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Spinning for gravity

Forces and motion · level 7, lesson 6 of 6

A question to start

Nobody knows yet whether people can stay healthy for years in Mars's 0.38 g. One idea: a spinning room where people sleep or exercise, giving them their full Earth weight for part of each day. How fast would it spin, and how would the floor have to tilt?

The idea

In space, a spinning station's floor pushes people inwards with \(m\omega^2 r\). That push is what they feel as weight, so the station gives an "artificial gravity" of \(a = \omega^2 r\).

On Mars there is real gravity too, pointing down, while the spin's push points sideways, to the axis. What a person feels is the combination. To feel exactly Earth's \(g_E\):

\[\sqrt{g_M^2 + a^2} = g_E \qquad\Rightarrow\qquad a = \sqrt{9.81^2 - 3.71^2} = 9.08 \text{ m/s}^2.\]

The floor must face along the combined pull, so it tilts like a very steep banked track, at an angle \(\tan\phi = a/g_M = 2.45\), about 68° from level. People inside stand "upright" on the tilted floor and feel perfectly normal, 1 g.

3.719.08 spin9.81 feltraxisfloor, outer edge up

The arrows show what a person feels: Mars's pull down, and an outward pull from the spin. (Really, the floor pushes them inwards to keep them going round; an outward pull is what that push feels like.) The floor faces the combined pull, with its outer edge high, like the banked track in the last lesson but far steeper.

How fast? It depends on the radius. People get dizzy and sick if they turn their heads while spinning fast; studies suggest most people adapt to about 2 to 4 turns a minute, a few to more. So a small room must spin fast, a large one slowly:

Room radius Turns a minute for 9.08 m/s² Time per turn
10 m 9.1 6.6 s
50 m 4.1 14.7 s
200 m 2.0 29.5 s

Worked examples

1. The 10 m room. \(\omega = \sqrt{a/r} = \sqrt{9.08/10} = 0.953\) rad/s. \(T = 2\pi/\omega = 6.6\) s, which is \(60/6.6 = 9.1\) turns a minute.

2. For 2 turns a minute. \(\omega = 2\pi \times 2/60 = 0.209\) rad/s. \(r = a/\omega^2 = 9.08/0.0438 = 207\) m.

3. In space instead. With no planet's gravity, the whole \(9.81\) m/s² must come from the spin, and the floor is flat along the rim. For 2 turns a minute: \(r = 9.81/0.0438 = 224\) m.

Try it

  1. Why does a spinning room on Mars need a tilted floor?
  2. What sideways acceleration is needed to feel 0.8 g in total on Mars?
  3. What floor tilt goes with it?
  4. A spinning ring in orbit, radius 50 m, gives 3.71 m/s² (Mars gravity). How long is each turn?
Answers
  1. Mars's gravity is down, the spin's is sideways. The felt pull is their combination, at an angle, and the floor must face it.
  2. \(\sqrt{7.85^2 - 3.71^2} = 6.92\) m/s².
  3. \(\tan\phi = 6.92/3.71 = 1.86\), \(\phi = 62°\) from level.
  4. \(\omega = \sqrt{3.71/50} = 0.272\) rad/s, \(T = 23\) s.

Common mistake

Adding the accelerations as plain numbers. \(3.71 + 6.10 = 9.81\) looks neat, but the two pulls are at right angles. They add like the sides of a right-angled triangle.

Where it leads