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

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

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

The number of attacking neutrons before fission is ${\mathrm{n}}_2=3$

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

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

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,

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

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,

$$\begin{gathered} \mathrm{Change}\mathrm{in}\mathrm{Earth}\text{'}\mathrm{s}\mathrm{momentum}=\mathrm{Impulse} \\ {\mathrm{m}}_{\mathrm{earth}}∆\mathrm{v}={\mathrm{F}}_{\mathrm{total}}\times ∆\mathrm{t}.........\left(1\right) \end{gathered}$$

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

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

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

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

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

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