Assume that Devlok (heaven) moves constantly at a speed of 0.9999c relative to the earth. In the earth's frame of reference, how much time elapses on the earth for one day to pass on Devlok?

Step-by-step solution

The time dilation formula is given by t = t₀ 1 - v^2 c^2 Here, the proper time t₀ = 1 day and the velocity v = 0.9999c . Calculating the term inside the square root: 1 - v^2 c^2 = (1 - 0.9999)(1 + 0.9999) = 0.0001 1.9999 2 10⁻⁴ Taking the square root: 1 - v^2 c^2 2 10⁻⁴ = 1.414 10⁻² Substituting this back into the time dilation formula: t = 1 1.414 10⁻² = 100 1.414 = 70.7 days