Answer to Question #82536 in Electricity and Magnetism for lee

Question

If the pull on the Earth due to the sun was caused by electrostatic forces instead of gravity, how much charge would need to be on the Earth and how much on the sun so that the Earth’s orbit remained the same?

Answer 1

Newton's law of universal gravitation - F_g=G (M_E M_s)/R^2 , where G – gravitational constant, M_E and M_S – mass Earth and mass Sun, respectively. R – distance from the Sun to Earth.

M_E=5,97∙〖10〗^24 kg

M_S=1,98∙〖10〗^30 kg

Coulomb’s law - F_c=1/(4πε_0 ) |q||q|/R^2 , where ε_0=8,85∙〖10〗^(-12) (f⁄m)

Suppose that the charge of the Earth and the Sun is the same, then

|q||q|=Q^2, F_g=F_c →G (M_E M_S)/R^2 = 1/(4πε_0 ) Q^2/R^2 , then the charge of the celestial bodies is equal - Q= √(4πε_0 M_E M_S )=2,97∙〖10〗^17 (C)