Book edition 4th edition

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

Pages 1135 pages

ISBN 9781118875865

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Short Answer

Suppose that all the people of the Earth go to the North Pole and, on a signal, all jump straight up. Estimate the recoil speed of the Earth. The mass of the Earth is $6 \times 10^{24} kg$ , and there are about 6 billion people ( $6 \times 10^{9}$ ).

The recoil speed of Earth is $5.88 \times 10^{- 12} m . s^{- 1}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of given data

The mass of the earth is $m_{earth} = 6 \times 10^{24} kg$

The number of people on earth is $n = 6 \times 10^{9}$

The number of attacking neutrons before fission is $n_{2} = 3$

Step 2: Concept Introduction

According to Newton’s Second law, we know that the rate of change of linear momentum can be expressed as the total applied force, such that, a change in the momentum of $m ∆ v$ in time $∆ t$ , will be expressed as,

$F = m \frac{∆ v}{∆ t}$

Rearranging the above equation, we get

$F ∆ t = m ∆ v$

Step 3: Determination of the total mass of all the people on the earth

If we assume that the average mass of each person on earth is 60 kg then the total mass of all the people will be,

$\begin{array}{rcl} m_{total} & = & n \times m_{average} \\ = & (6 \times 10^{9}) \times 60 kg \\ = & 3.6 \times 10^{11} kg \end{array}$

Step 4: Calculation of the recoil force

Initially, the Earth is at rest, so when all people jump simultaneously, the impulse will be equal to the total change in the Earth’s momentum, which can be expressed as,

$Change in Earth' s momentum = Impulse m_{earth} ∆ v = F_{total} \times ∆ t . . . . . . . . . (1)$

Where $∆ v$ is the change in Earth’s speed, $F_{total} \times ∆ t$ is the total impulse in time $∆ t$

Now, the force due to all people can be estimated as,

$\begin{array}{rcl} F_{total} & = & m_{total} \times g \\ = & (3.6 \times 10^{11} kg) \times 9.8 m . s^{- 2} \\ = & 3.53 \times 10^{12} kg . m . s^{- 2} \end{array}$

Now, substitute the given values and $∆ t = 1 s$ in the equation (1), and we get,

$(6 \times 10^{24} kg) \times v = (3.53 \times 10^{12} kg . m . s^{- 2}) \times 1 s v = \frac{(3.53 \times 10^{12} kg . m . s^{- 2}) \times 1 s}{(6 \times 10^{24} kg)} v = 5.88 \times 10^{- 12} m . s^{- 1}$

Thus, the recoil speed of Earth is $5.88 \times 10^{- 12} m . s^{- 1}$

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Matter & Interactions

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Short Answer

Suppose that all the people of the Earth go to the North Pole and, on a signal, all jump straight up. Estimate the recoil speed of the Earth. The mass of the Earth is $6 \times 10^{24} kg$ , and there are about 6 billion people ( $6 \times 10^{9}$ ).

Step by Step Solution

Step 1: Identification of given data

Step 2: Concept Introduction

Step 3: Determination of the total mass of all the people on the earth

Step 4: Calculation of the recoil force

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Matter & Interactions

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Short Answer

Suppose that all the people of the Earth go to the North Pole and, on a signal, all jump straight up. Estimate the recoil speed of the Earth. The mass of the Earth is 6×1024 kg, and there are about 6 billion people (6×109).

Step by Step Solution

Step 1: Identification of given data

Step 2: Concept Introduction

Step 3: Determination of the total mass of all the people on the earth

Step 4: Calculation of the recoil force

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Suppose that all the people of the Earth go to the North Pole and, on a signal, all jump straight up. Estimate the recoil speed of the Earth. The mass of the Earth is $6 \times 10^{24} kg$ , and there are about 6 billion people ( $6 \times 10^{9}$ ).