Sum

Suppose Swarglok (heaven) is in constant motion at a speed of 0.9999c with respect to the earth. According to the earth's frame, how much time passes on the earth before one day passes on Swarglok?

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#### Solution

Given:-

Speed of Swarglok, v = 0.9999c

Proper time interval, ∆t = One day on Swarglok

Suppose ∆t' days pass on Earth before one day passes on Swarglok.

Now,

\[∆ t' = \frac{∆ t}{\sqrt{1 - \frac{v^2}{c^2}}}\]

\[ = \frac{1}{\sqrt{1 - \frac{\left( 0 . 9999 \right)^2 c^2}{c^2}}}\]

\[ = \frac{1}{0 . 014141782} = 70 . 712\text{ days}\]

Thus,

∆t' = 70.7 days

Concept: Twin Paradox

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