Size of the Earth and the Distances to the Moon and Sun

Waterloo Collegiate Astronomy AssignmentSES4UISize of the Earth and the Distances to the Moon and SunObjectivesUsing observations of the Earth-Sun-Moon system and elementary geometry and trigonometry, the studentwill duplicate the methods of the ancient Greek astronomers in determining the size of the Earth, the size ofand distance to the Moon, and the size of and distance to the Sun. In doing so, the student will derive thevalue 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 touse a separate answer sheet to show you work and answer questions if you require more space.Materials‣ Scientific calculators‣ rulersIntroductionAs we look into the night sky, the Moon, planets, and stars appear to be set within a crystalline sphere. Weobserve the motion of these objects from night to night during the course of a year and discover predictablepatterns. The Sun, our star, reliably rises and sets each day. From a limited perspective, perhaps this is all weneed 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 notgive 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 yearsacross our visible limit—the “diameter” of our visible realm. We started with small steps on what has becomeknown as the "cosmic distance ladder." Our knowledge of distances started over 2300 years ago with the GreeksAristarchus of Samos (310-230 BC) and Eratosthenes (276-196 BC). Aristarchus suggested well beforeCopernicus that the Earth orbited the Sun, but his ideas were rejected. Erastosthenes enjoyed more successwith 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 theMoon and Sun. Most of these measurements were quite difficult for the ancient astronomers to carry out as allof this was done without telescope technology, and without precise measuring tools. In this exercise you willbe following the methods of Eratosthenes and Aristarchus. Although you will be following their methods, youwill be using more precise observations than were available to them. In fact, you will see that the simplemodel which these Hellenistic Greek thinkers had for the Sun-Earth-Moon system allows quite accuratedetermination 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 itselforbiting the Sun, these were all revolutionary ideas in the history of science. Consider your mind wiped clear ofwhat you have learned about astronomy and imagine how hard it would be to come up with these ideas bysimply 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 areview, in case it’s needed.1/3/2012 Size of the Earth and the Distances to the Moon and Sun.doc 1

ProcedureRung 1: The size of the EarthThis step was first performed by Eratosthenes and involves a very simple model for just the Sun andthe Earth: 1) the Earth is a sphere; 2) the Sun is very far away. The fact that the Earth is a sphere wasknown from elementary observations (the shape of the shadow on the Moon during an eclipse, the way atall ship disappeared as it sailed away), but what is the significance of assuming the Sun to be very faraway?Eratosthenes knew that at noon on June 21 (the summer solstice and the longest day of the year), the Sunwould shine into the bottom of a deep vertical pit in the city of Syene, meaning the Sun was at the zenith in thesky at this time. Being in Alexandria, which was located moreor less due north of Syene, he could measure the length of ashadow cast by an obelisk at the same time, noon on thesummer solstice. This measurement, along with the height ofthe obelisk, gave him the angle that the Sun appeared south ofthe zenith in Alexandria, an angle he determined to be 7.2degrees. The Sun was not as high in the sky in Alexandria as inSyene at the same time of day and the same time of year. Thisis in fact a measurement of the curvature of the Earth's spherebetween Syene and Alexandria, and once distance of this smallarc is known, the circumference of theentire Earth is determined. To this end (astronomers havingconsiderably more political clout in those days), he orderedsome soldiers to march off thedistance between Alexandria and Syene, a distance of 5000stadia (thought today to be equivalent to about 805 km or500 miles).The circumference of the Earth could be found by using the following proportionality:Syene to Alexandria 7.2Circumference of Earth 360What 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

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 twomeasurements, 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 kmscale, find out how many km (in the real world) is represented by 1 mm (on the map). You can also useGoogle 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 inkilometers.5

