Answer:
[tex]v = 0.7071c[/tex]
Explanation:
Given
Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.
[tex]\triangle{x'} = 70[/tex]
[tex]T_0 = 70\ years[/tex] i.e time of travel of the ship.
For the observer on earth, the time is:
[tex]T' = \gamma T_0[/tex]
The required speed so that it does not take more than 70 years is then calculated using:
[tex]\triangle x' = vT'[/tex]
Substitute [tex]T' = \gamma T_0[/tex]
[tex]\triangle x' = v\gamma T_0[/tex]
[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
So, we have:
[tex]\triangle x' = \frac{vT_0}{\sqrt{1 - v^2/c^2}}[/tex]
Make v the subject of formula.
Square both sides
[tex]\triangle x'^2 = \frac{v^2T^2_0}{1 - v^2/c^2}[/tex]
Cross Multiply
[tex](1 - \frac{v^2}{c^2}) *\triangle x'^2 = v^2T^2_0[/tex]
Divide both sides by [tex]\triangle x'^2[/tex]
[tex](1 - \frac{v^2}{c^2}) = \frac{v^2T^2_0}{\triangle x'^2}[/tex]
Divide through by [tex]v^2[/tex]
[tex](\frac{1}{v^2} - \frac{v^2}{v^2*c^2}) = \frac{v^2T^2_0}{v^2\triangle x'^2}[/tex]
[tex]\frac{1}{v^2} - \frac{1}{c^2} = \frac{T^2_0}{\triangle x'^2}[/tex]
Make [tex]\frac{1}{v^2}[/tex] the subject
[tex]\frac{1}{v^2} = \frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}[/tex]
Inverse both sides
[tex]v^2 = \frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}[/tex]
Take square root of both sides
[tex]v = \sqrt{\frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}}[/tex]
Substitute values for [tex]T_0[/tex] and [tex]\triangle x[/tex]
[tex]v = \sqrt{\frac{1}{\frac{70^2}{(70c)^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{70^2}{70^2*c^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{1}{c^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{2}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{c^2}{2}}[/tex]
[tex]v = c\sqrt{\frac{1}{2}}[/tex]
[tex]v = c * 0.7071[/tex]
[tex]v = 0.7071c[/tex]
