This polyhedral image was created using Stella 4d, a program you can try for yourself, for free, at http://www.software3d.com/Stella.php.
This polyhedral image was created using Stella 4d, a program you can try for yourself, for free, at http://www.software3d.com/Stella.php.
I propose that 384,400 km (238,855 miles), the average distance from the Earth to the Moon, be called a “moon unit.” Example: “The mileage of my car is over one moon unit.”
[Image found here.]
It is no secret than I am not a fan of our current president, Donald Trump. I’ve been watching him carefully, and have found exactly one point of agreement with the man: humans should colonize the planet Mars. The two of us differ, however, on the details. What follows is my set of reasons — not Trump’s — for supporting colonization of Mars.
First, we should not start with Mars. We should start, instead, by establishing a colony on Luna, our own planet’s moon. There are several reasons for this. First, as seen in this iconic 1969 photograph brought to us by NASA, we’ve been to the Moon before; it simply makes sense to start space-colonization efforts there.
At its furthest distance, the Moon is ~405,000 km away from Earth’s center, according to NASA. By contrast, at its closest approach to Earth in recent history, Mars was 55,758,006 km away from Earth. With the Moon less than 1% as far away as Mars at closest approach, Luna is the first logical place for an extraterrestrial colony. It need not be a large colony, but should at least be the size of a small town on Earth — say, 100 people or so. There are almost certainly problems we haven’t even discovered — yet — about establishing a sustainable reduced-gravity environment for human habitation; we already know about some of them, such as muscular atrophy and weakening of bones. Creating a lunar colony would demand of us that we solve these problems, before the much more challenging task of establishing a martian colony. (To find out more about such health hazards, this is a good place to start.) Once we have a few dozen people living on the Moon, we could then begin working in earnest on a martian colony, with better chances for success because of what we learned while colonizing the Moon.
An excellent reason to spend the billions of dollars it would take to colonize Mars (after the Moon) is that it is one of the best investment opportunities of the 21st Century. Space exploration has a fantastic record of sparking the development of new technologies that can help people anywhere. For example, the personal computers we take for granted today would not be nearly as advanced as they are without the enormous amount of computer research which was part of the “space race” of the 1960s. The same thing can be said for your cell phone, and numerous other inventions and discoveries. Even without a major space-colonization effort underway, we already enjoy numerous health benefits as a result of the limited exploration of space we have already undertaken. Space exploration has an excellent track record for paying off, big, in the long run.
Another reason for us to colonize Mars (after the Moon, of course) is geopolitical. The most amazing thing about the 20th Century’s Cold War is that anyone survived it. Had the United States and the Soviet Union simply decided to “nuke it out,” no one would be alive to read this, nor would I be alive to write it. We (on both sides) survived only because the USA and the USSR found alternatives to direct warfare: proxy wars (such as the one in Vietnam), chess tournaments, the Olympics, and the space race. In today’s world, we need safe ways to work out our international disagreements, just as we did then. International competition to colonize space — a new, international “space race” — would be the perfect solution to many of today’s geopolitical problems, particular if it morphs, over the years, into the sort of international cooperation which gave us the International Space Station.
Finally, there is the best reason to establish space colonies, and that is to increase the longevity of our species, as well as other forms of life on Earth. Right now, all our “eggs” are in one “basket,” at the bottom of Earth’s gravity well, which is the deepest one in the solar system, of all bodies with a visible solid surface to stand on. A 10-kilometer-wide asteroid ended the age of the dinosaurs 65 million years ago, and there will be more asteroid impacts in the future — we just don’t know when. We do know, however, that past and present human activity is causing significant environmental damage here, so we may not even need the “help” of an asteroid to wipe ourselves out. The point is, the Earth has problems. The Moon also has problems, as does the planet Mars — the two places are far from being paradises — but if people, along with our crops and animals, are located on Earth, the Moon, and Mars, we have “insurance” against a global disaster, in the form of interplanetary diversification. This would allow us to potentially repopulate the Earth, after the smoke clears, if Earth did suffer something like a major asteroid impact.
Since Moon landings ended in the 1970s, we’ve made many significant discoveries with space probes and telescopes. It’s time to start following them with manned missions, once again, that go far beyond low-Earth orbit. There’s a whole universe out there; the Moon and Mars could be our first “baby steps” to becoming a true spacefaring species.
[Later edit: Please see the first comment, below, for more material of interest added by one of my readers.]
During the Cold War, the usual way nations compete (direct warfare) was taken off the table by the invention of the hydrogen bomb. With the alternative being mutually-assured destruction, the two sides, led by the USA and the USSR, had to find other ways to compete. Some of those ways were harmful, such as proxy wars, as happened in Vietnam. Others, however, were helpful, such as the space race. The United States put men on the Moon in order to beat the Soviet Union there, as this iconic 1969 photograph makes evident (source: NASA).