Rung 2: The size of and distance to the MoonThe 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 ofthe Earth due to Eratosthenes and was first carried out by Aristarchus of Samos (310-230 BC). Once again amodel is required to make the determination of the relative sizes of the Earth and Moon, in particular a goodmodel for what is taking place during a lunar eclipse. It was surmised that during a lunar eclipse the fullMoon is passing through the shadow of the Earth. Aristarchus timed how long the Moon took to travelthrough Earth's shadow and compared this with the time required for the Moon to move a distance equal to itsdiameter (this could be done by timing how long a bright star was obscured by the Moon as the Moon passedbetween 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 Moonto the Earth by looking at the curvature of the Earth's shadow during a lunar eclipse. All of the abovemodel statements are still important to this conceptually simpler method. The measurement of the Earth'sshadow is not without its difficulties and uncertainties, however, the main reason being that the Sun is notinfinitely far away.6

Now use the curvature of the Earth's shadow during a lunar eclipse to determine the relative sizes of the Earthand Moon. Review the sketches above of an eclipse of the Moon by the Earth’s shadow, and state oneassumption 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 GalleryYou now have an approximation of the ratio of the radius of the Moon to the radius of the Earth. Use theradius of the Earth that you determined from Rung 1 of this exercise to find the radius of the Moon inkilometers. Be sure to watch out for whether or not you are asked to use the radius or the diameter of theMoon for the subsequent calculations.7

After making your measurements, give your opinion as to why this method would have been difficult for theancient 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 bymeasuring its angular diameter on the sky. The angle subtended by the full Moon on the sky is about 0.5degrees. Calculate the distance to the Moon using its absolute diameter and its angular diameter. For helpwith the geometry of this problem and with the trigonometry involved with this calculation, take a close look atthe logic used here:8

Rung 3: The Distance to the SunAristarchus 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 longerassume the Sun to be very far away in the sense that we did for the previous two rungs of our distance ladder (weessentially assumed the Sun to be infinitely far away). What we are relying on is that the Sun's rays do not hitboth 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 thedistance 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 atslightly differing angles; 3) the Moon orbits the Earth. Why is the second assumption in our model soimportant 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 thefollowing diagram:Note that when the moon is seen to be exactly in thefirst quarter phase, the Sun-Earth-Moon angle is a rightangle, or 90 degrees. If we can measure the anglebetween the Sun and the Moon when the Moon isprecisely in its first (or third) quarter phase, then wecan determine the distance to the Sun. With preciseobservations made at first-quarter lunar phase, aMoon-Earth-Sun angle of 89.853 degrees is measured(extremely close to 90 degrees!). We know the distancefrom the Earth to the Moon, we determined that in Rung2 of this exercise. The Earth-Moon distance plus themeasured Moon-Earth-Sun angle gives us one side andone angle of a right triangle, and we can usetrigonometry to determine the length of the hypotenuse--the distance to the Sun. Think for a moment aboutwhat might make this observation difficult to carry out, referring to what you learned in Rungs 1 and 2 andfrom the images given above. (Hint: Take a look at the images shown at the right. How difficult do you think itwould 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 todetermine 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 theSun in kilometers. We know that the angular diameter of the Sun is almost the same as the Moon, about 0.5degrees 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 theMoon from an estimate of its radius in kilometers.SummaryAs you wrap up your results for this exercise, check the actual quantities listed in the table on your answersheet and calculate the percentage error for each value.10

Size of the Earth and the Distances to the Moon and SunSome Review of MathematicsThis exercise starts us on the road to using a bit of math to describe our universe. Provided here is a briefrefresher 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 thatone ratio will be proportional to another ratio, and we know the values of three of the numbers, it makes findingthe unknown fourth number easy.Here is a simple example: A 3 3150 , A 30150 15 15When working with ratios, some of the units may cancel out: A 3mi 3150km , A 30km150km15mi15Here 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 xmix 15mi 15miAfter a bit more manipulation:15150 km x, x 750km3A little bit of trigonometry: The hardest part about trigonometry these days is trying to figure out whichbuttons to push on the calculator. A refresher for Sine, Cosine, and Tangent of an angle for a right-angletriangle, 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 theright-hand-side.Percentage Error: Calculating the percentage error is something we all (at least in theory) learned inmiddle-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 % errortrue valueImportant Additional Information: You may wish to do some outside reading on lunar eclipses to fullyunderstand 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