Multiple Choice
Carol is in the same reference frame with a clock. Bianca is flying past Carol and her clock at a high speed. Bianca sees Carol's clock ticking at one quarter the rate that Carol sees. How fast is Bianca flying relative to Carol?
35. Special Relativity
Consequences of Relativity
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The international space station travels in orbit at a speed of 7.67 km/s. If an astronaut and his brother start a stop watch at the same time, on Earth, and then the astronaut spends 6 months on the space station, what is the difference in time on their stopwatches when the astronaut returns to Earth? Note that 6 months is about 1.577 x 107 s, and c = 3 x 10 8 m/s.
A
1.00000000033 s
B
0.9999999993 s
C
0.9999999997 s
D
0.0052 s
Verified step by step guidance1
First, recognize that this problem involves time dilation, a concept from Einstein's theory of relativity. Time dilation occurs because the astronaut is moving at a high speed relative to his brother on Earth.
Use the time dilation formula: \( t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( t' \) is the time experienced by the moving observer (astronaut), \( t \) is the time experienced by the stationary observer (brother on Earth), \( v \) is the velocity of the moving observer, and \( c \) is the speed of light.
Substitute the given values into the formula: \( v = 7.67 \text{ km/s} = 7670 \text{ m/s} \), \( c = 3 \times 10^8 \text{ m/s} \), and \( t = 1.577 \times 10^7 \text{ s} \).
Calculate the factor \( \frac{v^2}{c^2} \) and then find \( \sqrt{1 - \frac{v^2}{c^2}} \). This will give you the time dilation factor.
Finally, calculate the difference in time \( \Delta t = t' - t \) to find how much less time has passed for the astronaut compared to his brother on Earth. This will give you the time difference on their stopwatches.
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