When a proton encounters a large atomic nucleus, to a good approximation, we can assume the large nucleus is fixed in space. The main force between the nucleus and proton is electrostatic. Assume a nucleus of charge Q is fixed at the origin, and a proton (mass m, charge q) approaches it moving along the x-axis. Far away from the nucleus the proton has a velocity v, but as the proton approaches the nucleus, it slows and comes to a stop at a so-called turning radius r. (See figure.) Energy is conserved during this process.
Note: The electric potential energy (or electrostatic energy) between two charged particles q1 and q2 a distance d apart is E = kq1q2/d.
Two protons are fired at the nucleus, the second with four times the velocity of the first. How would the electrostatic energy of the second proton at its turning radius compare with the electrostatic energy of the first proton at its turning radius?