First, consider a body like a ship whose weight is less than the weight of an equal volume of water. The body will float on the surface of the water, and displace some quantity of water. If we mark out a horizontal patch in the water at some depth directly below the floating body, with an area equal to the area of the body at its waterline, then the weight pressing down on this surface will be the weight of the floating body plus the weight of the water above that patch, but not including the water displaced by the body, as this water is no longer above the patch. We can compare this with the weight pressing down on an equal area at an equal depth, away from the location of the floating body. This of course does not include the weight of the floating body, but it does include all the water from this patch to the surface, with no water displaced. In order for both patches to be pressed down by the same weight, the weight of the water displaced by the floating body must equal the weight of the floating body. This is why the weight of a ship is referred to as its “displacement.”
Next consider a body whose weight is greater than the weight of an equal volume of water. Such a body will not float, but it can be suspended in the water from a cable. If the cable is attached to one arm of a balance, then in this way we can measure the apparent weight Wapparent of the body when submerged in water. The weight pressing down on a horizontal patch in the water at some depth directly below the suspended body will equal the true weight Wtrue of the suspended body, less the apparent weight Wapparent, which is canceled by the tension in the cable, plus the weight of the water above the patch, which of course does not include the water displaced by the body. We can compare this with the weight pressing down on an equal area at an equal depth, a weight that does not include Wtrue or –Wapparent, but does include the weight of all the water from this patch to the surface, with no water displaced. In order for both patches to be pressed down by the same weight, we must have
Wtrue − Wapparent = Wdisplaced
where Wdisplaced is the weight of the water displaced by the suspended body. So by weighing the body when suspended in the water and weighing it when out of the water, we can find both Wapparent and Wtrue, and in this way find Wdisplaced. If the body has volume V, then
Wdisplaced = ρwaterV
where ρwater (rhowater) is the density (weight per volume) of water, close to 1 gram per cubic centimeter. (Of course, for a body with a simple shape like a cube we could instead find V by just measuring the dimensions of the body, but this is difficult for an irregularly shaped body like a crown.) Also, the true weight of the body is
Wtrue = ρbodyV
where ρbody is the density of the body. The volume cancels in the ratio of Wtrue and Wdisplaced, so from the measurements of both Wapparent and Wtrue we can find the ratio of the densities of the body and of water:
This ratio is called the “specific gravity” of the material of which the body is composed. For instance, if the body weighs 20 percent less in water than in air, then Wtrue – Wapparent = 0.20 × Wtrue, so its density must be 1/0.2 = 5 times the density of water. That is, its specific gravity is 5.
There is nothing special about water in this analysis; if the same measurements were made for a body suspended in some other liquid, then the ratio of the true weight of the body to the decrease in its weight when suspended in the liquid would give the ratio of the density of the body to the density of that liquid. This relation is sometimes used with a body of known weight and volume to measure the densities of various liquids in which the body may be suspended.
10. Areas of Circles
To calculate the area of a circle, Archimedes imagined that a polygon with a large number of sides was circumscribed outside the circle. For simplicity, let’s consider a regular polygon, all of whose sides and angles are equal. The area of the polygon is the sum of the areas of all the right triangles formed by drawing lines from the center to the corners of the polygon, and lines from the center to the midpoints of the sides of the polygon. (See Figure 4, in which the polygon is taken to be a regular octagon.) The area of a right triangle is half the product of its two sides around the right angle, because two such triangles can be stacked on their hypotenuses to make a rectangle, whose area is the product of the sides. In our case, this means that the area of each triangle is half the product of the distance r to the midpoint of the side (which is just the radius of the circle) and the distance s from the midpoint of the side to the nearest corner of the polygon, which of course is half the length of that side of the polygon. When we add up all these areas, we find that the area of the whole polygon equals half of r times the total circumference of the polygon. If we let the number of sides of the polygon become infinite, its area approaches the area of the circle, and its circumference approaches the circumference of the circle. So the area of the circle is half its circumference times its radius.
