Answer:
The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame
Explanation:
From the question we are told that
The distance between earth and Retah is [tex]d = 20 \ light \ hours = 20 * 3600 * c = 72000c \ m[/tex]
Here c is the peed of light with value [tex]c = 3.0*10^8 m/s[/tex]
The time taken to reach Retah from earth is [tex]t = 25 \ hours = 25 * 3600 =90000 \ sec[/tex]
The velocity of the spacecraft is mathematically evaluated as
[tex]v_s = \frac{d }{t}[/tex]
substituting values
[tex]v_s = \frac{72000 * 3.0*10^{8} }{90000}[/tex]
[tex]v_s = 2.40*10^{8} \ m/s[/tex]
The time elapsed in the spacecraft’s frame is mathematically evaluated as
[tex]T = t * \sqrt{ 1 - \frac{v^2}{c^2} }[/tex]
substituting value
[tex]T = 90000 * \sqrt{ 1 - \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }[/tex]
[tex]T = 54000 \ s[/tex]
=> [tex]T = 15 \ hours[/tex]
So The time elapsed at the spacecraft’s frame is less that the time elapsed at earth's frame