/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q61P The masses of the Earth and Moon... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The masses of the Earth and Moon are \({\bf{5}}{\bf{.98 \times 1}}{{\bf{0}}^{{\bf{24}}}}\;{\bf{kg}}\) and \({\bf{7}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{22}}}}\;{\bf{kg}}\), respectively, and their centers are separated by \({\bf{3}}{\bf{.84 \times 1}}{{\bf{0}}^{\bf{8}}}\;{\bf{m}}\).

(a) Where is the CM of the Earth–Moon system located?

(b) What can you say about the motion of the Earth–Moon system about the Sun, and of the Earth and Moon separately about the Sun?

Short Answer

Expert verified

(a) The center of mass of the Earth-Moon system is\(4.66 \times {10^6}\;{\rm{m}}\)away from the Earth.

(b) The Moon will rotate around the Earth, and the Earth will rotate about the Sun.

Step by step solution

01

Motion of the Moon around the Sun

The movement of the Moon around the Sun is the addition of two motions:

(i) the movement of the Moon about the Earth-Moon CM, and

(ii) the movement of the Earth-Moon CM about the Sun.

02

Given information

The mass of the Earth is\({M_{\rm{E}}} = 5.98 \times {10^{24}}\;{\rm{kg}}\).

The mass of the Moon is\({M_{\rm{M}}} = 7.35 \times {10^{22}}\;{\rm{kg}}\).

The separation between the center of the Earth and the Moon is\({x_{\rm{M}}} = 3.84 \times {10^8}\;{\rm{m}}\).

03

Calculate the center of mass of the Earth-Moon system

(a)

Consider the center of the Earth as the origin. Then the distance of the center of the Earth from the CM is\({x_{\rm{E}}} = 0\).

The center of mass of the Earth-Moon system can be calculated as shown below:

\(\begin{array}{c}{x_{{\rm{cm}}}} = \frac{{{M_{\rm{E}}}{x_{\rm{E}}} + {M_{\rm{M}}}{x_{\rm{M}}}}}{{{M_{\rm{E}}} + {M_{\rm{M}}}}}\\{x_{{\rm{cm}}}} = \frac{{\left( {5.98 \times {{10}^{24}}\;{\rm{kg}}} \right)\left( 0 \right) + \left( {7.35 \times {{10}^{22}}\;{\rm{kg}}} \right)\left( {3.84 \times {{10}^8}\;{\rm{m}}} \right)}}{{\left( {5.98 \times {{10}^{24}}\;{\rm{kg}}} \right) + \left( {7.35 \times {{10}^{22}}\;{\rm{kg}}} \right)}}\\{x_{{\rm{CM}}}} = 4.66 \times {10^6}\;{\rm{m}}\end{array}\)

Thus, the center of mass of the Earth-Moon system is \(4.66 \times {10^6}\;{\rm{m}}\) away from the Earth.

04

Motion of the Earth and Moon system

(b)

As the CM lies very close to the Earth, the Moon rotates around the Earth while the Earth is rotating around the Sun. Here, the Earth is considered as the center of mass of the Earth-Moon system.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Calculate the impulse experienced when a 55-kg person lands on firm ground after jumping from a height of 2.8 m.

(b) Estimate the average force exerted on the person’s feet by the ground if the landing is stiff-legged, and again

(c) with bent legs. With stiff legs, assume the body moves 1.0 cm during impact, and when the legs are bent, about 50 cm. [Hint: The average net force on him, which is related to impulse, is the vector sum of gravity and the force exerted by the ground. See Fig. 7–34.] We will see in Chapter 9 that the force in (b) exceeds the ultimate strength of bone (Table 9–2).

FIGURE 7-34 Problem 24.

A pendulum consists of a mass M hanging at the bottom end of a massless rod of length l which has a frictionless pivot at its top end. A mass m, moving as shown in Fig. 7–35 with velocity v, impacts M and becomes embedded. What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear over the top of its arc?

FIGURE 7-35Problem 42.

Why is the CM of a 1-m length of pipe at its midpoint, whereas this is not true for your arm or leg?

You have been hired as an expert witness in a court case involving an automobile accident. The accident involved car A of mass 1500 kg which crashed into stationary car B of mass 1100 kg. The driver of car A applied his brakes 15 m before he skidded and crashed into car B. After the collision, car A slid 18 m while car B slid 30 m. The coefficient of kinetic friction between the locked wheels and the road was measured to be 0.60. Show that the driver of car A was exceeding the 55-mi/h (90 km/h) speed limit before applying the brakes.

A 12-kg hammer strikes a nail at a velocity of 7.5 m/s and comes to rest in a time interval of 8.0 ms.

(a) What is the impulse given to the nail?

(b) What is the average force acting on the nail?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.