Problem 8: In this problem you will measure the gravitational constant in a series of "observational experiments," making use of Newton's law of gravitation and second law of motion as well as Kepler's third law of planetary motion Part (a) Newton measured the centripetal acceleration of the moon in its orbit around Earth by comparing the force Earth exerts on the moon with the force Earth exerts on an apple. He obtained a value of a = 2.65×10³ m/s². If Newton had taken the mass of Earth to be Mp =5.97x1024 kg and the mean distance between the centers of Earth and the moon to be RME = 3.83x108 m, what value would he have obtained for the gravitational constant, in units of N-m?/kg? Numeric : Anumeric value is expected and not an expression. G = Part (b) Since measuring the cent period instead. Enter an expression for the gravitational constant, in terms of the distance between Earth and the moon, RME, Earth's mass ME, and the moon's period of rotation around Earth, T. Expression : G = acceleration of an orbiting body is rather difficult, an alternative approach is to use the body's rotational Select from the variables below to write your expression. Note that all variables may not be required. B. y. 7, 0, d, g, h, j, k, m, ME, n, P, RME, T Part (c) Using the expression you entered in part (b) and taking the rotational period of the moon to be T = 27.6 days, what value would Newton have calculated for the gravitation constant, in units of N-m2/kg2? Take Mp = 5.97x1024 kg and RME = 3.83×10% m. Numeric : A numeric value is expected and not an expression. G = Part (d) The gravitational constant may also be calculated by analyzing the motion of a rocket. Suppose a rocket is launched vertically from the surface or Earth at an initial speed of v;. Its initial distance from the center of Earth is R;, the radius of Earth. Its peak distance, where its speed is momentarily zero is, is Rf. For simplicity, ignore air resistance and Earth's rotation. Enter an expression for the gravitational constant, in terms of Vị, Rị, Rf, and ME- MultipleChoice: 1) v,? (1/R; - 1/Rf)/(2 MẸ ) 2) v,? / ( 2 MẸ ( 1 / Rf - 1 / R; )) 3) v? / ( 2 MẸ (1/ R¡ - 1 / Rf ) ) 4) v,? R¢ / ( 2 MẸ ) 5) v; / ( 2 MẸ ( 1/ R; - 1/ Rf )) 6) v? R; / ( 2 MẸ ) Part (e) Suppose a rocket is launched as described in part (d) with an initial speed of v; = 503 m/s and attains a peak altitude of H = 13.5 km above the surface of Earth. Taking Mp = 5.97x1024 kg and R; = 6.41x106 m, what is the measured value of the gravitational constant, in units of N-m?/kg? Numeric : A numeric value is expected and not an expression. G =
Problem 8: In this problem you will measure the gravitational constant in a series of "observational experiments," making use of Newton's law of gravitation and second law of motion as well as Kepler's third law of planetary motion Part (a) Newton measured the centripetal acceleration of the moon in its orbit around Earth by comparing the force Earth exerts on the moon with the force Earth exerts on an apple. He obtained a value of a = 2.65×10³ m/s². If Newton had taken the mass of Earth to be Mp =5.97x1024 kg and the mean distance between the centers of Earth and the moon to be RME = 3.83x108 m, what value would he have obtained for the gravitational constant, in units of N-m?/kg? Numeric : Anumeric value is expected and not an expression. G = Part (b) Since measuring the cent period instead. Enter an expression for the gravitational constant, in terms of the distance between Earth and the moon, RME, Earth's mass ME, and the moon's period of rotation around Earth, T. Expression : G = acceleration of an orbiting body is rather difficult, an alternative approach is to use the body's rotational Select from the variables below to write your expression. Note that all variables may not be required. B. y. 7, 0, d, g, h, j, k, m, ME, n, P, RME, T Part (c) Using the expression you entered in part (b) and taking the rotational period of the moon to be T = 27.6 days, what value would Newton have calculated for the gravitation constant, in units of N-m2/kg2? Take Mp = 5.97x1024 kg and RME = 3.83×10% m. Numeric : A numeric value is expected and not an expression. G = Part (d) The gravitational constant may also be calculated by analyzing the motion of a rocket. Suppose a rocket is launched vertically from the surface or Earth at an initial speed of v;. Its initial distance from the center of Earth is R;, the radius of Earth. Its peak distance, where its speed is momentarily zero is, is Rf. For simplicity, ignore air resistance and Earth's rotation. Enter an expression for the gravitational constant, in terms of Vị, Rị, Rf, and ME- MultipleChoice: 1) v,? (1/R; - 1/Rf)/(2 MẸ ) 2) v,? / ( 2 MẸ ( 1 / Rf - 1 / R; )) 3) v? / ( 2 MẸ (1/ R¡ - 1 / Rf ) ) 4) v,? R¢ / ( 2 MẸ ) 5) v; / ( 2 MẸ ( 1/ R; - 1/ Rf )) 6) v? R; / ( 2 MẸ ) Part (e) Suppose a rocket is launched as described in part (d) with an initial speed of v; = 503 m/s and attains a peak altitude of H = 13.5 km above the surface of Earth. Taking Mp = 5.97x1024 kg and R; = 6.41x106 m, what is the measured value of the gravitational constant, in units of N-m?/kg? Numeric : A numeric value is expected and not an expression. G =
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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