Q. An observer $A$ sees an asteroid with a radioactive element moving by at a speed $= 0.3\ c$ and measures the radioactivity decay time to be $T_{A}$. Another observer $B$ is moving with the asteroid and measures its decay time as $T_{B}$. Then $T_{A}$ and $T_{B}$ are related as below
A) $T_{B} < T_{A}$
B) $T_{B} = T_{A}$
C) $T_{B} > T_{A}$
D) Either (a) or (c) depending on whether the asteroid is approaching or moving away from $A$
Explanation
(a) $T = \frac{T_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$
$T$ is the time observed by the person on earth in relative motion $w.r.t.$ the asteroid. $T_{0}$ is measured by the person at rest
$\therefore T_{A} > T_{B}(T_{B} = T_{0})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.