In my last post I introduced you to dimensional analysis, an essential trick for theoretical physicists, and showed you how you could address and sometimes solve interesting and important problems with it while hardly doing any work. Today we’ll look at it differently, to see its historical role in Einstein’s relativity.
When we’re trying to figure out whether a confusing statement is really true or not, we have to speak precisely. Up to this stage, I haven’t been careful enough, and in this post, I’m going to try to improve upon that. There are a few small but significant points of clarification to make first. Then we’ll look in detail at what it means to “change coordinates” in such a way that would put the Sun in orbit around the Earth, instead of the other way round.
Sometimes, when you’re doing physics, you have to make a wild guess, do a little calculating, and see how things turn out.
In a recent post, you were able to see how Kepler’s law for the planets’ motions (R3=T2 , where R the distance from a planet to the Sun in Earth-Sun distances, and T is the planet’s orbital time in Earth-years), leads to the conclusion that each planet is subject to an acceleration a toward the Sun, by an amount that follows an inverse square law
a = (2π)2 / R2
where acceleration is measured in Earth-Sun distances and in Earth-Years.
That is, a planet at the Earth’s distance from the Sun accelerates (2π)2 Earth-distances per Earth-year per Earth-year, which in more familiar units works out (as we saw earlier) to about 6 millimeters per second per second. That’s slow in human terms; a car with that acceleration would take more than an hour to go from stationary to highway speeds.
What about the Moon’s acceleration as it orbits the Earth? Could it be given by exactly the same formula? No, because Kepler’s law doesn’t work for the Moon and Earth. We can see this with just a rough estimate. The time it takes the Moon to orbit the Earth is about a month, so T is roughly 1/12 Earth-years. If Kepler’s law were right, then R=T2/3 would be 1/5 of the Earth-Sun distance. But we convinced ourselves, using the relation between a first-quarter Moon and a half Moon, that the Moon-Earth distance is less than 1/10 othe Earth-Sun distance. So Kepler’s formula doesn’t work for the Moon around the Earth.
But perhaps objects that are orbiting the Earth satisfy a similar law,
R3=T2 for Earth-orbiting objects
except that now T should be measured not in years but in Moon-orbits (27.3 days, the period of the Moon’s orbit around the Earth) and R should be measured not in Earth-Sun distances but in Moon-Earth distances? That was Newton’s guess, in fact.
Newton had a problem though: the only object he knew that orbits the Earth was the Moon. How could he check if this law was true? We have an advantage, living in an age of artificial satellites, which we can use to check this Kepler-like law for Earth-orbiting objects, just the way Kepler checked it for the Sun-orbiting planets. But, still there was something else Newton knew that Kepler didn’t. Galileo had determined that all objects for which air resistance is unimportant will accelerate downward at 32 feet (9.8 meters) per second per second (which is to say that, as each second ticks by, an object’s speed will increase by 32 feet [9.8 meters] per second.) So Newton suspected that if he converted the Kepler-like law for the Moon to an acceleration, as we did for the planets last time, he could relate the acceleration of the Moon as it orbits the Earth to the acceleration of ordinary falling objects in daily life.
Kepler’s third law is so simple to state that (as shown last time) it is something that any grade school kid, armed with Copernicus’s data and a calculator, can verify. Yet it was 75 years from Copernicus’s publication til Kepler discovered this formula! Why did it take Kepler until 1618, nearly 50 years of age, to recognize such a simple relationship? Were people just dumber than high-school students back then?
Here’s a clue. We take all sorts of math for granted that didn’t exist four hundred years ago, and calculations which take an instant now could easily take an hour or even all day. (Imagine computing the cube root of 4972.64 to part-per-million accuracy by hand.) In particular, one thing that did not exist in Copernicus’ time, and not even through much of Kepler’s, was the modern notion of a logarithm.
Much of this work was done by Nicolai Copernicus himself, the most famous of those philosophers who argued for a Sun-centered universe rather than an Earth-centered universe during the millennia before modern science. He had all the ingredients we have, minus knowledge of Uranus and Neptune, and minus the clues we obtain from telescopes, which would have confirmed he was correct.
Copernicus knew, therefore, that although the planetary distances from the Sun and their cycles in the sky (which astrologers [not astronomers] have focused on for centuries) don’t seem to be related, the distances and their orbital times around the Sun are much more closely related. That’s what we saw in the last post.
Let me put these distances and times, relative to the Earth-Sun distance and the Earth year, onto a two-dimensional plot. [Here the labels are for Mercury (Me), Venus (V), Mars (Ma), Jupiter (J), Saturn (S), Uranus (U) and Neptune (N).] The first figure shows the planets out to Saturn (the ones known to Copernicus).
The second shows them out to Neptune, though it bunches up the inner planets to the point that you can’t really see them well.
You can see the planets all lie along a curve that steadily bends down and to the right.
Copernicus knew all of the numbers that go into Figure 1, with pretty moderate precision. But there’s something he didn’t recognize, which becomes obvious if we use the right trick. In the last post, we sometimes used a logarithmic axis to look at the distances and the times. Now let’s replot Figure 2 using a logarithmic axis for both the distances and the times.
Oh wow. (I’m sure that’s the equivalent of what Kepler said in 1618, when he first painstakingly calculated the equivalent of this plot.)
It looks like a straight line. Is it as straight as it looks?
And now we see three truly remarkable things about this graph:
First, the planet’s distances to the Sun and orbital times lie on a very straight line on a logarithmic plot.
Second, the slope of the line is 2/3 (2 grid steps up for every 3 steps right) rather than, say, 7.248193 .
