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3. Convert 5,000 stadia to kilometers. Again, you may use the following proportion: I stadia 0.1575 km 5,000 stadia D km (Hint: crossmukiplydivide) Thus, D= km. 4. Use the formula a = D/R to solve for R. Then us ing these values of a in radians (2), and D in kilometers (3), caleulate the value of the radius of the Earth, R, in kilometers. R km. 5. The radius of the Earth is known to be Rah =6378.1 km (accepted value). Was your calculation close to the accepted value? Find the percent error using this formula: % error = laccepted value – experimental value| /accepted value % error=

3. Convert 5,000 stadia to kilometers. Again, you may use the following proportion: I stadia 0.1575 km 5,000 stadia D km (Hint: crossmukiplydivide) Thus, D= km. 4. Use the formula a = D/R to solve for R. Then us ing these values of a in radians (2), and D in kilometers (3), caleulate the value of the radius of the Earth, R, in kilometers. R km. 5. The radius of the Earth is known to be Rah =6378.1 km (accepted value). Was your calculation close to the accepted value? Find the percent error using this formula: % error = laccepted value – experimental value| /accepted value % error=

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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BACKGROUND An ingenious solution to the Earth's circumference occured in 230 BC. Eratosthenes, a Greek geographer, mathematician, music theorist, poet, astronomer, and philosopher, was reading in the Library of Alexandria when he noticed an account for a deep well near Syene (now Aswan), some distance to the south (800 km) in which at high noon on the longest day of the year the bottom of the well was fully illuminated by the Sun. Eratosthenes exclaimed "Ah-ah!" (or something like that), "I can solve for the circumference of the Earth!". In his mind's eye, Eratosthenes could see that at Syene, at the moment when the bottom of the well was fully lit, the Sun must have been at the Zenith (directly overhead). Yet he knew that at the same moment in Alexandria vertical objects (like a tower, pole) cast shadows. Here is the experiment perfomed by Eratosthenes (see the picture below). • He erected a vertical pole at Alexandria (A) and measured the angle of its shadow at the moment when the well at Syene (S) was fully illuminated; • The angle of the pole equaled the angle formed by the points Alexandria (A), Earth's center (C), and Syene (S). That angle proved to be approximately a = 7°. • The distance between Syene and Alexandria is 5,000 stadia (the units of measure at the time). I stadia = 157.5 m=0.1575 km. Figure 1. The measure of Earth's circumference is based on the assumption that Syene is on the Tropic of Cancer and the same meridian as Alexandria. The image is adapted from a similar image prepared by Gico (Wikimedia).
Transcribed Image Text:BACKGROUND An ingenious solution to the Earth's circumference occured in 230 BC. Eratosthenes, a Greek geographer, mathematician, music theorist, poet, astronomer, and philosopher, was reading in the Library of Alexandria when he noticed an account for a deep well near Syene (now Aswan), some distance to the south (800 km) in which at high noon on the longest day of the year the bottom of the well was fully illuminated by the Sun. Eratosthenes exclaimed "Ah-ah!" (or something like that), "I can solve for the circumference of the Earth!". In his mind's eye, Eratosthenes could see that at Syene, at the moment when the bottom of the well was fully lit, the Sun must have been at the Zenith (directly overhead). Yet he knew that at the same moment in Alexandria vertical objects (like a tower, pole) cast shadows. Here is the experiment perfomed by Eratosthenes (see the picture below). • He erected a vertical pole at Alexandria (A) and measured the angle of its shadow at the moment when the well at Syene (S) was fully illuminated; • The angle of the pole equaled the angle formed by the points Alexandria (A), Earth's center (C), and Syene (S). That angle proved to be approximately a = 7°. • The distance between Syene and Alexandria is 5,000 stadia (the units of measure at the time). I stadia = 157.5 m=0.1575 km. Figure 1. The measure of Earth's circumference is based on the assumption that Syene is on the Tropic of Cancer and the same meridian as Alexandria. The image is adapted from a similar image prepared by Gico (Wikimedia).
3. Convert 5,000 stadia to kilometers. Again, you may use the following proportion: 1 stadia 0.1575 km 5,000 stadia Dkm (Hint: cros-multiply-divide) Thus, D = km. 4. Use the formula a = D/R to solve for R. Then using these values of a in radians (2), and D in kilometers (3), caleulate the value of the radius of the Earth, R, in kilometers. R = km. 5. The radius of the Earth is known to be REath = 6378.1 km (accepted value). Was your calculation close to the accepted value? Find the percent error us ing this formula: % eror = |accepted value – experimental value| / accepted value % error =
Transcribed Image Text:3. Convert 5,000 stadia to kilometers. Again, you may use the following proportion: 1 stadia 0.1575 km 5,000 stadia Dkm (Hint: cros-multiply-divide) Thus, D = km. 4. Use the formula a = D/R to solve for R. Then using these values of a in radians (2), and D in kilometers (3), caleulate the value of the radius of the Earth, R, in kilometers. R = km. 5. The radius of the Earth is known to be REath = 6378.1 km (accepted value). Was your calculation close to the accepted value? Find the percent error us ing this formula: % eror = |accepted value – experimental value| / accepted value % error =
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