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A spacecraft flies away from the earth with a speed of 4.60x10
6
m/s relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronised with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days (1 year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks?

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Answer:

Explanation:

Here we shall apply time dilation formula of relativity . Time recorded in spacecraft will be less .

365 = [tex]\frac{T}{\sqrt{1-\frac{v^2}{c^2} } }[/tex]

365 =  [tex]\frac{T}{\sqrt{1-\frac{(4.6\times10^6)^2}{(3\times 10^8)^2} } }[/tex]

365  = [tex]\frac{T}{.99988}[/tex]

T  = 364.95 days .

Difference of elapsed time

= .05 days .

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