The centers of a 10 kg lead ball and a 100 g lead ball are

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Problem 28P Chapter 6

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 28P

The centers of a 10 kg lead ball and a 100 g lead ball are separated by 10 cm. a. What gravitational force does each exert on the other? b. What is the ratio of this gravitational force to the weight of the 100 g ball?

Step-by-Step Solution:

Solution Step 1 of 4 a. What gravitational force does each exert on the other Gravitational force is the attraction force exerted by the two objects on each other with certain masses. Which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.Example: In Solar system. As shown in the figure below, G m1m 2 That is, F g r Where F is ghe gravitational force, m and m1are the 2asses of the two objects and r is 11 2 2 the distance between them. G=6.67 ×10 Nm /kg is universal gravitational constant. Step 2 of 4 In the given problem, two balls of mass m = 10 kg and m =100g separated by a distance 1 2 of r= 10cm. Mass m = 10 kg and m =100g 1 2 3 3 Using 1g = 1×10 kg m 2100×10 kg=0.1kg Radius, r =10 cm Using 1 cm = 1×10 m 2 r =10 ×10 m= 0.1m 11 2 G=6.67 ×10 Nm /kg Substituting in the above equation, 11 2 2 F g (6.67 ×10Nm /kg 2 (10 kg)(0.1kg) (0.1m) F =6.67×10 N 9 g Therefore, the gravitational force that both exert on each other is 6.67×10 N. 9

Step 3 of 4

Chapter 6, Problem 28P is Solved
Step 4 of 4

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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