# An astronaut travels away from the Earth at a speed of 0.89c and sends a light signal back to the...

## Question:

An astronaut travels away from the Earth at a speed of 0.89c and sends a light signal back to the Earth every 9.0 s as measured by his clock. An observer on the Earth finds that the arrival time between consecutive signals is {eq}\delta {/eq}t. Find {eq}\delta {/eq}t.

## Relativistic Time:

The change in the time observed by a stationary observer for a moving object at the speed of the light is known as relativistic time. The elapsed in the time depends upon the velocity of the moving object.

## Answer and Explanation:

**Given data**

Speed of the astronaut {eq}(v) = 0.89c {/eq}

Time measured by astronaut {eq}(T) = 9 \ s {/eq}

Now, the time measured from the earth would be

{eq}\delta t = \dfrac{T}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}} \\ \delta t = \dfrac{9}{\sqrt{1- \dfrac{(0.89c)^{2}}{c^{2}}}} \\ \delta t = 19.74 \ s {/eq}

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