# How large would a nucleus have to be for the gravitational force overcome the electrical repulsion?

As atomic nuclei get larger, the electrical repulsion between protons grows, eventually overcoming the weak nuclear force. My question is, how large would a nucleus have to be for the gravitational attraction to overcome the electrical repulsion?

• think neutron star – anna v Nov 25 '17 at 17:17
• @annav that isn't an answer, I mean the minimum size – mcchucklezz Nov 25 '17 at 17:18
• look at the coupling constants 1/137 electromagnetic and 6x10^-39 gravity.hyperphysics.phy-astr.gsu.edu/hbase/Forces/funfor.html Neutron stars happen when the electrostatic repulsion is overcome by the large gravitationally attracted mass.. order of magnitude should be ok – anna v Nov 25 '17 at 17:28
• @annav There are positive and negative charges collapsing in a neutron star. This is different from the OP's question about a nucleus with only positive charges. If we enlarge a nucleus, I don't think gravity would ever overcome the electrical repulsion. – safesphere Nov 25 '17 at 18:14

Since the EM force is much larger than the gravitational than we can safely assume that the size would be large (roughly of astronomical order). Suppose the giant nucleas has $n_p$ protons and $n_n$ neutrons.
The gravitational self energy of the spherical atom would be $$U_G = -\frac{3}{5}\frac{GM}{R}$$ where $R$ is the radius of the "Megatom". Its mass is $M = n_p m_p + n_n m_n \approx (n_p+n_n)m_p$.
The repulsive EM self energy of the Megatom is $$U_E = -\frac{3}{5}\frac{kQ}{R}$$ where $k = \frac{1}{4\pi \epsilon_0}$ and $Q = n_p e$.
For them to be of the same order we need to have $$G(n_p + n_n)m_p \approx k n_p e$$ We see that the $R$ cancels out. And the above equality gives a proton fraction of $$\frac{n_p}{n_p + n_n} \approx 10^{-40}$$