Answer:
0.9999986*c
Explanation:
The ship would travel 2.54*10^7 light years, which means that at a speed close to the speed of light the trip would take 2.54*10^7 years from the point of view of an observer on Earth. However from the point of view of a passenger of that ship it will take only 70 years if the speed is close enough to the speed of light.
[tex]\Delta t = \Delta t' * \sqrt{1 - (\frac{v}{c})^2}[/tex]
Where
Δt is the travel time as seen by a passenger
Δt' is the travel time as seen by someone on Earth
v is the speed of the ship
c is the speed of light in vacuum
We can replace the fraction v/c with x
[tex]\Delta t = \Delta t' * \sqrt{1 - x^2}[/tex]
[tex]\sqrt{1 - x^2} = \frac{\Delta t}{\Delta t'}[/tex]
[tex]1 - x^2 = (\frac{\Delta t}{\Delta t'})^2[/tex]
[tex]x^2 = 1 - (\frac{\Delta t}{\Delta t'})^2[/tex]
[tex]x = \sqrt{1 - (\frac{\Delta t}{\Delta t'})^2}[/tex]
[tex]x = \sqrt{1 - (\frac{70}{2.54*10^7})^2} = 0.9999986[/tex]
It would need to travel at 0.9999986*c
