If we're told that the time difference between Los Angeles and Boston is 3.5 hours, and the two cities are 3,000 miles apart ($3\times10^{3}$ miles if we're going to be scientific about this), then we've functionally been given the same pieces of information Eratosthenes had--distance over ground between two points (let's assume it's along a great circle) and what fraction of the earth's circumference that distance is. Let's start by turning those hours and miles into slightly more useful units. The hours we can actually leave alone--we're just going to be using them in a ratio, so they'll cancel on their own. Miles, though, aren't too helpful, so let's get rid of those $3\times10^{3}$ miles and call them $5\times10^{8}$ cm. Now all we need to do is set up a pretty simple pair of ratios (since we're relating 3.5 hours to 24 hours, we're going to have to relate our distance over ground to the circumference--$2\pi R$ rather than just the radius itself):\[\frac{3.5}{24}=\frac{5\times10^{8}}{2\pi R}\] Now all we need to do is solve for $R$ in there and we come out with a radius of $6\times10^8$ cm. Not bad--that's an error of just under 6%. So in the interest of preserving our egos, let's say Eratosthenes' error was 16.3%.
This is subject to a fair few errors--the biggest of these is probably that we aren't relying on the altitude of the sun like Eratosthenes was, but on a time difference, which assumes that the distance between Boston and Los Angeles is purely longitudinal (that is to say, the two cities are the same distance from the equator and you could travel between the two along a line of latitude). If it doesn't seem like this could matter, imagine if we had traveled between Anchorage and Honolulu. The two are also about $3\times10^3$ miles apart, or $5\times10^8$ cm. But they're only one time zone apart. Replace the 3.5 hours in that equation from earlier with one hour, and you get a radius of $2\times10^9$ cm, and suddenly we're off by an order of magnitude. Of course, we could also have used Montreal and Bogota, $3\times10^3$ miles apart but in the same time zone, which would mean plugging zero into the equation and blowing the whole thing up. But Boston and Los Angeles are close enough to the same latitude that the estimation stays reasonable
And now with a couple extra pieces of equipment and information, this gives us a chance to estimate a whole bunch of other useful constants. Let's start by figuring out the mass of the Earth. Aside from the radius we figured before, we'll just need a scale, a measuring cup with some water in it, and a rock. The scale is to weigh the rock, the water is to measure its volume (by seeing how much it displaces), and the rock is to accurately represent the average density of the entire Earth (estimating, remember?). Our rock came in at 90 grams and 30 milliliters (or cubic centimeters), giving us a density of 90 grams/cm$^3$. Meanwhile, we can figure the volume of the Earth with the radius we calculated, since our planet is more or less spherical. The volume of a sphere is $\frac{4}{3}\pi r^3$, so plugging $6\times10^8$ in for $r$ gives us about $9\times10^{26}$ cm$^3$. Mass is just density times volume, so we just multiple three by nine, multiply that by $10^{26}$, and come out with just under $3\times10^{27}$ grams. We're off by about half here--not anywhere near as nice as our radius estimation--but we've got the right order of magnitude.
So let's throw another celestial body into the picture. Kepler's third law tells us that that if you take the time it takes the moon to orbit the Earth (a month or so--let's call it 30 days), and cube it, that's going to equal the distance from the Earth to the moon squared (along with a few other constants--in this case, the mass of the Earth, Newton's gravitational constant G, pi squared, and a four for good measure). In equation from, that looks like this:\[p^2 = \frac{4\pi^2 a^3}{GM_E}\] If we're not too sure about this shady Kepler guy, or (more likely) aren't too sure about our memory, we can check to make sure the units on either side match up (as long as we can remember the insane units of $G$, which are $\frac{distance^3}{mass*time^2}$. \[time^2 = \frac{distance^3}{mass*\frac{distance^3}{mass*time^2}}\] That simplifies to $time^2 = time^2$, so it looks like we're good! We can put 30 days in seconds (it's about $3\times10^6$), and if we remember that G is $7\times10^{-8}$ in cgs units, we've got everything we need to solve for $a$. It comes out to $2\times10^{10}$ cm. Again, we're off by a bit half--but given that we already knew that our mass was half off, that means everything else here is close to perfect!
With that distance in hand, along with one more little fact and some high school math, we can figure out how big the moon is, too. Aristarchus had a pretty cool way of doing this a few years earlier even than Eratosthenes figured the size of the Earth, using the size of the full moon relative to the Earth's shadow during a lunar eclipse (figure how large the diameter of the moon is relative to the diameter of the Earth's shadow, and you know the size of the moon relative to the size of the Earth). We'll do things a bit differently, using a frequently-seen (and incredibly simple) mathematical tool in astronomy: the small angle approximation. A full moon covers up about half a degree on the sky, which means you can draw a right triangle from an earthbound observer to two opposite edges of the moon. The vertex of that triangle poking into the Earth has an angle of half a degree. The tangent of that angle will be "opposite over adjacent"--in this case, the diameter of the moon over the distance from the Earth to the moon. But with a sufficiently small angle (and half a degree is sufficiently small), the tangent of an angle is about equal to the angle itself in radians. The trickiest part of the process is just the degree-radian conversion, which is easy enough with a bit of cross-multiplication. \[\frac{.5}{36}=\frac{\theta}{2\pi}\] Solving for $\theta$ tells us that half a degree is 1/120 radians. Now we can plug that into a simple equation using that small angle approximation. \[1/120 = \frac{2r_{moon}}{2\times10^{10}}\] This comes out to a radius of about $10^{8}$ cm, again just under half off but the right order of magnitude.