We are all still reaping the benefits of the technological and scientific advances made during this period, and for this purpose. The most obvious example of such a benefit is the computer you are using to read this blog-post, for computer technology had to be advanced dramatically, on both sides, in order to escape the tremendously-challenging gravity-well of the Earth.
Wouldn’t it be wonderful if other conflicts in society took beneficial forms, as happened in this historical example?
This could happen in many ways, but the one that gave me the idea for this post is the conflict between teachers’ unions and school districts’ administrators, now taking place in school districts all over the place. I think it would be awesome if this previously-harmful competition changed, to take a helpful form: book drives, to help school libraries.
Please do not misunderstand, though: I’m not talking about taxpayer money, nor union dues. My idea need not, and should not, affect the budget of any school district, nor union budget. All that need happen is for individual people — teachers and administrators — to go home, look at their own bookshelves, and help students directly, by donating some of their already-paid-for books to school libraries.
While I make no claim to represent any organization, I am a teacher, and a member of the NEA (the National Education Association) in the United States, as well as my state and local NEA affiliates. In an effort to start this new, helpful way to compete, I will give books to the school library where I teach, next week, which is the second week of the new school year. That’s a lot easier than, well, putting men on the Moon.
This is something we can all do. All of us in the education profession, after all, already agree that we want students reading . . . and this is something we can easily do, to work together towards that goal. School libraries need hardcover books which are student-friendly, meaning that they appeal to a young audience, on a wide variety of subjects. Both fiction and non-fiction books are helpful.
Lastly, in the hope that this idea catches on, I will simply point out one fact: helping turn this idea into a reality is as easy as sharing a link to this blog-post.
Google’s search-suggestions for “is the moon,” shown above, clearly indicate support for the “magnet for ignorance” conjecture.
My favorite one from this list: “is the moon real”? I’ve looked into this, and there are apparently quite a few people utterly convinced that the Moon is a hologram, created by NASA, for reasons I have not been able to discern.
This image of binary polyhedra of unequal size was, obviously, inspired by the double dwarf planet at the center of the Pluto / Charon system. The outer satellites also orbit Pluto and Charon’s common center of mass, or barycenter, which lies above Pluto’s surface. In the similar case of the Earth / Moon system, the barycenter stays within the interior of the larger body, the Earth.
I know of one other quasi-binary system in this solar system which involves a barycenter outside the larger body, but it isn’t one many would expect: it’s the Sun / Jupiter system. Both orbit their barycenter (or that of the whole solar system, more properly, but they are pretty much in the same place), Jupiter doing so at an average orbital radius of 5.2 AU — and the Sun doing so, staying opposite Jupiter, with an orbital radius which is slightly larger than the visible Sun itself. The Sun, therefore, orbits a point outside itself which is the gravitational center of the entire solar system.
Why don’t we notice this “wobble” in the Sun’s motion? Well, orbiting binary objects orbit their barycenters with equal orbital periods, as seen in the image above, where the orbital period of both the large, tightly-orbiting rhombicosidodecahedron, and the small, large-orbit icosahedron, is precisely eight seconds. In the case of the Sun / Jupiter system, the sun completes one complete Jupiter-induced wobble, in a tight ellipse, with their barycenter at one focus, but with an orbital period of one jovian year, which is just under twelve Earth years. If the Jovian-induced solar wobble were faster, it would be much more noticeable.
[Image credit: the picture of the orbiting polyhedra above was made with software called Stella 4d, available at this website.]
This is a waxing moon, meaning the sunlit portion we can see is growing. The outer curve also makes this view of the moon shaped more like the letter “D,” compared to the letter “C.” For the useful mnemonic here, remember that “D” stands for “developing.” D-shaped moons are in the waxing part of their cycle of phases, growing larger for about two weeks.
Here is another gibbous moon, but it is shaped more like the letter “C” than the letter “D,” and, in this mnemonic, “C” stands for “concluding.” This moon’s sunlit portion is shrinking, moving away from fullness, towards the new moon state — in other words, it is a waning moon. All “C-shaped” moons, as viewed from Earth’s Northern hemisphere, are waning moons.
This crescent moon more closely resembles a “C” than a “D,” which is how I know, at a glance, that its phase cycle is concluding, and it is a waning crescent, soon to become invisible as a new moon.
This last picture shows the most difficult configuration to figure out: the points of the crescent near the moon’s North and South poles both point up. Having them both point down would pose the same problem. Here’s the solution, though: check to see which crescent-tip appears higher in the sky. In this case, it is the one on the left. That shifts the curve at the bottom of the moon (the one that is an actual moon-edge, rather than the terminator) slightly left-of-center, making the visible moon-edge more closely resemble a “C” than a “D.” This crescent moon, therefore, is a waning crescent.
