/*! 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

A mother pushes her child on a swing so that his speed is9.00m/s at the lowest point of this path. The swing is suspended 2.00 mabove the child鈥檚 center of mass.

(a) What is the magnitude of the centripetal acceleration of the child at the low point?

(b) What is the magnitude of the force the child exerts on the seat if his mass is 18.0 kg?

(c) What is unreasonable about these results?

(d) Which premises are unreasonable or inconsistent?

Find the mass of Jupiter based on data for the orbit of one of its moons, and compare your result with its actual mass.

When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were forced up the drain?

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components鈥攆riction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system鈥檚 weight).

(a) Show that\(\theta \)(as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway鈥攖hat is,\(\theta = {\tan ^{ - 1}}\,{v^2}/rg\)

(b) Calculate \(\theta \) for a \(12.0{\rm{ m}}/{\rm{s}}\) turn of radius \(30.0{\rm{ m}}\) (as in a race).

Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle鈥攖he ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle \(\theta \), the speed \(v\), and the radius of curvature \(r\) of the turn similar to that for the ideal banking of roadways.

Microwave ovens rotate at a rate of about 6rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Fluids

Read Explanation

Radiation

Read Explanation

Electrostatics

Read Explanation

Particle Model of Matter

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.