Waterloo Collegiate Astronomy Assignment SES4UI Size of the Earth and the Distances to the Moon and Sun Objectives Using observations of the Earth-Sun-Moon system and elementary geometry and trigonometry, the student will duplicate the methods of the ancient Greek astronomers in determining the size of the Earth, the size of and distance to the Moon, and the size of and distance to the Sun. In doing so, the student will derive the value of the astronomical unit, the fundamental baseline used for measuring the parallaxes of stars. Complete the Assignment by following the guides and filling in the material as required. You may want to use a separate answer sheet to show you work and answer questions if you require more space. Materials Scientific calculators rulers Introduction As we look into the night sky, the Moon, planets, and stars appear to be set within a crystalline sphere. We observe the motion of these objects from night to night during the course of a year and discover predictable patterns. The Sun, our star, reliably rises and sets each day. From a limited perspective, perhaps this is all we need to tell us when to plant our crops, when to harvest them, when we will be most successful at fishing, when we will be able to travel at night beneath the light of a full Moon. These observations, however, will not give us any information as to the size of our universe and our place in it. We did not immediately know that the Universe was incredibly huge, of order 26 billion or more light years across our visible limit the diameter of our visible realm. We started with small steps on what has become known as the "cosmic distance ladder." Our knowledge of distances started over 2300 years ago with the Greeks Aristarchus of Samos (310-230 BC) and Eratosthenes (276-196 BC). Aristarchus suggested well before Copernicus that the Earth orbited the Sun, but his ideas were rejected. Erastosthenes enjoyed more success with his measurement of the size of the Earth, employing an army to help in his measurements. Ancient Greeks were able to calculate the size of the Earth, Moon, and Sun, and to determine the distances to the Moon and Sun. Most of these measurements were quite difficult for the ancient astronomers to carry out as all of this was done without telescope technology, and without precise measuring tools. In this exercise you will be following the methods of Eratosthenes and Aristarchus. Although you will be following their methods, you will be using more precise observations than were available to them. In fact, you will see that the simple model which these Hellenistic Greek thinkers had for the Sun-Earth-Moon system allows quite accurate determination of distances when good observations are used. Remember that although today we accept for granted the fact that the Moon obits the Earth, which is itself orbiting the Sun, these were all revolutionary ideas in the history of science. Consider your mind wiped clear of what you have learned about astronomy and imagine how hard it would be to come up with these ideas by simply looking at what you see in the sky! The last couple of pages of this exercise have some practice equations for ratios and trigonometry as a review, in case it s needed. 1/3/2012 Size of the Earth and the Distances to the Moon and Sun.doc 1
Procedure Rung 1: The size of the Earth This step was first performed by Eratosthenes and involves a very simple model for just the Sun and the Earth: 1) the Earth is a sphere; 2) the Sun is very far away. The fact that the Earth is a sphere was known from elementary observations (the shape of the shadow on the Moon during an eclipse, the way a tall ship disappeared as it sailed away), but what is the significance of assuming the Sun to be very far away? Eratosthenes knew that at noon on June 21 (the summer solstice and the longest day of the year), the Sun would shine into the bottom of a deep vertical pit in the city of Syene, meaning the Sun was at the zenith in the sky at this time. Being in Alexandria, which was located more or less due north of Syene, he could measure the length of a shadow cast by an obelisk at the same time, noon on the summer solstice. This measurement, along with the height of the obelisk, gave him the angle that the Sun appeared south of the zenith in Alexandria, an angle he determined to be 7.2 degrees. The Sun was not as high in the sky in Alexandria as in Syene at the same time of day and the same time of year. This is in fact a measurement of the curvature of the Earth's sphere between Syene and Alexandria, and once distance of this small arc is known, the circumference of the entire Earth is determined. To this end (astronomers having considerably more political clout in those days), he ordered some soldiers to march off the distance between Alexandria and Syene, a distance of 5000 stadia (thought today to be equivalent to about 805 km or 500 miles). The circumference of the Earth could be found by using the following proportionality: Syene to Alexandria 7.2 Circumference of Earth 360 What value for the circumference of the Earth would Eratosthenes have determined from his measurements? What value for the radius of the Earth corresponds with this circumference? 2
