/*! 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} Q6.3-34 PE (a) Calculate the magnitude of t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Q6.3-34 PE (page 223)

(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon. (b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun. (c) Take the ratio of the Moon鈥檚 acceleration to the Sun鈥檚 and comment on why the tides are predominantly due to the Moon in spite of this number.

Short Answer

Expert verified

(a) Magnitude of acceleration to the moon,\({{\rm{a}}_{\rm{m}}}{\rm{ = 3}}{\rm{.43x1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\).

(b) Magnitude of acceleration to the sun,\({{\rm{a}}_{\rm{s}}}{\rm{ = 5}}{\rm{.93x1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\).

(c) Ratio of both acceleration of moon to sun\(\frac{{{{\rm{a}}_{\rm{m}}}}}{{{{\rm{a}}_{\rm{s}}}}}{\rm{ = 5}}{\rm{.78x1}}{{\rm{0}}^{{\rm{ - 3}}}}\).

Step by step solution

01

Definition of Gravity

Gravity is a universal phenomenon and is introduced by Newton and Derived the expression for gravitational force.

02

Given Data

Mass of the earth \({{\rm{M}}_{\rm{e}}}{\rm{ = 5}}{\rm{.979 x 1}}{{\rm{0}}^{{\rm{24}}}}{\rm{ kg}}\)

Mass of the moon \({{\rm{M}}_{\rm{m}}}{\rm{ = 7}}{\rm{.3477 x 1}}{{\rm{0}}^{{\rm{22}}}}{\rm{ kg}}\)

Mass of the sun \({{\rm{M}}_{\rm{s}}}{\rm{ = 1}}{\rm{.9891 x 1}}{{\rm{0}}^{{\rm{30}}}}{\rm{ kg}}\)

Distance from surface of the earth to moon

\(\begin{aligned}{}{{\rm{R}}_{\rm{m}}}{\rm{ = }}\left( {{\rm{3}}{\rm{.84 x 1}}{{\rm{0}}^{\rm{8}}}{\rm{-- 6371 x1}}{{\rm{0}}^{\rm{3}}}} \right)\\{{\rm{R}}_{\rm{m}}}{\rm{ = 3}}{\rm{.78 x 1}}{{\rm{0}}^{\rm{8}}}{\rm{ m}}\end{aligned}\)

Distance from surface of the earth to sun

\(\begin{aligned}{}{{\rm{R}}_{\rm{s}}}{\rm{ = }}\left( {{\rm{1}}{\rm{.496 x 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{-- 6371 x 1}}{{\rm{0}}^{\rm{3}}}} \right)\\{{\rm{R}}_{\rm{s}}}{\rm{ = 1}}{\rm{.496 x 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{ m}}\end{aligned}\)

Acceleration due to gravity due to moon\({{\rm{a}}_{\rm{m}}}{\rm{ = ?}}\)

Acceleration due to gravity due to sun \({{\rm{a}}_{\rm{s}}}{\rm{ = ?}}\)

03

Calculation of Magnitude of acceleration to moon

The gravitational force on the surface of the earth due to the moon is given by the equation

\({\rm{F = G}}\frac{{{{\rm{M}}_{\rm{e}}}{{\rm{M}}_{\rm{m}}}}}{{{{\rm{R}}_{\rm{m}}}^{\rm{2}}}}\)

Substituting the value of force in this equation, we get

\({{\rm{M}}_{\rm{e}}}{{\rm{a}}_{\rm{m}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{e}}}{{\rm{M}}_{\rm{m}}}}}{{{{\rm{R}}_{\rm{m}}}^{\rm{2}}}}\)

\({{\rm{a}}_{\rm{m}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{m}}}}}{{{{\rm{R}}_{\rm{m}}}^{\rm{2}}}}\)

\({{\rm{a}}_{\rm{m}}}{\rm{ = 6}}{\rm{.673 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}}\left( {\frac{{{\rm{7}}{\rm{.3477 \times 1}}{{\rm{0}}^{{\rm{22}}}}}}{{{{{\rm{(3}}{\rm{.78x1}}{{\rm{0}}^{\rm{8}}}{\rm{)}}}^{\rm{2}}}}}} \right)\)

\({{\rm{a}}_{\rm{m}}}{\rm{ = 3}}{\rm{.43x1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\)

04

Calculation of Magnitude of acceleration to sun

Similarly, the gravitational force exerted by the sun on the earth's surface is determined by the equation.

