Answer
$\beta = 3.7\times 10^{-5}$
Work Step by Step
We can find the escape speed of a projectile from the Earth’s surface:
$v = \sqrt{\frac{2GM}{r}}$
$v = \sqrt{\frac{(2)(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)}{6.37\times 10^6~m}}$
$v = 1.119\times 10^4~m/s$
We can find the speed parameter $\beta$:
$\beta = \frac{v}{c}$
$\beta = \frac{1.119\times 10^4~m/s}{3.0\times 10^8~m/s}$
$\beta = 3.7\times 10^{-5}$
