Question

A spaceship is moving at 0.74 c (0.74 times the speed of light) with respect to the earth. An observer on the spaceship measures the time interval between two events as 30 hrs. If the spaceship had been moving with a speed of 0.86 c with respect to the earth, what would the time interval between the events have been? Dt0.86 c = _____ hrs

Answer 1

Answer:

The time interval is 22.76 hr.

Explanation:

Given that,

Time = 30 hrs

Speed = 0.74c

We need to calculate the time interval measured in earth frame

Using formula of time dilation

$\Delta t=\dfrac{\Delta t_{0}}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Where, $\Delta t$ = time interval measured in earth frame

$\Delta t_{0}$=proper time interval

Put the value into the formula

$\Delta t=\dfrac{30}{\sqrt{1-\dfrac{0.74c^2}{c^2}}}$

$\Delta t=\dfrac{30}{\sqrt{1-(0.74)^2}}$

$\Delta t=\dfrac{30}{0.6726}$

$\Delta t=44.60\ hr$

We need to calculate the proper time interval when the velocity is 0.86c

Using formula of time dilation

$44.60=\dfrac{\Delta t_{0}}{\sqrt{1-\dfrac{0.86c^2}{c^2}}}$

$\Delta t_{0}=44.60\times\sqrt{1-(0.86)^2}$

$\Delta t_{0}=22.76\ hr$

Hence, The time interval is 22.76 hr.