Question

An armada of spaceships that is 1.60 ly long (in its rest frame) moves with speed 0.800c relative to the ground station in frame S. A messenger travels from the rear of the armada to the front with a speed of 0.910c relative to S. How long does the trip take as measured (a) in the messenger's rest frame, (b) in the armada's rest frame, and (c) by an observer in frame S?

Answer 1

(a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.

Step-by-step explanation:

Given that,

Length = 1.60 ly

Speed of spaceships= 0.800c

Speed of messenger = 0.910 c

(a). We need to calculate the velocity of armada

Using formula of velocity

`v'=(v-v_(m))/(1-(vv_(m))/(c^2))`

Put the value into the formula

`v'=(0.800c-0.910c)/(1-(0.800*0.910 c^2)/(c^2))`

`v'=(-0.11c)/(1-0.800*0.910)`

`v'=-0.404c`

We need to calculate the length

Using formula of length

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

Put the value into the formula

`l=1.60√(1-(-0.404)^2)`

`l=1.46\ ly`

We need to calculate the length of the trip

Using formula of time

`t=(l)/(|v'|)`

Put the value into the formula

`t=(1.46)/(0.404)`

`t=3.61\ year`

(b). If the armada's rest frame

We need to calculate the length

Using formula of length

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

Put the value into the formula

`l=1.60√(1-(0)^2)`

`l=1.60\ ly`

Using formula of time

`t=(l)/(|v'|)`

Put the value into the formula

`t=(1.60)/(0.404)`

`t=3.96\ year`

(c). If an observer in frame S

We need to calculate the length

Using formula of length

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

Put the value into the formula

`l=1.60√(1-(0.800)^2)`

`l=0.96\ ly`

We need to calculate the time

`t=(0.96)/(0.910-0.800)`

`t=8.73\ year`

Hence, (a). The trip take a time is 3.61 year.

(b). The trip take a time is 3.96 year.

(c). The trip take a time is 8.73 year.