Question

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

Answer 1

Mass of Earth = M kg
Mass of Moon = m kg
distance between them = r km
Gravitational attraction,F = (GMm)/r²

Charge is +q
Electrostatic repulsion force,F' = (K q×q) /r²
As, both the forces counterbalances each other,
F = F'
∴(GMm)/r² = (K q×q) /r²
⇒GMm = K q²
⇒q² = (GMm)/K
⇒q = √[(GMm)/K]  units

Answer 2

Given:

A shower of protons from outer space deposits $+q$ charges on earth and the moon such that the electrostatic force of repulsion balances the gravitational force of attraction.

To Find:

Value of $q$.

Solution:

From the universal law of gravitation, we know that the gravitational force between earth and the moon will be

   $F_G=G\frac{m_em_m}{r^2}$

where $G$ is the universal gravitational constant, $m_e$ is the mass of earth, $m_m$ is the mass of the moon and $r$ is the distance between their centers.

Similarly, from Coulomb's law, electrostatic force of repulsion,

   $F_E=k\frac{q^2}{r^2}$

where $k$ is the Coulomb's law constant.

Since the electrostatic and gravitational forces balance each other, they must have an equal magnitude in opposite directions.

   $F_E=F_G$

⇒ $k\frac{q^2}{r^2}=G\frac{m_em_m}{r^2}$

⇒ $q=\sqrt{\frac{Gm_em_m}{k} }$

Hence, the value of $q$ is equal to $\sqrt{\frac{Gm_em_m}{k} }$.