"The Alternate View" columns of John G. Cramer

*Alternate View Column AV-63*

Keywords: gravitational tidal
forces planetary systems

Published in the January-1994 issue of **Analog
Science Fiction & Fact Magazine**;

This column was written and
submitted 7/11/93 and is copyrighted ©1993 by John G. Cramer.

All rights
reserved. No part may be reproduced in any form without

the explicit
permission of the author.

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I have just returned from WesterCon, the annual West Coast SF convention which
is always held over the July 4th weekend. In several discussions I had there
with writers and fans, the subject of *tidal forces* came up. I
realized that this physics idea often arises in SF discussions in connection
with black holes or space construction or "world building" for story
backgrounds. It is also a subject about which there seems to be considerable
confusion. So I decided do this Alternate View column on gravitational tidal
forces. I will use a question-and-answer format for the discussion.

**Q: ***What makes the tides?*

**A:** The Earth, Moon, and Sun circle each other in a free-fall
dance, their orbits a delicate balance of mutual gravitational attraction and
centrifugal force. One might think that in such free fall all the forces would
be completely "used up", but this is not so. The left-over gravitational
forces from the Sun and Moon produce tides that move the waters of Earth's
oceans by dozens of feet and create changing stresses and displacements in the
Earth's crust. These left-overs are called "tidal forces".

**Q: ***What are tidal forces?*

**A:** An object is in a stable free-fall orbit when the
gravitational force acting on it is in precise balance with the centrifugal
force of its orbital motion. The center of mass of the Earth is in such a
free-fall orbit, but at all locations away from the center the gravitational
pull is slightly out of balance with the centrifugal force of the orbital
motion, and there is a small residual force. This is the tidal force. It is
called "tidal" because at the surface of the Earth this residual force moves
the water in the oceans and causes the ocean tides.

**Q: ***In what direction do tidal forces act?*

**A:** To answer this question, consider a space station orbiting
the Earth. The center of the space station is in a correct orbit, and no tidal
forces will act there. Above the center (in the direction away from the Earth)
the gravitational pull has fallen off a bit due to the inverse square law, so
the centrifugal force of the orbit isn't quite balanced by gravity and there is
an *upward* tidal force. Similarly, below the center there is a
*downward* tidal force. At locations away from the center but in the
horizontal plane (the same distance fro Earth as the center) the pull of
gravity is the correct magnitude to balance the centrifugal force, but it
points in a slightly wrong direction, leading to a net force toward the center
of the space station. If one placed a large globe of water at the center of a
space station (perhaps to make a spherical swimming pool) the action of tidal
forces would cause it to assume a slightly prolate or cigar shape, squeezed in
at the equator and stretched along the axis that passes through the center of
the Earth.

**Q: ***How do tidal forces depend on distance?*

**A:** The basic force of gravitational attraction is an "inverse
square law" force. We write **F _{g} = G m M / R^{2}** where

^{}

**Q: ***How does the tidal force depend on the other characteristics of
the object producing it?*

**A:** As shown above, the only other relevant characteristic of the
object producing the tidal force is its mass **M**. However, for a
spherical object with an average density (mass per unit volume) **d** and a
radius **r** we can view its mass as the product of its density and volume,
so that **M = d (4/3 pi r ^{3})**. Then the tidal force has the
form

**Q: ***Why is the Moon's influence on the tides greater than that of the
Sun?*

**A:** The average angular diameter of the Sun in the sky is 9.30
milliradians or 0.533^{o}, that of the Moon is 9.04 milliradians or
0.518^{o}. In other words, the angular diameters of the Sun and Moon
in the sky are almost exactly the same. (The values, of course, vary
periodically depending on the positions of the Earth and the Moon in their
slightly eccentric orbits.) The cosmic accident of nearly-equal angular
diameters makes possible solar eclipses in which the disc of the Moon precisely
covers that of the Sun. On this basis, one would expect the tidal effects of
Sun and Moon to be the same. However, the density of the Moon is 3.34
gm/cm^{3} while the mass density of the Sun is 1.41 gm/cm^{3}.
For this reason the Sun has only about 46% of the Moon's influence on the
tides.

