One day on a spacecraft corresponds to 2 days | vedclass

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(C) According to the theory of special relativity, the time dilation formula is given by $T = \frac{T_0}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $T$ is th

Vedclass · Accepted Answer

$2.6 	imes 10^8 \ m/s$

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(C) According to the theory of special relativity, the time dilation formula is given by $T = \frac{T_0}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $T$ is the time observed on Earth, $T_0$ is the proper time on the spacecraft, $v$ is the velocity of the spacecraft, and $c$ is the speed of light $(3 	imes 10^8 \ m/s)$.Given $T_0 = 1 \ day$ and $T = 2 \ days$, we have:$2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$Squaring both sides: $4 = \frac{1}{1 - \frac{v^2}{c^2}}$$1 - \frac{v^2}{c^2} = \frac{1}{4} = 0.25$$\frac{v^2}{c^2} = 1 - 0.25 = 0.75$$v = c \sqrt{0.75} = 3 	imes 10^8 	imes 0.866 = 2.598 	imes 10^8 \ m/s \approx 2.6 	imes 10^8 \ m/s$.

One day on a spacecraft corresponds to $2$ days on the earth. The speed of the spacecraft relative to the earth is

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