One of the many exciting phenomena of special relativity is time dilation. Imagine astronauts
in a spaceship that is passing by the Earth with a high velocity.
(a) Are clocks ticking slower for the people on Earth or for the astronauts on the spaceship?
(b) How fast must the spaceship travel such that the clocks go twice as slow?
In its own frame of reference (that is, in such a frame of reference moving with the body, in which this body is at rest all the time), time flows more slowly than in the laboratory frame of reference, from which the movement of the body is observed.
Hence
a) Astronauts' clocks on a spacecraft run slower.
b) The time interval on the spacecraft "\\Delta\\tau" is related to the time interval on the Earth "\\Delta t" by the formula
"\\Delta\\tau=\\sqrt{1- \\frac{v^2}{c^2}} \\cdot \\Delta t"
by the condition of the problem
"\\Delta\\tau= \\frac{\\Delta t}{2}"
Then we write
"\\frac{\\Delta t}{2}=\\sqrt{1- \\frac{v^2}{c^2}} \\cdot \\Delta t"
"\\frac{1}{2}=\\sqrt{1- \\frac{v^2}{c^2}}"
"\\frac{1}{4}=1- \\frac{v^2}{c^2}"
"\\frac{v^2}{c^2}=1- \\frac{1}{4}"
"\\frac{v^2}{c^2}=\\frac{3}{4}"
The spaceship should be casting at a speed
"v=c \\cdot \\frac{\\sqrt{3}}{2}=3 \\cdot 10^{8} \\cdot 0.866=2.598 \\cdot 10^{8}km\/s"
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