In modern terms, we define a number π = 3.14159 . . . such that the circumference of a circle of radius r is 2πr. The area of the circle is thus
1/2 × r × 2πr = πr2
The same argument works if we inscribe polygons within the circle, rather than circumscribing them outside the circle as in Figure 4. Since the circle is always between an outer polygon circumscribed around it and an inner polygon inscribed within it, using polygons of both sorts allowed Archimedes to give upper and lower limits for the ratio of the circumference of a circle to its radius—in other words, for 2π.
Figure 4. Calculation of the area of a circle. In this calculation a polygon with many sides is circumscribed about a circle. Here, the polygon has eight sides, and its area is already close to the area of the circle. As more sides are added to the polygon, its area becomes closer and closer to the area of the circle.
11. Sizes and Distances of the Sun and Moon
Aristarchus used four observations to determine the distances from the Earth to the Sun and Moon and the diameters of the Sun and Moon, all in terms of the diameter of the Earth. Let’s look at each observation in turn, and see what can be learned from it. Below, ds and dm are the distances from the Earth to the Sun and Moon, respectively; and Ds, Dm, and De denote the diameters of the Sun, Moon, and Earth. We will assume that the diameters are negligible compared with the distances, so in talking of the distance from the Earth to the Moon or Sun, it is unnecessary to specify points on the Earth, Moon, or Sun from which the distances are measured.
Observation 1
When the Moon is half full, the angle between the lines of sight from the Earth to the Moon and to the Sun is 87°.
When the Moon is half full, the angle between the lines of sight from the Moon to the Earth and from the Moon to the Sun must be just 90° (see Figure 5a), so the triangle formed by the lines Moon–Sun, Moon–Earth, and Earth–Sun is a right triangle, with the Earth–Sun line as the hypotenuse. The ratio between the side adjacent to an angle θ (theta) of a right triangle and the hypotenuse is a trigonometric quantity known as the cosine of θ, abbreviated cos θ, which we can look up in tables or find on any scientific calculator. So we have
dm/ds = cos 87° = 0.05234 = 1/19.11
and this observation indicates that the Sun is 19.11 times farther from the Earth than is the Moon. Not knowing trigonometry, Aristarchus could only conclude that this number is between 19 and 20. (The angle actually is not 87°, but 89.853°, and the Sun is really 389.77 times farther from the Earth than is the Moon.)
Figure 5. The four observations used by Aristarchus to calculate the sizes and distances of the Sun and Moon. (a) The triangle formed by the Earth, Sun, and Moon when the Moon is half full. (b) The Moon just blotting out the disk of the Sun during a total eclipse of the Sun. (c) The Moon passing into the shadow of the Earth during an eclipse of the Moon. The sphere that just fits into this shadow has a diameter twice that of the Moon, and P is the terminal point of the shadow cast by the Earth. (d) Lines of sight to the Moon spanning an angle of 2°; the actual angle is close to 0.5°.
Observation 2
The Moon just covers the visible disk
of the Sun during a solar eclipse.
This shows that the Sun and Moon have essentially the same apparent size, in the sense that the angle between the lines of sight from the Earth to opposite sides of the disk of the Sun is the same as for the Moon. (See Figure 5b.) This means that the triangles formed by these two lines of sight and the diameters of the Sun and Moon are “similar”; that is, they have the same shape. Hence the ratios of corresponding sides of these two triangles are the same, so
Ds/Dm = ds/dm
Using the result of observation 1 then gives Ds/Dm = 19.11, while the actual ratio of diameters is really close to 390.
Observation 3
The shadow of the Earth at the position of the Moon during a lunar eclipse is just wide enough to fit a sphere with twice the diameter of the Moon.
