Special relativity: How close to the speed of light is a proton with kinetic energy of 1TeV?

The final answer will be units independent as the speed will be in speed of light units, i.e., $1\ c$ .

The problem will be using Einstein's special relativity only. His general relativity will not be used.

The mass of a proton used in this problem is $1.672\,621\,898\times 10^{-27}\ \text{kg}$ . The experimental uncertainty is being ignored in this problem.

The mass equivalent of 1 electron-Volt (eV) used in this problem is $1.782\,661\,907\times 10^{-36}\ \text{kg}$ . Again, the experimental uncertainty is being ignored in this problem.

One TeV is $1\,000\,000\,000\,000$ eV. The Tevatron had a designed operating energy per proton or anti-proton of 1 TeV. For stability of the accelerator reasons, it usually operated near 936 Mev. This is from personal experience. In this problem, the designed per-proton energy is used.

One of the slightly less well known of Einstein's special relativity formulas is the one for mass increase for velocity:

$\text{mass}=\frac{\text{mass}_\text{rest}}{\sqrt{1-(\frac{\text{velocity} }{\text{velocityOfLight}})^2}}$

Rewritten with shorter variable names:

$\text{M}=\frac{m}{\sqrt{1-(\frac{v}{c})^2}}$

$v$ and $c$ have to be expressed in the same units, e.g., $c\ \text{is}\ 299792458 \text{ m/s}, \sim 186282.397 \text{ miles/s}, \sim 67061662.9 \text{ mph}$ or $\sim 1802617499785.25 \text{ furlongs/fortnight}$ .

We will also be using another of Einstein's special relativity formulas: $e=m c^2$ . by using them together.

$e=\frac{m c^2}{\sqrt{1-(\frac{v}{c})^2}}$

Because of the fact that the speed is very close to the speed of light, subtract the answer from the speed of light and give your answer in units of $\mu c$ (one millionth of $c$ ) with a precision of 3 significant digits. In the solution, I will provide more digits.