Answer
$\beta = 1.0\times 10^{-4}$
Work Step by Step
We can find the speed of the Earth in its orbit around the sun:
$v = \frac{distance}{time}$
$v = \frac{2\pi~r}{1~yr}$
$v = \frac{(2\pi)~(1.50\times 10^{11}~m)}{(365)(24)(3600~s)}$
$v = 2.989\times 10^4~m/s$
We can find the speed parameter $\beta$:
$\beta = \frac{v}{c}$
$\beta = \frac{2.989\times 10^4~m/s}{3.0\times 10^8~m/s}$
$\beta = 1.0\times 10^{-4}$