Let P be the point where the cone of shadow cast by the Earth ends. Then we have three similar triangles: the triangle formed by the diameter of the Sun and the lines from the edges of the Sun’s disk to P; the triangle formed by the diameter of the Earth and the lines from the edges of the Earth’s disk to P; and the triangle formed by twice the diameter of the Moon and the lines from a sphere with that diameter at the position of the Moon during a lunar eclipse to P. (See Figure 5c.) It follows that the ratios of corresponding sides of these triangles are all equal. Suppose that point P is at distance d0 from the Moon. Then the Sun is at distance ds + dm + d0 from P and the Earth is at distance dm + d0 from P, so
The rest is algebra. We can solve the second equation for d0, and find:
Inserting this result in the first equation and multiplying with DeDs(De – 2Dm) gives
The terms dmDs × (–2Dm) and 2DmdmDs on the right-hand side cancel. The remainder on the right-hand side has a factor De, which cancels the factor De on the left-hand side, leaving us with a formula for De:
If we now use the result of observation 2, that ds/dm = Ds/Dm, this can be written entirely in terms of diameters:
If we use the prior result Ds/Dm = 19.1, this gives De/Dm = 2.85. Aristarchus gave a range from 108/43 = 2.51 and 60/19 = 3.16, which nicely contains the value 2.85. The actual value is 3.67. The reason that this result of Aristarchus was pretty close to the actual value despite his very bad value for Ds/Dm is that the result is very insensitive to the precise value of Ds if Ds >> Dm. Indeed, if we neglect the term Dm in the denominator altogether compared with Ds, then all dependence of Ds cancels, and we have simply De = 3Dm, which is also not so far from the truth.
Of much greater historical importance is the fact that if we combine the results Ds/Dm = 19.1 and De/Dm = 2.85, we find Ds/De = 19.1/2.85 = 6.70. The actual value is Ds/De = 109.1, but the important thing is that the Sun is considerably bigger than the Earth. Aristarchus emphasized the point by comparing the volumes rather than the diameters; if the ratio of diameters is 6.7, then the ratio of volumes is 6.73 = 301. It is this comparison that, if we believe Archimedes, led Aristarchus to conclude that the Earth goes around the Sun, not the Sun around the Earth.
The results of Aristarchus described so far yield values for all ratios of diameters of the Sun, Moon, and Earth, and the ratio of the distances to the Sun and Moon. But nothing so far gives us the ratio of any distance to any diameter. This was provided by the fourth observation:
Observation 4
The Moon subtends an angle of 2°.
(See Figure 5d.) Since there are 360° in a full circle, and a circle whose radius is dm has a circumference 2πdm, the diameter of the Moon is
Aristarchus calculated that the value of Dm/dm is between 2/45 = 0.044 and 1/30 = 0.033. For unknown reasons Aristarchus in his surviving writings had grossly overestimated the true angular size of the Moon; it actually subtends an angle of 0.519°, giving Dm/dm = 0.0090. As we noted in Chapter 8, Archimedes in The Sand Reckoner gave a value of 0.5° for the angle subtended by the Moon, which is quite close to the true value and would have given an accurate estimate of the ratio of the diameter and distance of the Moon.
With his results from observations 2 and 3 for the ratio De/Dm of the diameters of the Earth and Moon, and now with his result from observation 4 for the ratio Dm/dm of the diameter and distance of the Moon, Aristarchus could find the ratio of the distance of the Moon to the diameter of the Earth. For instance, taking De/Dm = 2.85 and Dm/dm = 0.035 would give
(The actual value is about 30.) This could then be combined with the result of observation 1 for the ratio ds/dm = 19.1 of the distances to the Sun and Moon, giving a value of ds/De = 19.1 × 10.0 = 191 for the ratio of the distance to the Sun and the diameter of the Earth. (The actual value is about 11,600.) Measuring the diameter of the Earth was the next task.
12. The Size of the Earth
Eratosthenes used the observation that at noon on the summer solstice, the Sun at Alexandria is 1/50 of a full circle (that is, 360°/50 = 7.2°) away from the vertical, while at Syene, a city supposedly due south of Alexandria, the Sun at noon on the summer solstice was reported to be directly overhead. Because the Sun is so far away, the light rays striking the Earth at Alexandria and Syene are essentially parallel. The vertical direction at any city is just the continuation of a line from the center of the Earth to that city, so the angle between the lines from the Earth’s center to Syene and to Alexandria must also be 7.2°, or 1/50 of a full circle. (See Figure 6.) Hence on the basis of the assumptions of Eratosthenes, the Earth’s circumference must be 50 times the distance from Alexandria to Syene.
