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At a certain instant, the earth, the moon, and a

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 48P Chapter 13

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 48P

At a certain instant, the earth, the moon, and a stationary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 X 105 km in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? Ignore any gravitational effects due to the other planets or the sun.

Step-by-Step Solution:

Problem (a) Step 1: To find the magnitude of the net gravitational force 24 Mass of Earth M = E97 x 10 kg 22 Mass of Moon M = 7.M x 10 kg Mass of Satellite M = 1250 kg S Distance between Earth or Moon and Satellite R = 3.84 x 10 km or 3.84 x 10 m 8 Gravitational constant G = 6.67 x 10 -11N.m kg -2 Step 2: The force exerted on the satellite by Earth F E GM M F = 2 S E R 6.67*10 1* 5.97*1024*1250 FE 8 2 (3.84*10 ) FE= 3.376 N Step 3: The force exerted on the satellite by Moon F M GM M S FM R2 11 22 F = 6.67*10 *7.35*10 *1250 M (3.84 10 )2 * F = 0.042 N M Step 4: The vectors represent parallelogram. Therefore we can apply parallelogram law to find the resultant force or net gravitational force F. F = F 2 + F 2 + 2.F F cos E M E M Here = 60° (triangle is equilateral) F = 3.376 + 0.042 + 2(3.376)(0.042)cos60 F = 3.39 N or 3.4 N The net gravitational force is 3.4 N Step 5: Direction of force F sin = tan (1 M ) F EF cMs 1 0.042*sin60 = tan ( 3.376+0.04* cos60 tan (0.01071) = = 0.613 or 0.61° from Earth and spacecraft line Problem (b) Step 1: To find the amount of work to move the spacecraft to a point far from the earth and moon Gravitational potential energy of the satellite at current point U , 1 U = - G(M EM )M S 1 R Gravitational potential energy of the satellite at far po2t U , where R = U2 0

Step 2 of 2

Chapter 13, Problem 48P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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