\(\begin{aligned}{}{{\rm{M}}_{\rm{e}}}{{\rm{a}}_{\rm{s}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{e}}}{{\rm{M}}_{\rm{s}}}}}{{{{\rm{R}}_{\rm{s}}}^{\rm{2}}}}\\{{\rm{a}}_{\rm{s}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{s}}}}}{{{{\rm{R}}_{\rm{s}}}^{\rm{2}}}}\end{aligned}\)

\({{\rm{a}}_{\rm{s}}}{\rm{ = 6}}{\rm{.673 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}} \times \left( {\frac{{{\rm{1}}{\rm{.9891 \times 1}}{{\rm{0}}^{{\rm{30}}}}}}{{{{{\rm{(1}}{\rm{.496x1}}{{\rm{0}}^{{\rm{11}}}}{\rm{)}}}^{\rm{2}}}}}} \right)\)

\({{\rm{a}}_{\rm{s}}}{\rm{ = 5}}{\rm{.93x1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\)

05

Calculation of ratio of both acceleration of moon to sun

The ratio of the moon鈥檚 acceleration to the sun鈥檚 acceleration

\(\begin{aligned}{}\frac{{{{\rm{a}}_{\rm{m}}}}}{{{{\rm{a}}_{\rm{s}}}}}{\rm{ = }}\frac{{{\rm{3}}{\rm{.43x1}}{{\rm{0}}^{{\rm{ - 5}}}}}}{{{\rm{5}}{\rm{.93x1}}{{\rm{0}}^{{\rm{ - 3}}}}}}\\\frac{{{{\rm{a}}_{\rm{m}}}}}{{{{\rm{a}}_{\rm{s}}}}}{\rm{ = 5}}{\rm{.78x1}}{{\rm{0}}^{{\rm{ - 3}}}}\end{aligned}\)

If we take the difference of centripetal acceleration on two opposite sides of the moon, we get a difference of \({\rm{2}}{\rm{.2 x 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\) and for two opposite sides of the sun, this difference is \({\rm{1 x 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\). This difference tells us Centripetal force on earth is more influenced by the moon rather than the sun.

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 91影视!

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

Modern roller coasters have vertical loops like the one shown in Figure. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \({\bf{15}}.{\bf{0}}{\rm{ }}{\bf{m}}\) and the downward acceleration of the car is \({\bf{1}}.{\bf{50g}}\)?.

Space debris left from old satellites and their launchers is becoming a hazard to other satellites

(a) Calculate the speed of a satellite in an orbit900 km above Earth鈥檚 surface.

(b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite鈥檚 orbit at an angle of90掳relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it?

(c) Given the rivet is 3.00 mmin size, how long will its collision with the satellite last?

(d) If its mass is0.500 g, what is the average force it exerts on the satellite? (e) How much energy in joules is generated by the collision? (The satellite鈥檚 velocity does not change appreciably, because its mass is much greater than the rivet鈥檚.)

Calculate the mass of the Sun based on data for Earth鈥檚 orbit and compare the value obtained with the Sun鈥檚 actual mass.

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions鈥攊t then calculates the distance travelled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometres should the odometer read?

If centripetal force is directed toward the centre, why do you feel that you are 鈥榯hrown鈥� away from the centre as a car goes around a curve? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Particle Model of Matter

Read Explanation

Measurements

Read Explanation

Circular Motion and Gravitation

Read Explanation

Dynamics

Read Explanation

Wave Optics

Read Explanation
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.