Question

A shower of protons from outer space deposits equal charges q on the earth and the moon and the electrostatic repulsion then exactly counterbalances the gravitational attraction. How large is $q$?

Answer 1

Step 1: Given data

Charge: $+q$

Mass of earth: $Mkg$

Mass of moon: $mkg$

Distance between earth and moon: $rkm$

Step 2: Formula Used

1. The gravitational force between 2 bodies is given by $F_G=G\frac{Mm}{r^2}$, where $G$ is gravitational constant , $M$ and $m$ are masses of the 2 bodies, and $r$ is the distance between the centers of the bodies.
2. The electrostatic force between 2 charges is given by $F_E=k\frac{q_1\times q_2}{r^2}$, where $k$ is gravitational constant , $q_1$ and $q_2$ are charges of the 2 bodies, and $r$ is the distance between the charges.

Step 3: Find the charge

Equate the gravitational force and electrostatic repulsion force

$$\begin{gathered} F_E=F_G \\ \Rightarrow k\frac{q\times q}{r^2}=G\frac{Mm}{r^2} \\ \Rightarrow q^2=G\frac{Mm}{r^2}\times \frac{r^2}{k} \\ \Rightarrow q=\sqrt{G\frac{Mm}{k}} \end{gathered}$$

Thus, the value of $q$ is $\sqrt{G\frac{Mm}{k}}$.