Answer
The correct answer is:
(a) less than T
Work Step by Step
Let $\Delta t_0$ be the half-life measured by an observer moving with the asteroid.
$T = \frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$
$\Delta t_0 = T~\sqrt{1-\frac{v^2}{c^2}} \lt T$
The correct answer is:
(a) less than T
