Question

Suppose the Earth (mass 6.0 x 10 kg) has an excess of

positive charge and the Moon (mass 7.3x 1022 kg) has an equal

excess of positive charge. Calculate the size of the charge

required so that the electrostatic force between them

balances the gravitational force between them.

Answer 1

Mass of earth $m_1$ = $6.0\times 10 kg$

Mass of moon $m_2$ = $7.3\times 1022kg$

Let the charges on both earth and moon be equal to $q_1$ and $q_2$ .

as said in the question charge are equal $q_1=q_2=q$

now applying the condition electrostatic force balance gravitational force .

$=\dfrac{Gm_1m_2}{r^{2}}$ =$\dfrac{q_1q_2}{4{\pi}{\epsilon_o}r^{2}}$

$=G=6.674\times 10^{-11}m^{3}/kg/s^{2}$

$=\dfrac{1}{4{\pi}{\epsilon_o}}$ =$9\times10^{9}Nm^{2}/c^{2}$

$=q_1=q_2=q$

$=\dfrac{6.674\times10^{-11}\times 6.0\times 10\times 7.2\times1022}{9\times10^{9}}=q^{2}$

=$q=1.8\times 10^{-7}C$