1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Space Station Physics

  1. Mar 8, 2005 #1
    Junior level Physics problem I am having trouble with:

    Some plans for a future space station make use of a rotational force to simulate gravity. In order to be effective the centripetal acceleration at the outer rim of the station should equal about 1g, or 98.1 m/s^2. However humans can withstand a difference of only 1/100 g between their head and feet. Assume average human height is 2 m and calculate the minimum radius for a safe effective station. (Hint: The ratio of the centripetal acceleration of astronauts feet to the centripetal acceleration of the astronauts head must be at least 99/100.)

    My whole class is having trouble with this problem all help appreciated.
     
  2. jcsd
  3. Mar 8, 2005 #2
    Please Help! :cry:
     
  4. Mar 8, 2005 #3
    When af (a at your feet) =1.00*g = v^2/r
    then ah = 0.99*g =v^2/(r-2)

    so ah/af = 0.99 =(v^2/(r-2)) / v^2/r
     
  5. Mar 8, 2005 #4
    Use a centripital acceleration* equation to find the force of a large spinning circle, and one with a radius two meters smaller. That will give you two related equations.

    Then set them proportionally equal to eachother (ie, 99a = 100b ). Solve for your unknowns.

    *edit: acceleration, not force :)

    PS, if you know the answer, post it here. I want to know if I got it right :) It's a nice problem.
     
    Last edited: Mar 8, 2005
  6. Mar 8, 2005 #5

    xanthym

    User Avatar
    Science Advisor

    The Centripetal Acceleration of an object moving with Velocity "V" in a circle of Radius "r" is given by {V2/r}. Thus, for the spinning Space Station with an Outer Radius "r", the ratio of Centripetal Acceleration at {r - (2 meters)} where an astronaut's feet might be to the Centripetal Acceleration at {r} where an astronaut's head might be is given by the following fraction:

    [tex] (1) \ \ \ \ (Ratio.Feet.To.Head) = \left ( \frac { V_{ (r-2) }^{ 2 } } { r - 2 } \right ) / \left ( \frac {V_{ (r) }^{ 2 } } { r } \right ) \ \ = \ \ \frac { \frac { V_{ (r-2) }^{ 2 } } { r - 2 } } { \frac {V_{ (r) }^{ 2 } } { r } } [/tex]

    For rotational Period "T" (that is, the time in seconds for 1 complete rotation), we have:

    [tex] (2) \ \ \ \ V_{ (r-2) } = \frac { 2 \pi (r - 2) } {T} [/tex]

    [tex] (3) \ \ \ \ V_{ (r) } = \frac { 2 \pi (r) } {T} [/tex]

    so that:

    [tex] (4) \ \ \ \ \frac { V_{ (r-2) }^{ 2 } } { r - 2 } = \frac { 4 \pi^{2} (r - 2) } { T^{ 2 } } [/tex]

    [tex] (5) \ \ \ \ \frac {V_{ (r) }^{ 2 } } { r } = \frac { 4 \pi^{2} (r) } { T^{ 2 } } [/tex]

    Placing Eqs #4 & #5 into Eq #1 and simplifying:

    [tex] (6) \ \ \ \ (Ratio.Feet.To.Head) = \frac { r - 2 } { r } [/tex]

    We require the above ratio to be at least 99/100, so the Minimum acceptable "r" would be given by:

    [tex] (7) \ \ \ \ (Ratio.Feet.To.Head) = \frac { r - 2 } { r } = \frac {99} {100} [/tex]

    [tex] (8) \ \ \ \ (100)(r - 2) = (99)(r) [/tex]

    [tex] (9) \ \ \ \ \Longrightarrow \color{red} (Minimum.Acceptable."r") = (200 \ \ meters) [/tex]


    ~~
     
    Last edited: Mar 8, 2005
  7. Mar 8, 2005 #6

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    I also get 200 meters

    Code (Text):
    innerdiameter = 1
    outerdiameter = innerdiameter + 2
    Do
        velocity = Sqr(9.8 * innerdiameter)
        outeracceleration = velocity ^ 2 / outerdiameter
        innerdiameter = innerdiameter + 1
        outerdiameter = innerdiameter + 2
    Loop Until outeracceleration > 9.8 * 0.99
    answer = (innerdiameter + outerdiameter) / 2
     
  8. Mar 8, 2005 #7
    Thank you all very much for your quick responses. Using scholzie post I got an answer of 200 but I found xanthym easier too follow.
     
  9. Mar 8, 2005 #8
    Good, I also got 200m. I was trying to give you an idea of how to do it without giving you the answer :) Glad to see it worked out though.
     
  10. Mar 8, 2005 #9
    So did I. Hmm...

    Here's a more direct solution.

    w=v/r rad/s

    af=w^2*r
    ah=w^2*(r-2)

    ah/af=.99=(r-2)/r

    af = 1.00*g =w^2*200

    To be honest I'm not entirely sure why my first method worked. I think I subconsiously realized the limiting factor was r. Or maybe I was just lucky.
     
    Last edited: Mar 8, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Space Station Physics
  1. Space Station (Replies: 1)

  2. Space station (Replies: 10)

  3. Space station problem (Replies: 4)

Loading...