Question

An observer 'A' sees an asteroid with a radioactive element moving away at a speed \(0.3c\) and measures the radioactive 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

• 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

Hint:

Proper time is always the smallest and is measured in the rest frame of the object; moving observers measure a longer (dilated) time.

Answer 1

The Correct Option is C

Step 1: Identify proper time and dilated time.
Observer B is moving with the asteroid, so B measures the decay time in the asteroid’s rest frame.
This is called proper time \(T_B\).
Step 2: Time dilation concept.
Observer A sees the asteroid moving with speed \(v = 0.3c\).
According to special relativity:
\[ T_A = \gamma T_B \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \]
Step 3: Since \(\gamma>1\).
For any non-zero velocity, \(\gamma>1\).
So:
\[ T_A>T_B \]
Step 4: Choose correct relation.
\[ T_B<T_A \]
Final Answer:
\[ \boxed{\text{(C) } T_B<T_A} \]