Unit 6 - Problem 6 ==> Battle of the Forces

Given:

• $$ K_{electrostatic-constant} = 8.99 * 10^9 $$
• $$ G_{gravity-constant} = 6.67 * 10^-11 $$
• $$\mbox{Two bodies with same mass and charge of 1 coulomb each.}$$

Question:

• $$\mbox{At what mass are the force of gravity and the electrostatic force equal?}$$

Rationale:

• $$F_{gravity} = F{electrostatic-charge} $$
• $$ \frac{G * m^2}{r^2} = \frac{K * Q^2}{r^2} $$
• $$ m^2 = \frac{K * Q^2}{G} $$
• $$ m = \sqrt\frac{K * Q^2}{G} $$

Calculate

• $$ m = \sqrt\frac{(8.99 * 10^9)}{(6.67 * 10^-11)} $$
• $$ m = \sqrt{(1.347826 * 10^{20})} $$
• $$ m = 1.16 * 10^{10} $$

Comparison:

• $$ \mbox{mass of the earth} = 5.972 * 10^{24}\;\mbox{kg} $$
• $$ \frac{ 1.16 * 10^{10}}{5.972 * 10^{24}} = 0.19*10^{-14} $$
• $$ \mbox{Much smaller than the earth} $$
• $$ \mbox{mass of the earth's moon} = 0.07349 * 10^{24} $$
• $$ \frac{ 1.16 * 10^{10}}{0.7349 * 10^{24}} =15.7844604 * 10{-14} $$
• $$ \mbox{Much smaller than earth's moon} $$
• $$ \mbox{mass of a US Navy nuclear-powered aircraft carrier} = 196 * 10^6 \;\mbox{lbs} $$
• $$ \mbox{number of carriers} = \frac{1.16 * 10^{10}}{196 * 10^6} = 59.18367 $$
• $$ \mbox{That's more carrier's than the US Navy has ever had.} $$