Third, the line goes right through the point (1,1), where the first horizontal and first vertical lines cross.
What else can we learn just with simple observations? Since the stars’ daily motion is an illusion from the Earth’s spin, and since the stars do not visibly move relative to one another, our attention is drawn next to the motion of the objects that move dramatically relative to the stars: the Sun and the planets. Exactly once each year, the Sun appears to go around the Earth, such that the stars that are overhead at midnight, and thus opposite the Sun, change slightly each day. The question of whether the Earth goes round the Sun or vice versa is one we’ll return to.
Let’s focus today on the planets (other than Earth) — the wanderers, as the classical Greeks called them. Do some of them go round the Earth? Others around the Sun? Which ones have small orbits, and which ones have big orbits? In answering these questions, we’ll start to build up a clearer picture of the “Solar System” (in which we include the Sun, the planets and their moons, as well as asteroids and comets, but not the stars of the night sky.)
The Basic Patterns
If we make the assumption (whose validity we will check later) that the planets are moving in near-circles around whatever they orbit, then it’s not hard to figure out who orbits who. For each possible type of orbit, a planet will exhibit a different pattern of sizes and phases across its “cycle“ when seen through binoculars or a small telescope. Even with the naked eye, a planet’s locations in the sky and changes in brightness during its cycle give us strong clues. Simply by looking at these patterns, we can figure out who orbits who.
Once you’ve convinced yourself the Earth’s a spinningsphere of diameter about 8000 miles (13000 km), and you’ve estimated the Moon’s size and distance(diameter about 1/4 Earth’s, and distance about 30 times Earth’s diameter), it’s easy to convince yourself the Sun’s bigger than the Earth, and much further than the Moon. It just takes a couple of triangles, and a bit of Moon-gazing.
Since that’s all there is to it, you can guess that the ancient Greek astronomers, masters of geometry, already knew the Sun’s the larger of the two. That said, they never did quite figure out how big and far the Sun actually is; we need modern methods for that.
It’s Just a Phase
The Moon goes through a monthly cycle of phases, lasting about 291/2 Earth days, in which the part that glows brightly with reflected sunlight grows and shrinks, from crescent to full and back again. The phases arise because there are two simple ways of dividing the Moon in half:
At any moment, the half of the Moon that faces Earth — let’s call it the near half of the Moon — is the only half that we can potentially see. (We’d only be able to see the far half, facing away from Earth, if the Moon were transparent, or a big mirror was sitting beyond the Moon.)
At any moment, the half of the Moon that faces the Sun is brightly lit — let’s call it the lit half. The other half is dark, and its presence can only be detected by the fact that it can block stars that it moves in front of, and through a very dim glow in which it reflects sunlight that first reflected from the Earth (called “Earthshine.”)
The phases arise because the lit half and the near half aren’t the same, and the relationship between them changes from night to night. See the diagram below. When the Moon is more or less between the Sun and the Earth (it rarely passes exactly between, because its orbit is tilted by a few degrees out of the plane of the drawing below) then the Moon’s lit half is its far half, and the near half is unlit. We call this dark view of the Moon the “New Moon” because it is traditionally viewed as the start of the Moon’s monthly cycle.
When the Moon is on the opposite side of the Earth from the Sun (but again, rarely eclipsed by Earth’s shadow because of its tilted orbit), then its near side is its lit side, and that creates the “Full Moon”, a complete white disk in the sky.
At any other time, the near side of the Moon is partly lit and partly unlit. When the line between the Moon and Earth is perpendicular to the Earth-Sun line, then the lit side and unlit side slice the near side in half, and the Moon appears as a half-disk cut down the middle.
When I was a child, I wondered why half this half-lit phase of the Moon, midway between New Moon (invisible) and Full Moon (the bright full disk), was called “First Quarter”, when in fact the Moon at that time is half lit. Why not “First Half?” Two weeks later, the other half of the near-side of the Moon is lit, and why is that called “Third Quarter” and not, say, “Other Half”?
This turns out to have been an excellent question. The fact that a Half Moon is also a First Quarter Moon tells us that the Sun is large and far away!
Posted onFebruary 18, 2022|Comments Off on How to Figure Out the Distance to the Moon Yourself
Last time I described an easy way for you to determine the size of the Moon — easier than the famous techniques used by the classical Greeks. (We don’t need to know the Earth’s circumference, as they did, if we’re ok with a moderately precise estimate.) Once you’ve done that, there’s an simple method, well known since classical times, for figuring out how far away the Earth’s companion is. That’s what I’ll describe in this short post.
(What’s not so easy is to determine the distance and size of the Sun. The classical Greeks failed in their efforts. We’ll need a more modern approach… but that’s for next week.)
Size Versus Distance
Even the early classical Greeks knew something about the Sun, just from the fact that the Moon and Sun appear roughly the same size to our eyes — that is, they occupy about the same amount of sky. If the Sun is twice as far away as the Moon, its diameter must be twice as big, in order that it appear the same size. That’s illustrated in the figure below. If it is ten times as far away, its diameter must be ten times as big. If it’s four hundred times as far away, its diameter is four hundred times as big. (Spoiler: that last one’s the truth; but we’ll get to it later.)
You can run this logic in the other direction; if something perfectly blocks the Moon, then if it’s ten times closer than the Moon its diameter must be ten times smaller. If it’s a billion times closer than the Moon, it must be a billion times smaller.
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From the CMS experiment at the Large Hadron collider, a proton-proton collision that created a Higgs boson, which subsequently decayed to two particles of light (shown as green rods.)