91Ó°ÊÓ

Positive charges are needed to be placed ontheEarth and on theMoonto neutralize their gravitational attraction.

Using the concept of both gravitational and electrostatic force from Coulomb's law, we can get the value of the positive charge of the particle. Again, we can get the number of hydrogen ions using the concept of the net charge.

Formulae:

The magnitude ofelectrostatic force between any two particles,$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$. (i)

The magnitude of gravitational force$F=G\frac{Mm}{r^2}$,. (ii)

The number of electrons present,$n=q/e$. (ii)

The magnitudes of the gravitational and electrical forces must be the same. Thus, using equations (i) and (ii), we can get the value of the net positive charge as follows:$$\begin{gathered} k\frac{q^2}{r^2}=G\frac{Mn}{r^2}(\sin etheelectroststicforceisbeteentwosimilarcharges) \\ q=\sqrt{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}GMn} \end{gathered}$$
where q is the charge on either body, r is the Centre-to-Centre separation of Earth and Moon, G is the universal gravitational constant, M is the mass of Earth, and m is the mass of the Moon.

Now,onsubstituting the given values in the above equation,we get

$$\begin{gathered} q=\sqrt{\frac{\left(6.67×{10}^{-11}N{m}^2/k{g}^2\right)\left(7.36×{10}^{22}kg\right)\left(5.98×{10}^{24}kg\right)}{9.00×{10}^{9}N{m}^2/C^2}} \\ =5.7×{10}^{13}C \end{gathered}$$

Hence, the value of the positive charges is $5.7×{10}^{13}C$.

From the calculations of part (a), we can see that the distance r cancels because both the electric and gravitational forces are proportional to $1/r^2$. Thus, the value of lunar distance is not considered.

The charge on a hydrogen ion

$e=1.60×{10}^{-19}C$

Now, using the charge value from part (a) in equation (iii),the total number of hydrogen ionsis found to be

$$\begin{gathered} n=\frac{5.7×{10}^{13}C}{1.60×{10}^{-19}C} \\ =3.6×{10}^{32}ions \end{gathered}$$

Each positive ion has a mass of$m_1=1.67×{10}^{-27}kg$, so that the total mass needed

$$\begin{gathered} m=n{m}_l \\ =\left(3.6×{10}^{32}\right)\left(1.67×{10}^{-27}kg\right) \\ =6.0×{10}^{5}kg \end{gathered}$$

Hence, the total mass of the hydrogen ions needed is $6.0×{10}^{5}kg$.