How could you lose 10 pounds in one minute on a rotating space station?

Question

The space station itself is a torus shape, and is designed so that one gee of gravity will be along the outer edge, and zero gees will be in the middle. Seems like it'd be fairly simple, but the answer's been eluding me. >.>
Oh, and you can't amputate any body parts. ;)

The space station itself is a torus shape, and is designed so that one gee of gravity will be along the outer edge, and zero gees will be in the middle. Seems like it'd be fairly simple, but the answer's been eluding me. >.>
Oh, and you can't amputate any body parts. ;)

EDIT: I know you'd move closer to the center of the torus, but is there any way to calculate how much closer? Like, a percentage of the ratio or something along those lines?

Answer 1

you are correct that you move towards the centre.
To calculate how far you need to look at Centripetal force - (Centrifugal force is the apparent affect of Centripetal force)

centripetal acceleration can be written in terms of the angular velocity (w)
C = w^2.r
angular velocity w=velocity/radius
C = v^2/r
Force = mass * acceleration
Centripetal Force Cf = (mass * velocity squared )/ radius
ie. Cf =(m.v^2)/r or Cf = mass * angularvelocity^2 * r
as you can see there are 2 key terms - the radius and the velocity . both are needed to work out the ratio change

so if at the edge of the torus at radius r1 you experience a force Cf1 and at an inner radius r2 you experience a force Cf2

then Cf1-Cf2 = m.w^2 *( r1 - r2)

to convert the 10 pounds to Force, convert to Newtons
10lbs is 44.48 Newtons
if the person is 70kg or 154.4lbs
(there are 2.205 pounds per kilogram)

44.48 = 70* w^2 * (r1 - r2)


lets decide on an angular velocity of 0.1 radian per second
this is 5.7 degrees per second or 0.96 revolutions per minute


as you specified 1 gee we can calculate r1
1 gee is 9.8 m/s - so 1kg weighs 9.8 netwons

r1 = C/w^2 = 980 m

note that the velocity v = w * r = 98 m/s = approx 352 kmh

r1 - r2 = 44.48/(70 * 0.1^2)
r1 - r2 = 64

ie you move in 64 meters to change your weight by 10lbs in this example. so if you want to lose the weight in 1 minute you have to move at just over 1 meter per second towards the interior of the space station.

there is a good reference on this discussing the max rotational speeds tolerated by humans. http://www.spacefuture.com/archive/artificial_gravity_and_the_architecture_of_orbital_habitats.shtml

Answer 2

Climb closer to the center of the torus. You'd have to calculate the distance for the centrifugal force to reduce your apparent weight. Since the formula says the force varies with R (distance to center), you'd have to calculate what percentage 10 pounds is out of your weight at 1 gee and move that percentage of R closer to the middle

Answer 3

Moving closer to the center of rotation will decreasse the "gravity", but there is another way. If you can run at the correct speed you can run opposite to the rotation. This will remove some of the centripedal force, or gravity, from your body and it will not be rotating at the rate that creates one gravity.

As they say in the text books, I'll leave the computation of the correct rotation reduction necessary to cause a 10 pound loss as an exercise for the student.

In other words, I have no clue.

Answer 4

Move closer to the center, in such a way that the decrease in gee times your body mass equals 10 pounds.

If you want the full math:

10 pounds=44.5N

Your decrease in acceleration has to be

44.5N / m where m is your mass

The acceleration at the edge is

w^2 r = g (1)

Now you have to go to a distance "d" from the center such that

w^2 d = g - 44.5N / m (2)

Dividing (2) / (1)

d/r = 1 - 44.5N / (m g)

Since m g is wour weight (measured in Newtons)

d / r = 1 - 44.5N / f (where f is your weight in N)

The quantity d / r is a measure of the percentage change in your distance to the center, relative to the radius of the torus to get the desired change in weight.

Answer 5

Since centripetal acceleration is causing the g forces, why not just look at that formula? DUH!

Ac = rω²

Since ω² is constant, the g force is directly proportional to the radius. You figure it from there................

wires has an interesting concept, and a correct one at that. In this case, the g force would be proportional to the factor (Vs -V)²/Vs² where Vs is the velocity of the rim and V is the velocity of running in the opposite direction to the station rim.

If you wanted more exercise, youcould run the same direction and be extra heavy!

Answer 6

HAHA i literally just scared everyone in the room by laughing. well im guessing mr poo had to be the size of my shih tzu :)