**Q: ***What about the influence of other planets on Earth's tides? What
about the "Jupiter Effect"?*

**A:** The planet Jupiter has a density of 1.36 gm/cm^{3
}and in Earth's sky at closest approach has an angular diameter of 0.227
milliradians. Venus has a density of 5.24 gm/cm^{3 }and a
closest-approach angular diameter of 0.292 milliradians. The maximum tidal
influence of Venus is .0053% of that of the Moon and the maximum influence of
Jupiter is .0020%, effects on Earth's tides so small as to be essentially
unobservable.

The angular diameters of the other planets in the sky are even smaller, with consequently tiny tidal effects. The techno-myth that there should be very high tides and earthquakes when the planets are in the same part of the sky, the so-called Jupiter Effect, is therefore pure hokum.

**Q: ***Would tidal forces affect a space station in low-earth orbit
?*

**A:** Yes. There are two important effects for space stations.
First, the tidal forces will make the any elongated object tend toward an orbit
with its long axis pointing to the center of the Earth. Either the space
station has to be designed to orbit in this configuration, or an orientation
correction system must be included in the design to correct the orientation
drift that the tidal forces will produce.

There are also the tidal effects within a space station. About a decade ago, discussion of space-station experiments shifted terminology from "zero-gravity" to "microgravity". This terminology shift was an attempt more accurately reflect the fact that it is not possible to completely eliminate the last vestiges of Earth's gravity in an orbiting space vehicle. The small residual forces can have important effects on low-gravity experiments.

**Q: ***What about tidal effects on Earth-like planets of other star
systems?*

**A:** The cosmic accident of a very large moon is unlikely to be
repeated for earth-like planets elsewhere in the galaxy. More typical moons
will be likely to have small angular diameters in the sky and a correspondingly
diminished influence on the tides. Thus, the tides of such planets are likely
to be dominated by the influence of the local sun. Curiously, the bigger and
hotter the star is, the less will be its tidal influence. This is because in
order to be earth-like, a planet must be at a distance from the parent star
that gives it about the same amount of solar radiation intercepted by the
Earth. The radiation falls off as the inverse square of distance while the
tidal effects fall off as the inverse cube.

Thus, a very hot star would be rather far away from its earth-like planet and have a correspondingly smaller influence on its tides. For example, the tidal effect of Sirius A, a large hot type-A1 star, on an earth-like planet would be only 1.2% of the tidal influence of the Moon on the Earth. Similarly, Procyon A, type-F5, would have only 4.3% of the Moon's tide influence.

On the other hand, the tidal effects become much stronger for smaller cooler stars. Tau Ceti, type-G8, would exert 1.20 times the Moon's tidal influence on an earth-like planet. Epsilon Indi, type-K5, would exert 4.95 times the Moon's tidal influence on an earth-like planet. And Barnard's Star, a type-M5 star 5.9 light-years away and well publicized by the Hitchhiker's Guide books, would exert 377,000 times the Moon's tidal influence on an earth-like planet.

Such a huge tidal force would require that the planet would be tide-locked to its primary as Mercury is to the Sun and the Moon is to the Earth keeping the same face to its primary as it orbits it. Such a tide-locked planet would not very earth-like. In this case, there would be no tidal stresses as long as the orbit were perfectly circular, but there might be considerable tidal heating of the planet if it were in an elliptical orbit, as is the case with some of the moons of Jupiter.

**Q: ***What about the tidal effects near black holes?*

**A:** Because the tidal force is proportional to
**d/R ^{3}**, when

**Q: ***Is there any way to protect the crew of a space vehicle against
dangerous tidal forces?*

**A:** Dr. Robert Forward described a good technique for doing this
in his novels **Dragon's Egg** and **Starquake**. However, his solution
required an arrangement of six spheres of ultra-dense matter in a tight orbit
around the space station to cancel the severe tidal forces in the vicinity of a
neutron star. We would, unfortunately, have to solve the problem of producing
such matter before this solution would be feasible.

We should also note that tidal forces cannot be eliminated by passing a space vehicle through the symmetry point between a pair of orbiting black holes, as was suggested in at least one hard-SF novel. In that case, the tidal forces would add, not cancel, and their effect would be doubled.

*This page was created by John G. Cramer
on 7/12/96.*