$e$ in physics

According to $\displaystyle m = \frac{m_0}{ \sqrt{1 - \frac{v^2}{c^2}}} = \frac{m_0}{ \sqrt{1 - \beta^2}}$ the desired velocity $v$ satisfies the condition:

$\displaystyle m = \frac{m_0}{ \sqrt{1 - \frac{v^2}{c^2}}} = 2m_0$

$\implies v = \displaystyle \frac{\sqrt{3}}{2} c = \boxed{0.866\ c}$ .