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 Fg = G m M / R2 where G is Newton's gravitational constant, m and M are the masses of the gravitating bodies, and R is the distance separating their centers. Tidal forces result from imperfect cancellation of centrifugal and gravitational forces a distance L away from the center of gravity of the system and have the form Ft = G m M L / R3. The tidal force is therefore an "inverse cube law" force.
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 r3). Then the tidal force has the form Ft = (pi/6) G m L d (2r/R)3, so that Ft ~ d (2r/R)3. The factor 2r/R is the angular diameter of the object in the sky, i.e., the angle in radians between the lines of sight to the left edge of the object and to the right edge of the object. This provides several useful rules of thumb concerning tidal influences: (1) if two objects have the same angular diameter and the same density, they should have the same tidal influence; (2) the tidal influence of objects with different angular diameter should depend on the cube of the ratio of angular diameters: twice the angular diameter gives to eight times the tidal influence; and (3) the tidal influence of objects with different densities should depend directly on the ratio of densities.
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 0.533o, that of the Moon is 9.04 milliradians or 0.518o. 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/cm3 while the mass density of the Sun is 1.41 gm/cm3. 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/cm3 and in Earth's sky at closest approach has an angular diameter of 0.227 milliradians. Venus has a density of 5.24 gm/cm3 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/R3, when R becomes small and d becomes large the tidal forces can become very large indeed. A 6 solar mass black hole would have an event horizon with a radius of just 27 km and an density of about 1.45 × 1014 gm/cm3. If we assume that a 6 foot tall human being could survive a head-to-toe stretching force of about 100 pounds, he could survive at a distance of about 5300 km but would be literally pulled apart by tidal forces if he ventured much closer.
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.