Question

Astronomers discover a planet orbiting around a star similar to our sun that is 35 light years away. How fast must a rocket ship go if the round trip is to take no longer than 70 years in time for the astronauts aboard

Answer 1

Answer:

$v = 0.7071c$

Explanation:

Given

Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.

$\triangle{x'} = 70$

$T_0 = 70\ years$ i.e time of travel of the ship.

For the observer on earth, the time is:

$T' = \gamma T_0$

The required speed so that it does not take more than 70 years is then calculated using:

$\triangle x' = vT'$

Substitute $T' = \gamma T_0$

$\triangle x' = v\gamma T_0$

$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

So, we have:

$\triangle x' = \frac{vT_0}{\sqrt{1 - v^2/c^2}}$

Make v the subject of formula.

Square both sides

$\triangle x'^2 = \frac{v^2T^2_0}{1 - v^2/c^2}$

Cross Multiply

$(1 - \frac{v^2}{c^2}) *\triangle x'^2 = v^2T^2_0$

Divide both sides by $\triangle x'^2$

$(1 - \frac{v^2}{c^2}) = \frac{v^2T^2_0}{\triangle x'^2}$

Divide through by $v^2$

$(\frac{1}{v^2} - \frac{v^2}{v^2*c^2}) = \frac{v^2T^2_0}{v^2\triangle x'^2}$

$\frac{1}{v^2} - \frac{1}{c^2} = \frac{T^2_0}{\triangle x'^2}$

Make $\frac{1}{v^2}$ the subject

$\frac{1}{v^2} = \frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}$

Inverse both sides

$v^2 = \frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}$

Take square root of both sides

$v = \sqrt{\frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}}$

Substitute values for $T_0$ and $\triangle x$

$v = \sqrt{\frac{1}{\frac{70^2}{(70c)^2} + \frac{1}{c^2}}}$

$v = \sqrt{\frac{1}{\frac{70^2}{70^2*c^2} + \frac{1}{c^2}}}$

$v = \sqrt{\frac{1}{\frac{1}{c^2} + \frac{1}{c^2}}}$

$v = \sqrt{\frac{1}{\frac{2}{c^2}}}$

$v = \sqrt{\frac{c^2}{2}}$

$v = c\sqrt{\frac{1}{2}}$

$v = c * 0.7071$

$v = 0.7071c$