NEET-XII-Physics
47: The Special Theory of Relativity
- #11Suppose 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?Ans : 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\text{'}=\frac{∆t}{\sqrt{1-{\displaystyle \frac{{v}^{2}}{{c}^{2}}}}}``
`` =\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.9999\right)}^{2}{c}^{2}}{{c}^{2}}}}}``
`` =\frac{1}{0.014141782}=70.712\,\mathrm{\,days\,}``
`` ``
Thus,
∆t' = 70.7 days
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