Later addition: as a commenter pointed out, below, this method does not work from Earth’s Southern hemisphere — in fact, in that half of the world, the “D”/”C” rule must be completely reversed, in order to work. To accomplish this, “D” could stand for “diminishing,” and “C” could stand for “commencing,” instead.
[Image/copyright note: I did not take these photographs of the moon. They were found with a Google-search, and I chose images with no apparent signs of copyright. I am assuming, on that basis, that these images are not copyrighted — but, if I am wrong, I will replace them with other images, upon request.]
The images on the faces of this polyhedron are based on information sent from NASA’s Lunar Reconnaisance Orbiter, as seen at http://lunar.gsfc.nasa.gov/lola/feature-20110705.html and tweeted by @LRO_NASA, which has been happily tweeting about its fifth anniversary in a polar lunar orbit recently. I have no idea whether this is actually an A.I. onboard the LRO, or simply someone at NASA getting paid to have fun on Twitter.
To get these images from near the Lunar South Pole onto the faces of a rhombic enneacontahedron, and then create this rotating image, I used Stella 4d: Polyhedron Navigator. There is no better tool available for polyhedral research. To check this program out for yourself, simply visit www.software3d.com/Stella.php.
It isn’t difficult to find rankings for the most massive objects in the solar system, rankings of objects in terms of increasing distance from the sun, or rankings of objects by radius. However, ranking objects by surface gravitational field strength is another matter, and is more complicated, for it is affected by both the mass and radius of the object in question, but in different ways. If two objects have different masses, but the same radius, the gravitational field strength will be greater for the more massive object. However, increasing the radius of an object decreases its surface gravitational field strength, in an inverse-square relationship.
Gravitational field strength is measured in N/kg, which are equivalent to m/s², the units for acceleration. The terms “gravitational field strength” and “acceleration due to gravity,” both of which are symbolized “g,” are actually synonymous. I prefer “gravitational field strength” because referring to acceleration, when discussing the weight of a stationary object on the surface of a planet, can cause confusion.
Use of the numbers given below is easy: given the mass of a thing (an imaginary astronaut, for example), in kilograms, simply multiply this figure by the given gravitational field strength, and you’ll have the weight of the thing, in newtons, on the surface of that planet (or other solar system object). If, for some odd reason, you want the weight in the popular non-metric unit known as the “pound,” simply divide the weight (in newtons) by 4.45, and then change the units to pounds.
How is surface gravitational field strength determined? To explain that, a diagram is helpful.
The large green circle represents a planet, or some other solar system object, and the blue thing on its surface, which I’ll call object x, can be pretty much anything on the solar system object’s surface. There are two formulas for Fg, the force of gravity pulling the planet and the thing on its surface toward each other. One is simply Fg= mxg, a form of Newton’s Second Law of Motion, where “g” is the gravitational field strength, and mx is the mass of the object at the surface. The other formula is more complicated: Fg= (Gmxmp)/r². This is Newton’s Law of Universal Gravitation, where “G” (not to be confused with “g”) is the universal gravitational constant, 6.67259 x 10-11 Nm²/kg², and mp and r are the mass and radius of the planet (or other solar system object). Because they each equal Fg, the expressions mxg and (Gmxmp)/r² can be set equal to each other, yielding the equation mxg = (Gmxmp)/r², which becomes g = (Gmp)/r² after mx is cancelled. The mass of the object on the surface is not needed — “g” is simply a function of mp and r.
There is a problem, however, with the idea of “surface” gravitational field strength — and that is the fact that the five largest objects in the solar system, the sun and the gas giants, all lack visible solid surfaces. One cannot stand on Jupiter — if you tried, you’d simply fall inside the planet. Therefore, for Jupiter, picture a solid platform floating at the top of the visible clouds there, and place the test object on this solid platform. Under those conditions, multiplying the test object’s mass by the Jovian value of “g” will, indeed, yield the weight of the object there, as it could be measured by placing it on a bathroom scale, at rest on the floating platform. For the other gas giants, as well as the sun, the idea is the same.
The objects included in the list below are the sun, all eight major planets, all dwarf planets (and dwarf planet candidates) with known values of “g,” all major satellites, some minor satellites, and a few of the largest asteroids. Many more objects exist, of course, but most have values for “g” which are not yet known.
Here are the top five:
Sun/Sol, 274.0 N/kg
Jupiter, 24.79 N/kg
Neptune, 11.15 N/kg
Saturn, 10.44 N/kg
Earth/Terra, 9.806 65 N/kg
The top five, alone, make me glad I undertook this project, for I did not realize, before doing this, that our planet has the highest surface gravitational field strength of any object in the solar system with a visible solid surface.
