If the various true satellites of the Solar System (see Chapter 7) keep one face to their primaries at all times, as is very likely true, their sidereal day would be equal to their period of revolution about their primary.

If this is so I can prepare a table (not quite like any I have ever seen) listing the sidereal period of rotation for each of the 32 major bodies of the Solar System: the Sun, the Earth, the eight other planets (even Pluto, which has a rotation figure, albeit an uncertain one), the Moon, and the 21 other true satellites. For the sake of direct corn parison I'll give the period in minutes and list them in the order of length. After each satellite I shall put the name of the primary in parentheses and give a number to represent the position of that satellite, counting outward from the primary.

Sidereal Day

Body (minutes)

Venus 324,000

Mercury 129,000

Iapetus (Satum-8) 104,000

Moon (Earth-1) 39,300

Sun 35,060

Hyperion (Saturn-7) 30,600

Callisto (Jupiter-5) 24,000

Titan (Satum-6) 23,000

Oberon (Uranus-5) 19,400

Titania (Uranus-4) 12,550

Ganymede (Jupiter-4) 10,300

Pluto 8650

Triton (Neptune-1)

Rhea (Saturn-5) 6500

Umbriel (Uranus-3) 5950

Europa (Jupiter-3) 5100

Dione (Satum-4) 3950

Ariel (Uranus-2) 3630

Tethys (Saturn-3) 2720 lo (Jupiter-2) 2550

Miranda (Uranus-1) 2030

Enceladus (Saturn-2) 1975

Deimos (Mars-2) 1815

Mars 1477

Earth 1436

Mimas (Satum-1) 1350

Neptune 948

Amaltheia (Jupiter-1) 720

Uranus 645

Saturn 614

Jupiter 590

Phobos 460

These figures-represent the time it takes for stars to make a complete circuit of the skies from the frame of reference of an observer on the surface of the body in question. If you divide each figure by 720, you get the number of minutes it would take a star (in the region of the body's celestial equator) to travel the width of the Sun or Moon as seen from the Earth.

On Earth itself, this takes about 2 minutes and no more, believe it or not. On Phobos (Mars's inner satellite), it takes only a little over half a minute. The stars will be whirling by at four times their customary rate, while a bloated Mars hangs motionless in the sky. What a sight that would be to see.

On the Moon, on the other hand, it would take 55 minutes for a star to cover the apparent width of the Sun.

Heavenly bodies could be studied over continuous sustained intervals nearly thirty times as long as is possible on the Earth. I have never seen this mentioned as an advantage for a Moon-based telescope, but, combined with the absence of clouds or other atmospheric, interference, it makes a lunar observatory something for which astron omers ought to be willing to undergo rocket trips.

On Venus, it would take 450 minutes or 7'h hours for a star to travel the apparent width of the Sun as we see it. What a fix astronomers could get on the heavens there - if only there were no clouds.

Almost every book on astronomy I have ever seen, large or small, contains a little table of the Solar System. For each planet, there's given its diameter, its distance from the sun, its time of rotation, its albedo, its density, the number of its moons, and so on.

Since I am morbidly fascinated by numbers, I jump on such tables with the perennial hope of finding new items of information. Occasionally, I am rewarded with such things as surface temperature or orbital velocity, but I never really get enough.

So every once in a while' when the ingenuity-circuits in my brain are purring along with reasonable smoothness, I deduce new types of data for myself out of the material on band, and while away some idle hours. (At least I did this in the long-gone days when I had idle hours.)

I can still do it, however, provided I put the results into formal essay-form; so come join me and we will just moon around together in this fashion, and see what turns up.

Let's begin this way, for instance…

According to Newton, every object in the universe attracts every other object in the universe with a force (i) that is proportional to the product of the masses (ml and M2) of the two objects divided by the square of the distance (d) between them, center to center. We multiply by the gravitational constant (g) to convert the propor tionality to an equality, and we have-. f = 9MIM2 (Equation 1) d2

This means, for instance, that there is an attraction be tween the Earth and the Sun, and also between the Earth and the Moon, and between the Earth and each of the various planets and, for that matter, between the Earth and any meteorite or piece of cosmic dust in the heavens.

Fortunately, the Sun is so overwhelmingly massive corn 'Pared with everything else in the Solar System that in calculating the orbit of the Earth, or of any other planet, an excellent first approximation is attained if only the planet and the Sun are considered, as though they were alone in the Universe. The effect of other bodies can be calculated later for relatively minor refinements.

In the same way, the orbit of a satellite can be worked out first by supposing that it is alone in the Universe with its primary.

It is at this point that something interests me. If the Sun is so much more massive than any planet, shouldwt it exert a considerable attraction on the satellite even though it is at a much greater distance from that satellite than the primary is? If so, just how considerable is "considerable"?

