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.
How would the turning radius be affected if the initial velocity v were increased by a factor of 4?