Figure 6. The observation used by Eratosthenes to calculate the size of the Earth. The horizontal lines marked with arrows indicate rays of sunlight at the summer solstice. The dotted lines run from the Earth’s center to Alexandria and Syene, and mark the vertical direction at each place.
Syene is not on the Earth’s equator, as might be suggested by the way the figure is drawn, but rather close to the Tropic of Cancer, the line at latitude 231⁄2°. (That is, the angle between lines from the Earth’s center to any point on the Tropic of Cancer and to a point due south on the equator is 231⁄2°.) At the summer solstice the Sun is directly overhead at noon on the Tropic of Cancer rather than on the equator because the Earth’s axis of rotation is not perpendicular to the plane of its orbit, but tilted from the perpendicular by an angle of 231⁄2°.
13. Epicycles for Inner and Outer Planets
Ptolemy in the Almagest presented a theory of the planets according to which, in its simplest version, each planet goes on a circle called an epicycle around a point in space that itself goes around the Earth on a circle known as the planet’s deferent. The question before us is why this theory worked so well in accounting for the apparent motions of the planets as seen from Earth. The answer for the inner planets, Mercury and Venus, is different from that for the outer planets, Mars, Jupiter, and Saturn.
First, consider the inner planets, Mercury and Venus. According to our modern understanding, the Earth and each planet go around the Sun at approximately constant distances from the Sun and at approximately constant speeds. If we do not concern ourselves with the laws of physics, we can just as well change our point of view to one centered on the Earth. From this point of view, the Sun goes around the Earth, and each planet goes around the Sun, all at constant speeds and distances. This is a simple version of the theory due to Tycho Brahe, which may also have been proposed by Heraclides. It gives the correct apparent motions of the planets, apart from small corrections due to the facts that planets actually move on nearly circular elliptical orbits rather than on circles, the Sun is not at the centers of these ellipses but at relatively small distances from the centers, and the speed of each planet varies somewhat as the planet goes around its orbit. It is also a special case of the theory of Ptolemy, though one never considered by Ptolemy, in which the deferent is nothing but the orbit of the Sun around the Earth, and the epicycle is the orbit of Mercury or Venus around the Sun.
Now, as far as the apparent position in the sky of the Sun and planets is concerned, we can multiply the changing distance of any planet
from the Earth by a constant, without changing appearances. This can be done, for instance, by multiplying the radii of both the epicycle and the deferent by the same constant factor, chosen independently for Mercury and Venus. For instance, we could take the radius of the deferent of Venus to be half the distance of the Sun from the Earth, and the radius of its epicycle to be half the radius of the orbit of Venus around the Sun. This will not change the fact that the centers of the planets’ epicycles always stay on the line between the Earth and the Sun. (See Figure 7a, which shows the epicycle and deferent for one of the inner planets, not drawn to scale.) The apparent motion of Venus and Mercury in the sky will be unchanged by this transformation, as long as we don’t change the ratio of the radii of each planet’s deferent and epicycle. This is a simple version of the theory proposed by Ptolemy for the inner planets. According to this theory, the planet goes around its epicycle in the same time that it actually takes to go around the Sun, 88 days for Mercury and 225 days for Venus, while the center of the epicycle follows the Sun around the Earth, taking one year for a complete circuit of the deferent.
Specifically, since we do not change the ratio of the radii of the deferent and epicycle, we must have
rEPI/rDEF = rP/rE
where rEPI and rDEF are the radii of the epicycle and deferent in Ptolemy’s scheme, and rP and rE are the radii of the orbits of the planet and the Earth in the theory of Copernicus (or equivalently, the radii of the orbits of the planet around the Sun and the Sun around the Earth in the theory of Tycho). Of course, Ptolemy knew nothing of the theories of Tycho or Copernicus, and he did not obtain his theory in this way. The discussion above serves to show only why Ptolemy’s theory worked so well, not how he derived it.
To Explain the World: The Discovery of Modern Science Page 30
