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In the future, an experimental spacecraft leaves space dock for a test flight. After flying in a very large circle at constant speed, the spacecraft returns to the space dock. The observers at the stationary space dock note that the flight lasted for 7.24 hours. However, the clock aboard the spacecraft records only 3.69 hours passing during the flight. at what speed did the spacecraft fly?

Respuesta :

Answer:

2.58 x 10āø m/s

Explanation:

Time dilation fomula will be applicable here, which is given below.

t = \frac{T}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}

Where T is dilated time or time observed by clock in motion , t is stationary time , v is velocity of clock in motion and c is velocity of light .

c is 3 times 10āø msā»Ā¹ , T is 7.24 h , t is 3.69 h. Put these values in the formula

7.24 = \frac{3.69}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}\\

\frac{v^2}{c^2}=0.744\\\\

v=2.58\times 10^8

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