Finally, knowing that the moon is made up of more or less the same material as the Earth, we can figure its mass the same way we did the Earth's. Calculating its volume using the radius we just figured, we get \[\frac{4\pi(10^8)^3}{3} = 3\times10^{25} \: cm^3\]Multiply that by the density of 3 grams/cm$^3$ from earlier, and we end up with a mass of $10^{26}$ grams, an error of about 1/3.
I may have been able to work through this alone, but fortunately I didn't have to, so credit must go to my outstanding colleagues Anne Madoff, Louise Decoppet, and Jennifer Shi.
This is subject to a fair few errors--the biggest of these is probably that we aren't relying on the altitude of the sun like Eratosthenes was, but on a time difference, which assumes that the distance between Boston and Los Angeles is purely longitudinal (that is to say, the two cities are the same distance from the equator and you could travel between the two along a line of latitude). If it doesn't seem like this could matter, imagine if we had traveled between Anchorage and Honolulu. The two are also about $3\times10^3$ miles apart, or $5\times10^8$ cm. But they're only one time zone apart. Replace the 3.5 hours in that equation from earlier with one hour, and you get a radius of $2\times10^9$ cm, and suddenly we're off by an order of magnitude. Of course, we could also have used Montreal and Bogota, $3\times10^3$ miles apart but in the same time zone, which would mean plugging zero into the equation and blowing the whole thing up. But Boston and Los Angeles are close enough to the same latitude that the estimation stays reasonable
And now with a couple extra pieces of equipment and information, this gives us a chance to estimate a whole bunch of other useful constants. Let's start by figuring out the mass of the Earth. Aside from the radius we figured before, we'll just need a scale, a measuring cup with some water in it, and a rock. The scale is to weigh the rock, the water is to measure its volume (by seeing how much it displaces), and the rock is to accurately represent the average density of the entire Earth (estimating, remember?). Our rock came in at 90 grams and 30 milliliters (or cubic centimeters), giving us a density of 90 grams/cm$^3$. Meanwhile, we can figure the volume of the Earth with the radius we calculated, since our planet is more or less spherical. The volume of a sphere is $\frac{4}{3}\pi r^3$, so plugging $6\times10^8$ in for $r$ gives us about $9\times10^{26}$ cm$^3$. Mass is just density times volume, so we just multiple three by nine, multiply that by $10^{26}$, and come out with just under $3\times10^{27}$ grams. We're off by about half here--not anywhere near as nice as our radius estimation--but we've got the right order of magnitude.
So let's throw another celestial body into the picture. Kepler's third law tells us that that if you take the time it takes the moon to orbit the Earth (a month or so--let's call it 30 days), and cube it, that's going to equal the distance from the Earth to the moon squared (along with a few other constants--in this case, the mass of the Earth, Newton's gravitational constant G, pi squared, and a four for good measure). In equation from, that looks like this:\[p^2 = \frac{4\pi^2 a^3}{GM_E}\] If we're not too sure about this shady Kepler guy, or (more likely) aren't too sure about our memory, we can check to make sure the units on either side match up (as long as we can remember the insane units of $G$, which are $\frac{distance^3}{mass*time^2}$. \[time^2 = \frac{distance^3}{mass*\frac{distance^3}{mass*time^2}}\] That simplifies to $time^2 = time^2$, so it looks like we're good! We can put 30 days in seconds (it's about $3\times10^6$), and if we remember that G is $7\times10^{-8}$ in cgs units, we've got everything we need to solve for $a$. It comes out to $2\times10^{10}$ cm. Again, we're off by a bit half--but given that we already knew that our mass was half off, that means everything else here is close to perfect!
With that distance in hand, along with one more little fact and some high school math, we can figure out how big the moon is, too. Aristarchus had a pretty cool way of doing this a few years earlier even than Eratosthenes figured the size of the Earth, using the size of the full moon relative to the Earth's shadow during a lunar eclipse (figure how large the diameter of the moon is relative to the diameter of the Earth's shadow, and you know the size of the moon relative to the size of the Earth). We'll do things a bit differently, using a frequently-seen (and incredibly simple) mathematical tool in astronomy: the small angle approximation. A full moon covers up about half a degree on the sky, which means you can draw a right triangle from an earthbound observer to two opposite edges of the moon. The vertex of that triangle poking into the Earth has an angle of half a degree. The tangent of that angle will be "opposite over adjacent"--in this case, the diameter of the moon over the distance from the Earth to the moon. But with a sufficiently small angle (and half a degree is sufficiently small), the tangent of an angle is about equal to the angle itself in radians. The trickiest part of the process is just the degree-radian conversion, which is easy enough with a bit of cross-multiplication. \[\frac{.5}{36}=\frac{\theta}{2\pi}\] Solving for $\theta$ tells us that half a degree is 1/120 radians. Now we can plug that into a simple equation using that small angle approximation. \[1/120 = \frac{2r_{moon}}{2\times10^{10}}\] This comes out to a radius of about $10^{8}$ cm, again just under half off but the right order of magnitude.
Finally, knowing that the moon is made up of more or less the same material as the Earth, we can figure its mass the same way we did the Earth's. Calculating its volume using the radius we just figured, we get \[\frac{4\pi(10^8)^3}{3} = 3\times10^{25} \: cm^3\]Multiply that by the density of 3 grams/cm$^3$ from earlier, and we end up with a mass of $10^{26}$ grams, an error of about 1/3.
I may have been able to work through this alone, but fortunately I didn't have to, so credit must go to my outstanding colleagues Anne Madoff, Louise Decoppet, and Jennifer Shi.
Great job introducing the problem in its historical context, estimating errors, and linking to auxiliary concepts like "great circle".
ReplyDeleteSome minor comments:
Near the end of your post some of the exponents don't appear typeset correctly.
You do a great job of explaining things in words, but in the future feel free to use diagrams as well.