Chapter 16, Problem 60GP

To determine

To find: The value of charge required to maintain the moon orbit around the Earth

Answer to Problem 60GP

  $5.7\times {10}^{13}\text{C}$

Explanation of Solution

Given:

Mass of earth $=5.98\times {10}^{24}\mathrm{k}\mathrm{g}({\mathrm{M}}_{\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}})$

Mass of moon $=7.35\times {10}^{22}\mathrm{k}\mathrm{g}({\mathrm{M}}_{\mathrm{m}\mathrm{o}\mathrm{o}\mathrm{n}})$

Mass of orbit $=3.84\times {10}^8\mathrm{k}\mathrm{m}({\mathrm{r}}_{\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}})$

Formula used:

To maintain moon orbit around the earth Coulomb electric force must be equal to the Newtonian gravitational force, so charge is given by: ${\mathrm{F}}_{\mathrm{E}}={\mathrm{F}}_{\mathrm{G}}\Rightarrow \frac{\mathrm{K}{\mathrm{Q}}^2}{\cancel{{\mathrm{r}}^2{}_{\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}}}}=\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{m}\mathrm{o}\mathrm{o}\mathrm{n}}{\mathrm{M}}_{\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}{\cancel{{\mathrm{r}}^2{}_{\mathrm{o}\mathrm{r}\mathrm{b}\mathrm{i}\mathrm{t}}}}$

  $\Rightarrow \mathrm{Q}=\sqrt{\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{m}\mathrm{o}\mathrm{o}\mathrm{n}}{\mathrm{M}}_{\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}{\mathrm{K}}}$

Calculation:

  $\Rightarrow \mathrm{Q}=\sqrt{\frac{\mathrm{G}{\mathrm{M}}_{\mathrm{m}\mathrm{o}\mathrm{o}\mathrm{n}}{\mathrm{M}}_{\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}}}{\mathrm{K}}}$

  $\begin{array}{l}\mathrm{Q}=\sqrt{\frac{\left(6.67\times {10}^{-11}\mathrm{N}{\mathrm{m}}^2/\mathrm{k}{\mathrm{g}}^2\right)\left(7.35\times {10}^{22}\mathrm{k}\mathrm{g}\right)\left(5.98\times {10}^{24}\mathrm{k}\mathrm{g}\right)}{\left(9\times {10}^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2\right)}}\\ \mathrm{Q}=5.71\times {10}^{13}\mathrm{C}\end{array}$

Conclusion:

Charge on Earth and moon must be $\text{5}{\text{.7×10}}^{\text{13}}\text{C}$ to maintain orbit of moon around the Earth.