# Would the gravitational attractive force between two protons counterbalance their electrical repulsion?

Would you expect the gravitational force to counterbalance the electrical repulsion?

Calculate the ratio between the electric and gravitational forces acting between two protons. Would this ratio change according to the distance between the protons?

Thanks so much, this material has left very confused and would VERY MUCH appreciate some insight into this!

Relevance

The electrostatic repulsion between two protons at separation 'r' is given by:

Fe = ke^2/r^2

k = 8.99 x 10^9 Nm^2/C^2

e = 1.6 x 10^-19 C (proton charge)

The gravitational attraction between two protons at separation 'r' is:

Fg = G(mp)^2/r^2

G = 6.67 x 10^-11 m^3/kg/s^2

mp = 1.67 x 10^-27 kg (proton mass)

Take the ratio of (1) and (2) making the r's cancel:

Fg/Fe = (1.86 x 10^-64)/(2.5 x 10^-44) = 7.42 x 10^-21

For protons the gravitational attraction at **any** radius is about 22 orders of magnitude smaller than the electrostatic repulsion. So, no, the gravitational attraction is definitely not enough to counterbalance the electrostatic repulsion.

An interesting question is, then: How do protons stay clustered together inside the nucleus of an atom when they are all repelling one another?

The answer to this question lies in the nuclear 'strong' force.

I'll leave it at that.