# Space Station Physics

1. Mar 8, 2005

### LHS Students

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. Mar 8, 2005

3. Mar 8, 2005

### kevinalm

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

4. Mar 8, 2005

### scholzie

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
5. Mar 8, 2005

### xanthym

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:

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

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

$$(2) \ \ \ \ V_{ (r-2) } = \frac { 2 \pi (r - 2) } {T}$$

$$(3) \ \ \ \ V_{ (r) } = \frac { 2 \pi (r) } {T}$$

so that:

$$(4) \ \ \ \ \frac { V_{ (r-2) }^{ 2 } } { r - 2 } = \frac { 4 \pi^{2} (r - 2) } { T^{ 2 } }$$

$$(5) \ \ \ \ \frac {V_{ (r) }^{ 2 } } { r } = \frac { 4 \pi^{2} (r) } { T^{ 2 } }$$

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

$$(6) \ \ \ \ (Ratio.Feet.To.Head) = \frac { r - 2 } { r }$$

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

$$(7) \ \ \ \ (Ratio.Feet.To.Head) = \frac { r - 2 } { r } = \frac {99} {100}$$

$$(8) \ \ \ \ (100)(r - 2) = (99)(r)$$

$$(9) \ \ \ \ \Longrightarrow \color{red} (Minimum.Acceptable."r") = (200 \ \ meters)$$

~~

Last edited: Mar 8, 2005
6. Mar 8, 2005

### tony873004

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

7. Mar 8, 2005

### LHS Students

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.

8. Mar 8, 2005

### scholzie

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.

9. Mar 8, 2005

### kevinalm

So did I. Hmm...

Here's a more direct solution.

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