Question

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.

Answer 1

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$.