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 = [tex]\frac{t_p}{ \sqrt{1- (v/c)^2} }[/tex]
where t_p is the person's own time in an immobile reference frame,
[tex]t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }[/tex]
let's calculate
we assume that the speed of the space station is constant
[tex]t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }[/tex]
[tex]t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}[/tex]
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 =[tex]\frac{1}{1.3333 \ 10^{-5}}[/tex]
#_time = 7.5 10⁴ s
