface gravity of 0.86 G. The formula used for the calculations is t = 2 * pi * (r/a)^0.5, leading to a rotational period of 612 seconds and a linear velocity of 2957 km/hr. Participants confirm the calculations are correct, noting only minor rounding errors could exist due to significant figures. Additionally, it is mentioned that the period is typically denoted as capital T. Overall, the calculations are validated and deemed accurate.
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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/hrSo 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?
 
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Bizmuth said:

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/hrSo 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:
Thank you!
 
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