Book edition 4th edition

Author(s) Ruth W. Chabay, Bruce A. Sherwood

Pages 1135 pages

ISBN 9781118875865

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Short Answer

The angle between the gravitational force on a planet by a star and the momentum of the planet is $61 °$ at a particular instant. At this instant the magnitude of the planet’s momentum is role="math" localid="1654013162020" $3.1 \times 10^{29} kgm / s$ , and the magnitude of the gravitational force on the planet is role="math" localid="1654013174728" $1.8 \times 10^{23} N$ . (a) What is the parallel component of the force on the planet by the star? (b) What will be the magnitude of the planet’s momentum after 8h?

$(a) 8.7 \times 10^{22} Np (b) 3.15184 \times 10^{29} kg . m / s$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of the given data

The given data is listed below as,

$\circ$ The angle between the planet’s gravitational force and momentum is, $θ = 60 °$

$\circ$ The magnitude of the planet’s initial momentum is, $p_{i} = 3.1 \times 10^{29} kg . m / s$

$\circ$ The magnitude of the planet’s gravitational force is $\vec{F} = 1.8 \times 10^{23} N$

Step 2: Significance of the parallel force

The parallel force acts in the opposite or same direction at the different points of a particular object.

The equation of the parallel component of the force is expressed as-

$\vec{F_{∥}} = |\vec{F}| \cos θ p$ .... $(1)$

Here, $F_{∥}$ is the parallel force, $\vec{|F|}$ is the absolute value of the gravitational force, $p$ is the unit vector and $θ$ is the angle between the momentum and gravitational force

Step 3: Determination of the parallel component of the force on the planet

(a) For, $\vec{|F|} = 1.8 \times 10^{23} N$ and $θ = 0$ in equation (1).

$F_{∥} = (1.8 \times 10^{23} N \times \cos 0 °) p = (1.8 \times 10^{23} N) p$

Thus, the parallel component of the force on the planet by the star is role="math" localid="1654016428235" $(1.8 \times 10^{23} N) p$

Step 4: Determination of the magnitude of the planet’s final momentum

$P_{f} = P_{i} + F_{net} ∆ t Here, P_{f} is the final momentum, P_{i} is the initial momentum, F_{net} is the net force exerted by the planet and ∆ t is the difference in the time period For p_{i} = 3.1 \times 10^{29} kg . m / s, p_{i} = 1.8 \times 10^{23} N and ∆ t = 8 h - 0 = 8 h p_{f} = (3.1 \times 10^{29} kg . m / s) + (1.8 \times 10^{23} N) \times (8 h \frac{3600 s}{1 h}) = (3.1 \times 10^{29} kg . m / s) + 5.184 \times 10^{27} N . s \times \frac{1 kg . m / s^{2}}{1 N} = 3.15184 \times 10^{29} kg . m / s Thus, the magnitude of the planet ’ s momentum after 8 h is 3.15184 \times 10^{29} kg . m / s$ The equation of the magnitude of the planet’s momentum after 8h is expressed as,

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Matter & Interactions

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Short Answer

Step by Step Solution

Step 1: Identification of the given data

Step 2: Significance of the parallel force

Step 3: Determination of the parallel component of the force on the planet

Step 4: Determination of the magnitude of the planet’s final momentum

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Matter & Interactions

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Short Answer

Step by Step Solution

Step 1: Identification of the given data

Step 2: Significance of the parallel force

Step 3: Determination of the parallel component of the force on the planet

Step 4: Determination of the magnitude of the planet’s final momentum

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