Question

"A spaceship carrying a light clock moves at a speed of 0.960 c relative to an observer on Earth. If the clock on the ship advances by 1.00 s as measured by the space travelers aboard the ship, how long did that advance take as measured by the observer on Earth?"

Answer 1

Answer:

The ship advance take 3.6 sec wrt. earth observer.

Explanation:

Given :

Speed of ship $v = 0.960c$

Time wrt. ship aboard $\Delta t_{o} = 1$ sec

From the lorentz transformation,

    $\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^{2} }{c^{2} } } }$

Where $\Delta t =$ improper time means time wrt. earth observer, $\Delta t_{o} =$ proper time interval means time wrt. rest frame of ship observer.

And always improper time is grater than proper time which we call time dilation.

Put the values,

   $\Delta t = \frac{1}{\sqrt{0.0784} }$

   $\Delta t = 3.57$ ≅ 3.6 sec

Thus, the ship advance take 3.6 sec wrt. earth observer.