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The radius of the earth’s orbit around the sun (assumed to

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

Solution for problem 28E Chapter 3

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 28E

The radius of the earth’s orbit around the sun (assumed to be circular) is 1.50 X 108 km, and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in m/s? (b) What is the radial acceleration of the earth toward the sun, in m/s2? (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = 5.79 X 107 km, orbital period = 88.0 days).

Step-by-Step Solution:

Solution 28E Step 1: Equation for speed in centripetal motion, v = distance /time = 2 r / T Where, 2 r - circumference of the circular path T - Time taken to cover this circular path Step 2: a) In the case of earth, radius of earth’s orbit, r - 1.50 × 10 km 1 km = 1000 m Therefore, r - 1.50 × 10 × 1000 m = 1.50 × 10 m 11 Therefore, the circumference of the orbit of the earth, C = 2 r = 2 × 3.14 × 1.50 × 10 m 11 C = 9.42×10 m11 Step 3: a ) Time required to cover this orbit, T = 365 days We know that, 1 day = 24 hours Therefore, 365 days = 365 × 24 hours = 8760 hours 1 hour = 60 minutes Therefore, 8760 hours = 8760 × 60 minutes = 525600 minutes 1 minute = 60 seconds Therefore, 525600 minutes = 525600 × 60 seconds = 3.1536 × 10 seconds 7 Step 4: 11 7 a) Speed or magnitude of velocity of earth, v = (9.42×10 m) / (3.1536 × 10 seconds) v = 2.987 × 10 m/s 4 Step 5: 2 b) E quation for centripetal acceleration, a = v /r Where, v - magnitude of velocity of centripetal motion r - radius of the orbit In the case of earth, v = 2.987 × 10 m/s (from step 4) and r = 1.50 × 10 m 11 Therefore, a = (2.987 × 10 m/s) / (1.50 × 10 m) = (8.922 × 10 m /s ) / (1.50 × 10 m) 8 2 2 11 -3 2 a = 5.948 × 10 m/s Radial acceleration of earth towards the sun is, a = 5.948 × 10 m/s -3 2

Step 6 of 9

Chapter 3, Problem 28E is Solved
Step 7 of 9

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 28E from chapter: 3 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: EARTH, orbit, radius, sun, days. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 28E from 3 chapter was answered, more than 2016 students have viewed the full step-by-step answer. The answer to “The radius of the earth’s orbit around the sun (assumed to be circular) is 1.50 X 108 km, and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in m/s? (b) What is the radial acceleration of the earth toward the sun, in m/s2? (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = 5.79 X 107 km, orbital period = 88.0 days).” is broken down into a number of easy to follow steps, and 81 words.

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