The free-fall acceleration at the surface of planet 1 is .

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

Problem 30P 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 30P

The free-fall acceleration at the surface of planet 1 is . The radius and the mass of planet 2 are twice those of planet 1. What is the free-fall acceleration on planet 2?

Step-by-Step Solution:

Solution Step 1 of 4 The acceleration due to gravity at the surface is g and it is given by, GM g = R2 Where G is the universal gravitational constant, M is the mass and R is the radius. Step 2 of 4 Given data, Mass of two planets, M = M a1d M = 2M 2 Radius of the planet 1, R = R 1 Radius of the planet 2, R = 2R 2 Acceleration due to gravity on the surface of planet 1, g = 21 m/s To find Acceleration due to gravity on the surface of planet 2, g = 2 Using the above data we can find the acceleration due to gravity at the surface of the planet 1, GM g1= 2 R Where G is the universal gravitational constant, M is the mass and R is the radius of planet 1 .

Step 3 of 4

Chapter 6, Problem 30P 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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