×
Get Full Access to University Physics - 13 Edition - Chapter 9 - Problem 77p
Get Full Access to University Physics - 13 Edition - Chapter 9 - Problem 77p

×

# The earth, which is not a uniform sphere, has a moment of

ISBN: 9780321675460 31

## Solution for problem 77P Chapter 9

University Physics | 13th Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants

University Physics | 13th Edition

4 5 1 290 Reviews
18
4
Problem 77P

The earth, which is not a uniform sphere, has a moment of inertia of 0.3308?MR?2 about an axis through its north and south poles. It takes the earth 86,164 s to spin once about this axis. Use Appendix F to calculate (a) the earth’s kinetic energy due to its rotation about this axis and (b) the earth’s kinetic energy due to its orbital motion around the sun. (c) Explain how the value of the earth’s moment of inertia tells us that the mass of the earth is concentrated toward the planet’s center.

Step-by-Step Solution:
Step 1 of 3

Solution 77P 1 2 The earth’s kinetic energy due to rotation about its axis is = I , wher2 I is the moment of inertia of the earth and is angular velocity. 2 Given that, I = 0.3308MR Time period to spin about its axis = 86.164 s Angular velocity = 2/86,164 rad/s = 7.3 × 10 5 rad/s 24 Now, the mass of earth is M = 5.97 × 10 kg Radius of earth R = 6.37 × 10 m 6 24 6 2 2 So, I = 0.3308 × (5.97 × 10 ) kg × (6.37 × 10 ) m I = 8.01 × 10 37 kg.m 2 (a) Now, substituting the values of I and in the expression for kinetic energy K = I , 2 2 1 37 2 5 2 K = 2 × 8.01 × 10 kg.m × (7.3 × 10 rad/s) 27 K = 214 × 10 J K = 2.14 × 10 29 J This is the earth’s kinetic energy due to its rotation about this axis. (b) The earth takes 365 days for a complete revolution around the sun. 7 Therefore period of revolution = 365 days = 365 × 24 × 60 × 60 s = 3.15 × 10 s 11 Orbital radius of earth r =o1.50 × 10 m The linear velocity is v = r 2 o3.15×107 11 v = 1.50 × 10 m × 2×3.147 s 3.15×10 4 v = 3.0 × 10 m/s Therefore, the kinetic energy of orbital motion K =o 1 × 5.97 × 10 24 kg × (3.0 × 10 ) J 2 K = 26.6 × 10 J 32 o 33 K =o2.66 × 10 J This is the earth’s kinetic energy due to its orbital motion around the sun. 2 (c) For earth, the I/MR value is 0.3308. For a perfect sphere, the value is 0.4. Smaller is the I/MR value than 0.4, more the mass of the body will be concentrated toward the center of the planet.

Step 2 of 3

Step 3 of 3

#### Related chapters

Unlock Textbook Solution