To put it another way, suppose we picture a tug of war going on for each satellite, with its planet on one side of the gravitational rope and the Sun on the other. In this tug of war, how well is the Sun doing?

I suppose astronomers have calculated such things, but I have never seen the results reported in any astronomy text, or the subject even discussed, so I'll de it for myself.

Here's how we can go about it. Let us call the mass of a satellite m, the mass of its primary (by which, by the way, I mean the planet it circles) m,, and the mass of the Sun m.. The distance from the satellite to its primary will be d, and the distance from the satellite to the Sun will be d.. The gravitational force between the satellite and its primary would be J, and that between the satellite and the Sun would be fg-and that's the whole business. I promise to use no other symbols in this chapter.

From Equation 1, we can say that the force of attraction between a satellite and its primary would be:

fp = gmmp (Equation 2) while that between the same satellite and the Sun would be: gmm f,, = ds2 (Equation 3)

What we are interested in is how the gravitational force between satellite and primary compares with that between satellite and Sun. In other words we want the ratio which we can call the "tug-of-war value." To get that we must divide equation 2 by equation 3. The result of such a division would be: fl,/f. = (M,/M.) (d./d,) 2 (Equation 4)

In making the division, a number of simplications have taken place. For one thing the gravitational constant has dropped out, which means we won't have to bother with an inconveniently small number and some inconvenient units. For another, the mass of the satellite has dropped out. (In other words, in obtaining the tug-of-war value, it doesn't matter how big or little a particular satellite is.

The result would be the same in any case.)

What we need for the tug-of-war value is the ratio of the mass of the planet to that of the sun (mlm,,) and the square of the ratio of the distance from satellite to Sun to the distance from satellite to primary (dld,)2.

There are only six planets that have satellites and these, in order of decreasing distance from the Sun, -are: Nep tune, Uranus, Saturn, Jupiter, Mars, and Earth. (I place Earth at the end, instead of at the beginning, as natural chauvinism would dictate, for my own reasons. YouR find out.)

For these, we will first calculate the mass-ratio and the results turn out as follows:

Neptune 0.000052

Uranus 0.000044

Saturn 0.00028

Jupiter 0.00095

Mars 0.00000033

Earth 0.0000030

As you see, the mass ratio is really heavily in favor of the Sun. Even Jupiter, which is by far the most massive planet, is not quite one-thousandth as massive as the Sun.

In fact, all the planets together (plus satellites, planetoids, comets, and meteoric matter) make up, no more than 1/750 of the mass of the Sun.

So far, then, the tug of war is all on the side of the Sun.

However, we must next get the distance ratio, and that favors the planet heavily, for each satellite is, of course, far closer to its primary than it is to the Sun. And what's more, this favorable (for the planet) ratio must be squared, making it even more favorable, so that in the end we can be reasonably sure that the Sun will lose out in the tug of war. But we'll check, anyway.

Let's take Neptune first. It has two satellites, Triton and Nereid. The average distance of each of these from the Sun is, of necessity, precisely the same as the average dis tance of Neptune from the Sun, which is 2,797,000,000 miles. The average distance of Triton from Neptune is, however only 220,000 miles, while the average distance of Nerei@ from Neptune is 3,460,000 miles.

If we divide the distance from the Sun by the distance frbm Neptune for each satellite and square the result we get 162,000,000 for Triton and 655,000 for Nereid. We multiply each of these figures by the mass-ratio of Neptune to the Sun, and that gives us the tug-of-war value, which is:

Triton 8400

Nereid 34

The conditions differ markedly for the two satellites.

The gravitational influence of Neptune on its nearer satel lite, Triton, is overwhelmingly greater than the influence of the Sun on the same satellite. Triton is @y in Nep tune's grip. The outer satellite, Nereid, however, is at tracted by Neptune considerably, but not overwhelmingly, more strongly than by the Sun. Furthermore, Nereid has a highly eccentric orbit, the most eccentric of any satellite in the system. It approaches to within 800,000 miles of Neptune at one end of its orbit and recedes to as far as 6 million miles at the other end. When most distant from Neptune, Nereid experiences a tug-of-war value as low as For a variety of reasons (the eccentricity of Nereid's orbit, for one thing) astronomers generally suppose that Nereid is not a true satellite of Neptune, but a planetoid captured by Neptm on the occasion of a too-close ap, proach.

Neptune's weak hold on Nereid certainly seems to sup port this. In fact, from the long astronomic view, the asso ciation between Neptune and Nereid may be a temporary one. Perhaps the disturbing effect of the solar pull will eventually snatch it out of Neptune's grip. Triton, on the other hand, will never leave Neptune's company short of some catastrophe on tL System-wide scale.