The next five include the rest of the major planets, plus one Jovian moon.
Venus, 8.87 N/kg
Uranus, 8.69 N/kg
Mars, 3.711 N/kg
Mercury, 3.7 N/kg
Io, 1.796 N/kg
The third set of five are all planetary moons, starting with earth’s own moon. The others are Jovian moons, except for Titan, which orbits Saturn.
Moon/Luna, 1.622 N/kg
Ganymede, 1.428 N/kg
Titan, 1.352 N/kg
Europa, 1.314 N/kg
Callisto, 1.235 N/kg
The fourth set of five begins with the largest dwarf planet, Eris, and includes two other dwarf planets as well.
Eris, 0.827 N/kg (dwarf planet)
Triton, 0.779 N/kg (Neptune’s largest moon)
Pluto, 0.658 N/kg (dwarf planet)
Haumea, 0.63 N/kg (dwarf planet)
Titania, 0.38 N/kg (largest moon of Uranus)
The fifth set of five includes the remaining dwarf planets with known values of “g.”
Oberon, 0.348 N/kg (moon of Uranus)
1 Ceres, 0.28 N/kg (dually classfied: dwarf planet and largest asteroid)
Charon, 0.278 N/kg (largest moon of Pluto)
Ariel, 0.27 N/kg (moon of Uranus)
90482 Orcus, 0.27 N/kg (probable dwarf planet)
The sixth set of five are dominated by Saturnian moons.
Rhea, 0.265 N/kg (Saturnian moon)
4 Vesta, 0.25 N/kg (2nd largest asteroid)
Dione, 0.233 N/kg (Saturnian moon)
Iapetus, 0.224 N/kg (Saturnian moon)
Umbriel, 0.2 N/kg (moon of Uranus)
The seventh set of five are mostly asteroids.
704 Interamnia, 0.186 N/kg (5th most massive asteroid)
2 Pallas, 0.18 N/kg (3rd most massive asteroid)
Tethys, 0.147 N/kg (Saturnian moon)
52 Europa, 0.14 N/kg (7th most massive asteroid)
3 Juno, 0.12 N/kg (large asteroid, w/~1% of mass of the asteroid belt)
Starting with the eighth group of five, I have much less certainty that something may have been omitted, although I did try to be thorough. My guess is that most future revisions of this list will be necessitated by the discovery of additional dwarf planets. Dwarf planets are hard to find, and there may be hundreds of them awaiting discovery.
Enceladus, 0.114 N/kg (Saturnian moon)
Vanth, 0.11 N/kg (moon of probable dwarf planet 90482 Orcus)
10 Hygiea, 0.091 N/kg (4th most massive asteroid)
15 Eunomia, 0.08 N/kg (large asteroid, with ~1% of mass of asteroid belt)
Miranda, 0.079 N/kg (moon of Uranus)
Here is the ninth group of five:
Nereid, 0.072 N/kg (Neptunian moon; irregular in shape)
Proteus, 0.07 N/kg (Neptunian moon; irregular in shape)
Mimas, 0.064 N/kg (Saturnian moon / smallest gravitationally-rounded object in the solar system)
Puck, 0.028 N/kg (6th largest moon of Uranus)
Amalthea, 0.020 N/kg (5th largest Jovian moon)
Finally, here are “g” values for the two tiny moons of Mars, included because they are nearby, and are the only moons Mars has to offer. A more exhaustive search would reveal many asteroids and minor satellites with “g” values greater than either Martian moon, but smaller than Amalthea, the last solar system object shown in the last set of five.
Phobos, 0.0057 N/kg
Deimos, 0.003 N/kg
Have you ever wondered why the number seven appears in all the places it does? We have seven days in the week. Churches teach about the seven deadly sins, and “seven heavens” is a common phrase. There are seven wonders of the ancient world, and seven of the modern world. The number seven has appeared in many other socially significant ways, in societies all over the world, for millennia.
It is no coincidence, I think, that the ancients were able to see seven lights in the sky which are either visible in daylight, or move against the background of “fixed” stars at night. They ascribed great significance to what went on in the sky, since they viewed “the heavens” as the realm of the gods in which they believed. The evidence for this lives on today, in the names of the seven days of the week, and numerous other sets of seven, all over the world.
It is possible to see the planet Uranus without a telescope, but it is very dim, and you have to know exactly where to look. No one noticed it until after the invention of the telescope. If Uranus were brighter, and had been seen in numerous ancient societies, I have no doubt that we would have eight days in the week, etc., rather than seven.