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.
7. Just Mooning Around
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.