Question

Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay aboard the space station to be a full second youngerthan people on the ground? Please show and explain how you would set-up the problem,before you actually try to solve it. If you cannot solve it exactly, please try to offer an estimate.(5 pts)

Answer 1

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

            t = $\frac{t_p}{ \sqrt{1- (v/c)^2} }$

where t_p is the person's own time in an immobile reference frame,

           $t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }$

let's calculate

we assume that the speed of the space station is constant

              $t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }$

             $t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}$

             t_ =  0.99998666657   s

             

therefore the time change is

             Δt = t - t_p

             Δt = 1 - 0.9998666657                  

              Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

               #_time = 1 / Δt

               #_time =$\frac{1}{1.3333 \ 10^{-5}}$

               #_time = 7.5 10⁴ s