How to Calculate Artificial Gravity in a Rotating Space Station?

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SUMMARY

This discussion focuses on calculating the necessary parameters to achieve artificial gravity of 9.81 m/s² in a rotating space station. A diameter of 224 meters and an angular speed of 2 revolutions per minute (0.21 RAD/s) are proposed to create this gravity. The calculations confirm that a person weighing 70 kg experiences a centrifugal acceleration of approximately 9.88 m/s², which is nearly equal to Earth's gravity. The discussion also addresses the implications of jumping within the rotating environment and the relevance of the barrel's weight in the calculations.

PREREQUISITES
  • Understanding of angular velocity and its units (RAD/s)
  • Familiarity with centrifugal and centripetal forces
  • Basic knowledge of Newton's laws of motion
  • Ability to perform calculations involving tangential velocity and acceleration
NEXT STEPS
  • Learn about the effects of varying radius on centrifugal acceleration
  • Research the engineering requirements for constructing a rotating space station
  • Explore the dynamics of jumping in a rotating frame of reference
  • Investigate the power requirements for maintaining rotation at specified speeds
USEFUL FOR

Aerospace engineers, physicists, and anyone interested in the mechanics of artificial gravity in space environments will benefit from this discussion.

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This is not actually homework, I'm just trying to calculate the speeds/forces needed to create 9,81 m/s2 artificial gravity (outward) within a rotating barrel in space.

Homework Statement


I read on the internet that a space station would require a diameter of 224m and an angular speed of 2 rev/min = 0,21 RAD/s to create an equal to Earth gravity. According to my calculations I think one would need exactly double that, so a radius of 224m. I've assumed a man weighing 70kg is standing in the space station and I want to calculate the acceleration he feels (to see if its equal to 9,81m/s2)

Homework Equations


Tangential velocity Vt=r•ω
Centrifugal force F= ω2•r•m
Centrifugal acceleration a = F/m
Centripetal force F=m•(v2/r)

The Attempt at a Solution


Vt=r•ω=224•0,21=47 m/s
Centrifugal force F= ω2•r•m=0,212•224•70=691,5 N
a=F/m=691,5/70=9,88 m/s2
Centripetal force F=m•(v2/r)=70•(472/224)=690,3 N

1. Are my calculations correct?
2. The centrifugal force is (almost) equal to the centripetal force, does this mean he will stay in place even if he's upside down in the barrel (meaning the artificial gravity works)? 'he is in balance'.
3. What happens if he jumps? Will he have a new 'balance location' in the barrel space since the forces are in balance or will he return to the wall of the barrel (the way gravity works)?
4. In the calculations I didn't take the weight of barrel itself into consideration, is it relevant to do so and if so, how do I go about adding that?
5. Also, how do I go about calculating the power of the engine needed to turn this thing around at 2 rev/min with X amount of total weight, I assume the thing must be crazy powerful.
6. Also, are there any other relevant velocities/accelerations/forces I could calculate?
 
Last edited:
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Don't believe everything you read on the internet. What you need to do is create a force acting outward against the occupant of the barrel equal to 9.8 m/s2. The speed you need to spin an object at is dependent on the radius or diameter you choose for the object. No diameter or radius is the right or wrong one, you have to match your speed to your radius. I'm not going to solve your "not really homework" question. If you want verification of your calculations ask your teacher.
 

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