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

As a nucleus 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 Nov 25 '17 at 17:17
• @annav that isn't an answer, I mean the minimum size 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 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. 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 $$\frac{n_p}{n_p + n_n} \approx 10^{-40}$$