Imagine you have a friend living in Mazatlan who has noticed that each year on June 21 (summer solstice) the rays of the Sun fall straight down his water well. You, meanwhile, are living in Denver in a small house on a productive wheat farm. In the middle of one of the access roads sits an ancient obelisk. After hearing from your friend last year, you decide to test the ways of the ancient Greek astronomers yourself. You know that the longitude of the two places is only slightly different, and that by measuring the shadow of the obelisk in Denver at the time the Sun is directly overhead in Mazatlan, you will be able to calculate the radius of the Earth (and truly impress your friends, family, and neighbours). The two of you get on your cell phones as the precise time of the summer solstice approaches. You click a picture at exactly the right moment. The image you took is shown above. Using a similar triangles construction like the one shown in the figure at the left, we can directly determine the radius of the Earth. The length of the shadow is proportional to the distance between Denver and Mazatlan as the height of the obelisk is proportional to the radius of the Earth. We can write this in equation form as: Radius of Earth Height of Obelisk Denver to Mazatlan Length of Shadow 3
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Finding the ratio for the right-hand side of this equation: Using a ruler, measure the length of the shadow and the height of the obelisk. Find the ratio of these two measurements, as shown in the equation. Fill in that value in the answer sheet. Finding the ratio for the left-hand side of this equation: The radius of the Earth is the unknown we are solving for. To find the distance from Denver to Mazatlan, find the scaling factor for the map. At the bottom left on the map is a scale in km and miles. Using the km scale, find out how many km (in the real world) is represented by 1 mm (on the map). You can also use Google on the internet (or better maps) to determine this distance. Putting the two together: Using the above equation and your measurements from the images, calculate the radius of the Earth in kilometers. 5
Rung 2: The size of and distance to the Moon The distance to the Moon was determined by first finding the size of the Moon relative to the size of the Earth. This determination of the relative sizes of the Earth and Moon predated the estimate of the absolute size of the Earth due to Eratosthenes and was first carried out by Aristarchus of Samos (310-230 BC). Once again a model is required to make the determination of the relative sizes of the Earth and Moon, in particular a good model for what is taking place during a lunar eclipse. It was surmised that during a lunar eclipse the full Moon is passing through the shadow of the Earth. Aristarchus timed how long the Moon took to travel through Earth's shadow and compared this with the time required for the Moon to move a distance equal to its diameter (this could be done by timing how long a bright star was obscured by the Moon as the Moon passed between it and the Earth). He found that the shadow was about 8/3 the diameter of the Moon. The model we use here is extended from the one given in Rung 1 because we need to add the Moon: 1) The Earth is a sphere; 2) The Sun is very far away; 3) The Moon orbits the Earth in such a way that eclipses occur. How is the second statement important to the measurement of the relative size of the Moon to the Earth? What evidence do you think the ancient Greek astronomers had that the Moon orbits the Earth? Although Aristarchus used a timing method, we can get a crude estimate of the relative size of the Moon to the Earth by looking at the curvature of the Earth's shadow during a lunar eclipse. All of the above model statements are still important to this conceptually simpler method. The measurement of the Earth's shadow is not without its difficulties and uncertainties, however, the main reason being that the Sun is not infinitely far away. 6
Now use the curvature of the Earth's shadow during a lunar eclipse to determine the relative sizes of the Earth and Moon. Review the sketches above of an eclipse of the Moon by the Earth s shadow, and state one assumption we ve made that will probably lead to an error in our determination of the radius of the Earth. Image copyrighted by Gregory Terrance and used with his permission. For more of his images see CCD Image Gallery You now have an approximation of the ratio of the radius of the Moon to the radius of the Earth. Use the radius of the Earth that you determined from Rung 1 of this exercise to find the radius of the Moon in kilometers. Be sure to watch out for whether or not you are asked to use the radius or the diameter of the Moon for the subsequent calculations. 7
After making your measurements, give your opinion as to why this method would have been difficult for the ancient Greek astronomers to perform, and why the timing method would be more favored. Once we have the absolute diameter of the Moon we can easily determine its distance from the Earth by measuring its angular diameter on the sky. The angle subtended by the full Moon on the sky is about 0.5 degrees. Calculate the distance to the Moon using its absolute diameter and its angular diameter. For help with the geometry of this problem and with the trigonometry involved with this calculation, take a close look at the logic used here: 8
