3. how much equal charge should be placed on earth and the moon so that the electrical repulsion balances…

Explanation:

Step1: Write Coulomb's law and gravitational - force condition

Let the charge on Earth and the Moon be $q_1 = q_2=q$. Coulomb's law is $F_e=k\frac{q_1q_2}{r^2}$, and we want $F_e = F_g$. So $F_g=k\frac{q^2}{r^2}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $F_g=1.98\times 10^{20}\ N$, and $r = 3.84\times 10^{8}\ m$.

Step2: Solve for $q$

Rearrange the equation $F_g=k\frac{q^2}{r^2}$ to get $q^2=\frac{F_g r^2}{k}$. Then $q=\sqrt{\frac{F_g r^2}{k}}$. Substitute the given values: [ \begin{align*} q&=\sqrt{\frac{1.98\times 10^{20}\times(3.84\times 10^{8})^2}{9\times 10^{9}}}\ &=\sqrt{\frac{1.98\times 10^{20}\times14.7456\times 10^{16}}{9\times 10^{9}}}\ &=\sqrt{\frac{1.98\times14.7456\times 10^{20 + 16}}{9\times 10^{9}}}\ &=\sqrt{\frac{29.296288\times 10^{36}}{9\times 10^{9}}}\ &=\sqrt{3.255143111\times 10^{27}}\ &=\sqrt{32.55143111\times 10^{26}}\ &\approx5.7\times 10^{13}\ C \end{align*} ]

Answer:

$5.7\times 10^{13}\ C$