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In the classic science fiction 2001: Space Odyssey (written in 1968 before men stepped on the Moon) by Arthur C. Clarke, the artificial gravities on the Space Station One and the spaceship Discovery are generated with the centrifugal force. Some technical figures of the centrifuges being used are given in the novel.

I wonder whether the figures given are realistic in that they satisfy the laws of physics in our universe, or are simply fictional. I believe that they should be realistic or the author won't care to give such figures to 'challenge' the readers. Anyway, I still carried out the calculation to verify them. Here it goes.

1. Centrifugal Force and Artificial Gravity

You can read the Wikipedia entry on centrifugal force to learn a lot about it. To create an artificial gravity as described in the novel, we use a rotating disk or drum. For our calculations, we need to know the following formula:

\[ F=ma_\mathrm{centrifugal}=m \omega^2 r \]

Where $m$ is the mass of the object that experiences the force, $a_\mathrm{centrifugal}$ is the centrifugal acceleration, $\omega$ and $r$ are the angular velocity and the radius of the rotating disk or drum respectively. Therefore, to generate an artificial gravity that will accelerate free-falling objects by an amount of $a_\mathrm{gravity}$, we need to rotate the disc or drum such that it generates a centrifugal acceleration that is equal to

\[ a_\mathrm{gravity}=a_\mathrm{centrifugal}= \omega^2 r \]

In other words, to generate an acceleration of $a_\mathrm{centrifugal}$, we need to rotate the disk with an angular velocity of

\[ \omega=\sqrt{\frac{a_\mathrm{centrifugal}}{r}} \]

2. Artificial Gravity on Space Station One

We will now check if the figures given in the novel satisfy this equation. Note that we are generally satisfied with three significant digits during the calculations.

2.1. Gravities on Earth and on Moon

We will use the following figures in the calculations:

• Acceleration due to Earth's gravity $a_\mathrm{Earth} = g = 9.81\mathrm{m}\mathrm{s}^{-2}$. (Refer to acceleration due to gravity [en.wikipedia.org].)
• Acceleration due to Moon's gravity $a_\mathrm{Moon} = 1.62\mathrm{m}\mathrm{s}^{-2}$. (Refer to Moon [en.wikipedia.org].)

2.2. Artificial Gravity on Space Station One

The figures needed for Space Station One are described in the following parts of the novel:

• "... A few minutes later he caught his first glimpse of Space Station One, only a few miles away. The sunlight glinted and sparkled from the polished metal surfaces of the slowly revolving, three-hundred-yard-diameter disk. ..."
  — From Chapter 8: Orbital Rendezvous, or page 52 in the millennial edition of the book (ISBN: 0-451-45799-4) published by Roc. (I'm assuming this edition from now on.)
• "... Space Station One revolved once a minute, and the centrifugal force generated by this slow spin produced an artificial gravity equal to the Moon's. ..."
  — From Chapter 9: Moon Shuttle, or page 58.

So we know that the radius of the rotating disk is

\[ 300/2\ \mathrm{yard} = 150 \cdot 0.9144\ \mathrm{m} = 137.16\ \mathrm{m} \]

To generate an aritificial gravity equal to the Moon's, we need

\[ \omega_\mathrm{Moon} = \sqrt{\frac{a_\mathrm{Moon}}{r}} = \sqrt{\frac{1.62}{137.16}}\ \mathrm{rad}\cdot\mathrm{s}^{-1} = 0.1087\ \mathrm{rad}\cdot\mathrm{s}^{-1} = 0.1087\ \cdot \frac{60}{2\pi}\ \mathrm{rpm} = 1.038\ \mathrm{rpm} \]

So we need 1.038 rpm, while the novel gives about 1 rpm, which is close enough.

What if we need an artificial gravity equal to the Earth's? We need

\[ \omega_\mathrm{Earth} = \sqrt{\frac{a_\mathrm{Earth}}{r}} = \sqrt{\frac{9.81}{137.16}}\ \mathrm{rad}\cdot\mathrm{s}^{-1} = 0.2674\ \mathrm{rad}\cdot\mathrm{s}^{-1} = 0.2674\ \cdot \frac{60}{2\pi}\ \mathrm{rpm} = 2.553\ \mathrm{rpm} \]

There is a faster way to calculate $\omega_\mathrm{Earth}$ if we know $\omega_\mathrm{Moon}$ for the same rotating disk. In particular,

\[ \omega_\mathrm{Earth} = \sqrt{\frac{a_\mathrm{Earth}}{r}} = \sqrt{\frac{a_\mathrm{Earth}}{a_\mathrm{Moon}}} \cdot \sqrt{\frac{a_\mathrm{Moon}}{r}} = \sqrt{\frac{9.81}{1.62}} \cdot \omega_\mathrm{Moon} = 2.46\omega_\mathrm{Moon} \]

2.3. Artificial Gravity on Discovery

The figures needed for Discovery are described in the following part of the novel:

• "... The equatorial region of the pressure sphere—the slice, as it were, from Capricorn to Cancer—enclosed a slowly rotating drum, thirty-five feet in diameter. As it made one revolution every ten seconds, this carrousel or centrifuge produced an artificial gravity equal to that of the Moon. ..."
  — From Chapter 17: Cruise Mode, or page 124.

So we know that the radius of the rotating drum is

\[ 35/2\ \mathrm{feet} = 17.5 \cdot 0.3048\ \mathrm{m} = 5.334\ \mathrm{m} \]

This is considerably smaller than the Space Station One, and thus we expect to get a higher $\omega$. To generate an aritificial gravity equal to the Moon's, we need

\[ \omega_\mathrm{Moon} = \sqrt{\frac{a_\mathrm{Moon}}{r}} = \sqrt{\frac{1.62}{5.334}}\ \mathrm{rad}\cdot\mathrm{s}^{-1} = 0.5511\ \mathrm{rad}\cdot\mathrm{s}^{-1} = 0.5511\ \cdot \frac{60}{2\pi}\ \mathrm{rpm} = 5.263\ \mathrm{rpm} \]

So we need 5.263 rpm, while the novel gives about 6 rpm, which is close enough.

What if we need an artificial gravity equal to the Earth's? We need

\[ \omega_\mathrm{Earth} = 2.46 \omega_\mathrm{Moon} = 2.46 \cdot 5.263\ \mathrm{rpm} = 12.94\ \mathrm{rpm} \]

This equals to making a revolution in 4.64 seconds, which looks quite fast for such a big drum.

3. Remarks

Although using a centrifuge to generate an artificial gravity seems feasible, it is not an ideal solution. Human beings have evolved to live best under Earth's gravity. Note that we may not live comfortably under the aritificial gravity generated with centrifugal force. For example, it is suggested that the Coriolis effect due to the rotation may cause dizziness and disorientation.

In the novel, Arthur C. Clarke wrote that "... it had been discovered, [artificial gravity equal to the Moon's] was a good compromise between Earth gravity and no gravity at all; ...", and "... [An artificial gravity equal to the Moon's] was enough to prevent the physical atrophy which would result from the complete absence of weight, and it also allowed the routine functions of living to be carried out under normal—or nearly normal—conditions. ..." Just note that as of this writing, scientists are still doing research on this topic and do not know the exact effects yet. For example, the Mars Gravity Biosatellite Program has been initiated to study the effects of Martian gravity on mammals.

Nonetheless, the calculations above show that the author did write this part of the novel carefully. Besides, the calculations here involve only high-school physics and should be quite educational. Has any textbook included this to attract the students?