Rung 3: The Distance to the Sun Aristarchus also came up with a method for finding the distance to the Sun relative to the distance to the Moon. This method once again relies on a model for the Sun-Earth-Moon system. In particular, we can no longer assume the Sun to be very far away in the sense that we did for the previous two rungs of our distance ladder (we essentially assumed the Sun to be infinitely far away). What we are relying on is that the Sun's rays do not hit both the Moon and the Earth at the same angle, that the rays are not parallel over the Earth-Moon distance. This difference in the angle that the Sun's rays hit the Moon versus the angle they hit the Earth is very small, and this is the reason why this was a particularly difficult measurement for Greek astronomers to perform. It is important to understand that even though Aristarchus' estimate of the distance to the Sun relative to the distance to the Moon was wrong by a factor of 20, his basic method was correct. Our model now reads: 1) the Earth is a sphere; 2) the Sun may be far away, but close enough that its rays hit the Earth and Moon at slightly differing angles; 3) the Moon orbits the Earth. Why is the second assumption in our model so important in determining the distance to the Sun? This new model and its use in determining the distance to the Sun can be understood by studying the following diagram: Note that when the moon is seen to be exactly in the first quarter phase, the Sun-Earth-Moon angle is a right angle, or 90 degrees. If we can measure the angle between the Sun and the Moon when the Moon is precisely in its first (or third) quarter phase, then we can determine the distance to the Sun. With precise observations made at first-quarter lunar phase, a Moon-Earth-Sun angle of 89.853 degrees is measured (extremely close to 90 degrees!). We know the distance from the Earth to the Moon, we determined that in Rung 2 of this exercise. The Earth-Moon distance plus the measured Moon-Earth-Sun angle gives us one side and one angle of a right triangle, and we can use trigonometry to determine the length of the hypotenuse--the distance to the Sun. Think for a moment about what might make this observation difficult to carry out, referring to what you learned in Rungs 1 and 2 and from the images given above. (Hint: Take a look at the images shown at the right. How difficult do you think it would be to determine the exact time of the first- or third-quarter Moon?) Use the appropriate trigonometry formula, the angles given above, and the distance to the Moon from Rung 2 to determine the distance to the Sun. 9
Sin( ) = opposite/hypotenuse (or A/C) Cos( ) = adjacent/hypotenuse (or B/C) Tan( ) = opposite/adjacent (or A/B) Now that we have an estimate for the distance to the Sun, it is also possible to determine the diameter of the Sun in kilometers. We know that the angular diameter of the Sun is almost the same as the Moon, about 0.5 degrees on the sky, the most dramatic evidence of this being during a total solar eclipse as shown here: Using the distance to the Sun determined above and the angular diameter of the Sun, which is 0.5 degrees, determine the radius of the Sun in kilometers. Hint: recall the way in which you calculated the distance to the Moon from an estimate of its radius in kilometers. Summary As you wrap up your results for this exercise, check the actual quantities listed in the table on your answer sheet and calculate the percentage error for each value. 10
Size of the Earth and the Distances to the Moon and Sun Some Review of Mathematics This exercise starts us on the road to using a bit of math to describe our universe. Provided here is a brief refresher course for the math that will be needed in this exercise. Working with ratios: Finding the ratio of two numbers means dividing one by the other. When we know that one ratio will be proportional to another ratio, and we know the values of three of the numbers, it makes finding the unknown fourth number easy. Here is a simple example: A 3 3 150, A 30 150 15 15 When working with ratios, some of the units may cancel out: A 3mi 3 150km, A 30km 150km 15mi 15 Here is a slightly more difficult example, where the unknown is in the denominator of one of the ratios: 150 km 3 3 mi, 150 km x mi x 15mi 15mi After a bit more manipulation: 15 150 km x, x 750km 3 A little bit of trigonometry: The hardest part about trigonometry these days is trying to figure out which buttons to push on the calculator. A refresher for Sine, Cosine, and Tangent of an angle for a right-angle triangle, in this case the angle represented by the Greek letter alpha, having sides of lengths A, B, and C: Sin( ) = opposite/hypotenuse (or A/C) Cos( )=adjacent/hypotenuse (or B/C) Tan( ) = opposite/adjacent (or A/B) To solve the problems when the angle is known (such as 0.5 degrees for the angular size of the Sun and Moon), manipulate the equation so that the unknown is on the left-hand-side and the known values are all on the right-hand-side. Percentage Error: Calculating the percentage error is something we all (at least in theory) learned in middle-school or earlier. In calculating this error, we compare the value we obtained to the accepted (true) my value true value value. The formula to use is: 100 % error true value Important Additional Information: You may wish to do some outside reading on lunar eclipses to fully understand what is happening when you get to the end of "Rung 2" of the exercise. 1/3/2012 Size of the Earth and the Distances to the Moon and Sun.doc 11