Under what circumstances, if any, is the gravitational attraction between two protons equal to their electric repulsion?
Answers
Answered by
0
Fgrav = Gm(p)^2/r^2
Felec = kq(p)^2/r^2
For them to be equal you'd have to have Gm(p)^2 = kq(p)^2.
Since this isn't true, the answer is under no circumstances.
In fact Gm(p)^2 = 1.87E-64 N-m^2 while kq(p)^2 = 2.31e-28 N-m^2, so electric repulsion force is roughly 10^36 times stronger. That's why Fgrav is ignored in problems dealing with force and potential energy of two charges. If the charges are made up of electrons (same charge magnitude and only 1/1800 the mass of a proton) the ratio is even greater, roughly 10^42.
Felec = kq(p)^2/r^2
For them to be equal you'd have to have Gm(p)^2 = kq(p)^2.
Since this isn't true, the answer is under no circumstances.
In fact Gm(p)^2 = 1.87E-64 N-m^2 while kq(p)^2 = 2.31e-28 N-m^2, so electric repulsion force is roughly 10^36 times stronger. That's why Fgrav is ignored in problems dealing with force and potential energy of two charges. If the charges are made up of electrons (same charge magnitude and only 1/1800 the mass of a proton) the ratio is even greater, roughly 10^42.
Similar questions