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(A) The ratio of electrostatic force $F_e$ to gravitational force $F_G$ is given by:$\frac{F_e}{F_G} = \frac{\frac{K q_1 q_2}{r^2}}{\frac{G m_1 m_2}{r

Vedclass · Accepted Answer

$10^{20}$

Vedclass · Answer

(A) The ratio of electrostatic force $F_e$ to gravitational force $F_G$ is given by:$\frac{F_e}{F_G} = \frac{\frac{K q_1 q_2}{r^2}}{\frac{G m_1 m_2}{r^2}} = \frac{K}{G} \cdot \frac{q_1 q_2}{m_1 m_2}$Given $\frac{F_e}{F_G} = 2.4 	imes 10^{39}$,$q_1 = q_2 = 1.6 	imes 10^{-19} \; C$,$m_1 = 9.11 	imes 10^{-31} \; kg$,and $m_2 = 1.67 	imes 10^{-27} \; kg$.Substituting the values:$2.4 	imes 10^{39} = \frac{K}{G} \cdot \frac{(1.6 	imes 10^{-19})^2}{(9.11 	imes 10^{-31}) 	imes (1.67 	imes 10^{-27})}$$\frac{K}{G} = \frac{2.4 	imes 10^{39} 	imes (9.11 	imes 1.67) 	imes 10^{-58}}{(1.6)^2 	imes 10^{-38}}$$\frac{K}{G} = \frac{2.4 	imes 15.2137 	imes 10^{-19}}{2.56 	imes 10^{-38}} = \frac{36.51288}{2.56} 	imes 10^{19} \approx 14.26 	imes 10^{19} = 1.426 	imes 10^{20}$Thus,the ratio is approximately $10^{20}$.

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