Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

Answers without the blur.

Just sign up for free and you're in.

Short Answer

The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. The radius of the Moon is about $0.250 R_{E}$ $(R_{E} = Earths radius) = 6.37 \times 10^{6} m.$ . Find the ratio of their average densities $ρ_{M o o n / ρ_{E a r t h}}$ .

Thus, the required ratio of their average densities $\frac{ρ_{M o o n}}{ρ_{E a r t h}} = 0.6667$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Concept

An object at a distance h above the Earth’s surface experiences a gravitational force of magnitude mg, where g is the free-fall acceleration at that elevation:

$g = \frac{G M_{E}}{r^{2}} = \frac{G M_{E}}{{(R {}_{E}+ h)}^{2}}$

And force is given by:

$F = \frac{G M_{E} m_{a}}{{(R {}_{E}+ r)}^{2}}$

$M_{E}$ is the mass of the Earth and $R_{E}$ is its radius. Therefore, the weight of an object decreases as the object moves away from the Earth’s surface.

Step 2: Find the ratio of their average densities

Let free fall acceleration of moon is $g_{M o o n}$ .

Let free fall acceleration of earth is $g_{E a r t h}$ .

Radius of moon $R_{M o o n}$

it is Given that in question:

$g_{M o o n} = \frac{g_{E a r t h}}{6}$

According to Concept, we have

$\frac{G m_{M o o n}}{{R_{M o o n}}^{2}} = \frac{G m_{E a r t h}}{6 \times {R_{E a r t h}}^{2}} \frac{ρ {}_{M o o n}\times \frac{4}{3} π {R^{3}}_{M o o n}}{{R_{M o o n}}^{2}} = \frac{1}{6} \times \frac{ρ {}_{E a r t h}\times \frac{4}{3} π {R^{3}}_{E}}{{R_{E}}^{2}} . . . . . . . (M = ρ V; V = \frac{4}{3} π R^{3}) ρ {}_{M o o n}\times R_{M o o n} = \frac{ρ {}_{E a r t h}\times R_{E}}{6} \frac{ρ_{M o o n}}{ρ_{E a r t h}} = \frac{R_{E}}{6 R_{M o o n}} \frac{ρ_{M o o n}}{ρ_{E a r t h}} = \frac{R_{E}}{6 \times 0.250 R_{E}} . . . . . . . . . (R_{M o o n} = 0.250 R_{E}) \frac{ρ_{M o o n}}{ρ_{E a r t h}} = \frac{1}{1.5} \frac{ρ_{M o o n}}{ρ_{E a r t h}} = 0.6667$

Thus the required ratio of their average densities $\frac{ρ_{M o o n}}{ρ_{E a r t h}} = 0.6667$

Find Study Materials for

Create Study Materials

Select your language

Physics For Scientists & Engineers

Answers without the blur.

Short Answer

The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. The radius of the Moon is about $0.250 R_{E}$ $(R_{E} = Earths radius) = 6.37 \times 10^{6} m.$ . Find the ratio of their average densities $ρ_{M o o n / ρ_{E a r t h}}$ .

Step by Step Solution

Step 1: Concept

Step 2: Find the ratio of their average densities

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Physics For Scientists & Engineers

Answers without the blur.

Short Answer

The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. The radius of the Moon is about 0.250RE (RE=Earths radius)=6.37×106 m.. Find the ratio of their average densities ρMoon/ρEarth.

Step by Step Solution

Step 1: Concept

Step 2: Find the ratio of their average densities

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics

The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. The radius of the Moon is about $0.250 R_{E}$ $(R_{E} = Earths radius) = 6.37 \times 10^{6} m.$ . Find the ratio of their average densities $ρ_{M o o n / ρ_{E a r t h}}$ .