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Found in: Page 128

### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

# 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 ${\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{24}}}{\mathbf{}}{\mathbf{kg}}$, and there are about 6 billion people (${\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{9}}}$).

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

See the step by step solution

## Step 1: Identification of given data

The mass of the earth is ${\mathrm{m}}_{\mathrm{earth}}=6×{10}^{24}\mathrm{kg}$

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

The number of attacking neutrons before fission is ${\mathrm{n}}_{2}=3$

## 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 ${\mathbf{m}}{\mathbf{∆}}{\mathbf{v}}$ in time ${\mathbf{∆}}{\mathbf{t}}$, will be expressed as,

${\mathbf{F}}{\mathbf{=}}{\mathbf{m}}\frac{\mathbf{∆}\mathbf{v}}{\mathbf{∆}\mathbf{t}}$

Rearranging the above equation, we get

${\mathbf{F}}{\mathbf{∆}}{\mathbf{t}}{\mathbf{=}}{\mathbf{m}}{\mathbf{∆}}{\mathbf{v}}$

## 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,

$\begin{array}{rcl}{\mathrm{m}}_{\mathrm{total}}& =& \mathrm{n}×{\mathrm{m}}_{\mathrm{average}}\\ & =& \left(6×{10}^{9}\right)×60\mathrm{kg}\\ & =& 3.6×{10}^{11}\mathrm{kg}\end{array}$

## 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,

$\mathrm{Change}\mathrm{in}\mathrm{Earth}\text{'}\mathrm{s}\mathrm{momentum}=\mathrm{Impulse}\phantom{\rule{0ex}{0ex}}{\mathrm{m}}_{\mathrm{earth}}∆\mathrm{v}={\mathrm{F}}_{\mathrm{total}}×∆\mathrm{t}.........\left(1\right)$

Where $∆\mathrm{v}$ is the change in Earth’s speed, ${\mathrm{F}}_{\mathrm{total}}×∆\mathrm{t}$ is the total impulse in time $∆\mathrm{t}$

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

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{total}}& =& {\mathrm{m}}_{\mathrm{total}}×\mathrm{g}\\ & =& \left(3.6×{10}^{11}\mathrm{kg}\right)×9.8\mathrm{m}.{\mathrm{s}}^{-2}\\ & =& 3.53×{10}^{12}\mathrm{kg}.\mathrm{m}.{\mathrm{s}}^{-2}\end{array}$

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

$\left(6×{10}^{24}\mathrm{kg}\right)×\mathrm{v}=\left(3.53×{10}^{12}\mathrm{kg}.\mathrm{m}.{\mathrm{s}}^{-2}\right)×1\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{v}=\frac{\left(3.53×{10}^{12}\mathrm{kg}.\mathrm{m}.{\mathrm{s}}^{-2}\right)×1\mathrm{s}}{\left(6×{10}^{24}\mathrm{kg}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{v}=5.88×{10}^{-12}\mathrm{m}.{\mathrm{s}}^{-1}$

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