Question

An astronaut leaves Earth in a spaceship at a speed of 0.96 c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?

Answer 1

t=4.2 years

Step-by-step explanation:

velocity v= 0.96c

destination star distance = 14.4 light year

According to the theory of relativity length contraction

`l= \frac{l_0}{\sqrt{1-(v^2)/(c^2)}}`

`l_0= l{\sqrt{1-(v^2)/(c^2)}}`

putting values we get

`l_0= 14.4{\sqrt{1-((0.96c)^2)/(c^2)}}`

`l_0= 4.032 light years`

now distance the trip covers

D= vt

`4.032* c= 0.96c* t`

`t= (4.032c)/(0.96c)`

t= 4.2 years

so the trip will take 4.2 years