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 motionPart (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 ac = 2.72×10-3 m/s2. If Newton had taken the mass of Earth to be ME = 6.01×1024 kg and the mean distance between the centers of Earth and the moon to be RME = 3.85×108 m, what value would he have obtained for the gravitational constant, in units of N⋅m2/kg2? G = 6.685 * 10( - 11 )G = 6.685E-11 ✔ Correct! Part (b) Since measuring the centripetal acceleration of an orbiting body is rather difficult, an alternative approach is to use the body’s rotational 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. G = 4 π2 RME3/( T2 ME ) ✔ Correct! Part (c) Using the expression you entered in part (b) and taking the rotational period of the moon to be T = 27.3 days, what value would Newton have calculated for the gravitation constant, in units of N⋅m2/kg2? Take ME = 6.01×1024 kg and RME = 3.85×108 m. G = 6.92 * 10( - 11 )G = 6.92E-11 ✔ Correct! 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 vi. Its initial distance from the center of Earth is Ri, 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 vi, Ri, Rf, and ME. vi2 / ( 2 ME ( 1 / Ri - 1 / Rf ) ) ✔ Correct! Part (e) Suppose a rocket is launched as described in part (d) with an initial speed of vi = 491 m/s and attains a peak altitude of H = 13.3 km above the surface of Earth. Taking ME = 6.01×1024 kg and Ri = 6.43×106 m, what is the measured value of the gravitational constant, in units of N⋅m2/kg2?

(e) Initial speed of the rocket vi=491 m/s, radius of the Earth Ri=6.43*106m, mass of the Earth…

Answered: Part (d) The gravitational constant…

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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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_c = 2.72×10^-3 m/s². If Newton had taken the mass of Earth to be M_E = 6.01×10²⁴ kg and the mean distance between the centers of Earth and the moon to be R_ME = 3.85×10⁸ m, what value would he have obtained for the gravitational constant, in units of N⋅m²/kg²?

G = 6.685 * 10^( - 11 )
G = 6.685E-11 ✔ Correct!

Part (b) Since measuring the centripetal acceleration of an orbiting body is rather difficult, an alternative approach is to use the body’s rotational period instead. Enter an expression for the gravitational constant, in terms of the distance between Earth and the moon, R_ME, Earth’s mass,M_E, and the moon’s period of rotation around Earth, T.

G = 4 π² R_ME³/( T² M_E ) ✔ Correct!

Part (c) Using the expression you entered in part (b) and taking the rotational period of the moon to be T = 27.3 days, what value would Newton have calculated for the gravitation constant, in units of N⋅m²/kg²? Take M_E = 6.01×10²⁴ kg and R_ME = 3.85×10⁸ m.

G = 6.92 * 10^( - 11 )
G = 6.92E-11 ✔ Correct!

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_i. Its initial distance from the center of Earth is R_i, the radius of Earth. Its peak distance, where its speed is momentarily zero is, is R_f. For simplicity, ignore air resistance and Earth’s rotation. Enter an expression for the gravitational constant, in terms of v_i, R_i, R_f, and M_E.

v_i² / ( 2 M_E ( 1 / R_i - 1 / R_f ) ) ✔ Correct!

Part (e) Suppose a rocket is launched as described in part (d) with an initial speed of v_i = 491 m/s and attains a peak altitude of H = 13.3 km above the surface of Earth. Taking M_E = 6.01×10²⁴ kg and R_i = 6.43×10⁶ m, what is the measured value of the gravitational constant, in units of N⋅m²/kg²?

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