Answer
$v=c( 1-9\times10^{-9})$.
Work Step by Step
First, from the total energy, calculate the speed of the proton. Use equation 26–6b and solve for v/c.
$$\frac{v}{c}=\sqrt{1-{\left( \frac{mc^2}{E}\right)^2}}$$
The second term under the square root is so small for this ultrarelativistic particle that we need to use the binomial expansion, or else our calculator will just say that v/c = 1.
$$\frac{v}{c}\approx 1-\frac{1}{2}\left( \frac{mc^2}{E}\right)^2$$
$$\frac{v}{c}\approx 1-\frac{1}{2}\left( \frac{9.38\times10^8 eV}{7.0\times10^{12}eV}\right)^2$$
$$\frac{v}{c}\approx 1-9\times10^{-9}$$
$$v=c( 1-9\times10^{-9})$$
This proton is about 2.7 m/s slower than the speed of light.
