Question

Suppose the force between the Earth and Moon were electrical instead of gravitational, with the Earth having a positive charge and the Moon having a negative one. If the magnitude of each charge were proportional to the respective body’s mass, find the Q/m ratio required for the moon to follow its present orbit of 3.84 x 108 m radius with its period of 27.3 days. The Earth’s mass is 5.98 x 1024 kg, and the moon’s mass is 7.3 x 1022 kg.

Answer 1

hint:
we can write this equation:
\[\Large \begin{gathered}
{M_M}{\left( {\frac{{2\pi }}{T}} \right)^2}d = K\frac{{{Q_M}{Q_H}}}{{{d^2}}}, \hfill \\
{M_M} = {\text{moon}}\;{\text{mass}}, \hfill \\
{Q_M} = {\text{moon}}\;{\text{charge}}, \hfill \\
{Q_H} = {\text{earth}}\;{\text{charge}}, \hfill \\
T = {\text{period}} \hfill \\
d = {\text{mean}}\;{\text{distance}}\;{\text{moon - earth}} \hfill \\
K = {\text{Coulomb's}}\;{\text{constant}} = 9 \times {10^9}\frac{{N{m^2}}}{{{C^2}}} \hfill \\
\end{gathered} \]

Answer 2

if\[ q_{e} = \lambda m_e \ and \ q_m = \lambda m_m, \ where \ \lambda = q/m = constant\]

\[\frac {k q_e q_m}{r^2_{orbit}} = m_m \omega^2 r_{orbit}\]
\[\frac {k \lambda m_e \lambda m_m}{r^2_{orbit}} = m_m \omega^2 r_{orbit}\]
\[\lambda = \sqrt{\frac{\omega^2 r^3_{orbit}}{k \ m_e}} = \frac{2 \pi }{T}\sqrt{\frac{ r^3_{orbit}}{k \ m_e}}\]
where T is the period, 27.3 days