Answer to Question #131143 in Astronomy | Astrophysics for Dheeraj

Question

How much imbalance of protons to electrons numbers have to be done to the bulk Earth and Moon in order to make the electrostatic interaction dominate the gravity by a bare minimum magnitude?

Answer 1

There should be an imbalance of $3.55×10^{32}$ number of charges on moon as well as on Earth

. Explanation

$G=6.67×10^{-11}Nm^{2}Kg^{24}$

$Mass$ $Of$ $Earth(M)=5.97×10^{24}Kg$

$Mass$ $Of$ $Moon(m)=7.3×10^{22}Kg$

mean distance between earth and moon $d=3.84×10^8m$

Gravitational force between earth and moon is given as.

$\implies F_G=\frac{GMm}{d^{2}}$

$\implies=\frac{(6.67×10^{-11})×(5.97×10^{24})×(7.3×10^{22})}{(3.84×10^{8})^2}$

$F_G=1.97×10^{20}N$

Let there be n+be charges on Earth and n+be charges on moon

so charge of earth $Q_e=n×1.6×10^{-19}$

charge on moon $Q_m=n×1.6×10^{-19}C$

$K=9×10^{9}Nm^{2}C^{-2}$

Then the electrostatic force between the earth and moon

. $F_E=\frac{KQ_mQ_e}{d^{2}}$

$F_E=\frac{(9.9×10^{9})×(n×1.6×10^{-19})×(n×1.6×10^{-19}}{(3.84×10^{8})^2}.$

For electrostatic interaction to be just greater than the gravitational interaction,the magnitude of the gravitational force and electrostatic force should be equal

$\implies \mid F_E\mid =\mid F_G\mid$

$\implies n^{2}×1.56×10^{-45}=1.97×10^{20}N$

$\implies n^{2}=1.26×10^{65}$

$n=3.55×10^{32}$

So there should be an imbalance of $3.55×10^{32}$ Number of charges on moon as well as on Earth