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Q6.3-34 PE

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College Physics (Urone)
Found in: Page 223
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

(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’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of Gravity

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

Step 2: 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 \({{\rm{a}}_{\rm{s}}}{\rm{ = ?}}\)

Step 3: 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

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

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

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

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

Step 4: 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.

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

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

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

Step 5: Calculation of ratio of both acceleration of moon to sun

The ratio of the moon’s acceleration to the sun’s 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.

Most popular questions for Physics Textbooks

An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?

Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

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 orbit 900 km above Earth’s surface.

(b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90° 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 mm in size, how long will its collision with the satellite last?

(d) If its mass is 0.500 g, what is the average force it exerts on the satellite? (e) How much energy in joules is generated by the collision? (The satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)

A non-rotating frame of reference placed at the centre of the Sun is very nearly an inertial one. Why is it not exactly an inertial frame?

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—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s 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—that 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—the 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.
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