- #1
stunner5000pt
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- Homework Statement
- A space traveler moving at a speed of 0.70c with respect to Earth makes a trip to a distant star that is stationary relative to Earth. He measures the length of this trip to be 6.5ly. What would be the length of this same trip (in ly) as measured by a traveler moving at a speed of 0.90c with respect to Earth?
- Relevant Equations
- [tex] \Delta t = \gamma t_{0} [/tex]
If we took the perspective of the space traveller themselves, they are stationary and the whole universe goes past them at 0.7c. THen th elapsed time of 6.5 yr looking outside is
[tex] \Delta t =6.5 \frac{1}{\sqrt{1-0.7^2}} = 9.11 yrs [/tex]
THen, when the traveller looks at the person travelling at 0.9c
the time taken should be
[tex] 9.11 \times \frac{1}{\sqrt{1-0.9^2}} = 3.97 yrs [/tex]
but the answer is 4.01 yr. Is this just down to rounding or is there something missing in the solution above?
thanks in advance for your help
[tex] \Delta t =6.5 \frac{1}{\sqrt{1-0.7^2}} = 9.11 yrs [/tex]
THen, when the traveller looks at the person travelling at 0.9c
the time taken should be
[tex] 9.11 \times \frac{1}{\sqrt{1-0.9^2}} = 3.97 yrs [/tex]
but the answer is 4.01 yr. Is this just down to rounding or is there something missing in the solution above?
thanks in advance for your help