Answer
$3\times10^7kg$
Work Step by Step
Use equation 26–5b to find the kinetic energy, and equation 26–7 to find the rest energy.
$$KE= (\gamma-1)m_{Enterprise}c^2=m_{convert}c^2$$
$$m_{convert}= (\gamma-1)m_{Enterprise}$$
$$ = (\frac{1}{\sqrt{1-0.10^2}}-1)(6\times10^9kg)$$
$$ = 3\times10^7kg$$
