Question

An armada of spaceships that is 1.00 ly long (as measured in its rest frame) moves with speed 0.800c relative to a ground station in frame S.A messenger travels from the rear of the armada to the front with a speed of 0.950c relative to S. How long does the trip take as measured (a) in the rest frame of the messenger, (b) in the rest frame of the armada, and (c) by an observer in the ground frame S?

Short answer

(a) The trip time in the rest frame of messenger is $1.25\text{\hspace{0.17em}y}$.

(b) The trip time in the rest frame of armada is $1.60\text{\hspace{0.17em}y}$.

(c) The trip time for observer in ground frame S is $4\text{\hspace{0.17em}y}$.

Step-by-step solution

Identification of given data

The length of the spaceships is $L_0=1\text{\hspace{0.17em}ly}$

The speed of spaceships is $u=0.800c$

The speed of messenger is $v^{\text{'}}=0.950c$

The speed of armada of messenger relative to ground station S is found by using the formula for relativistic relative velocity. The length of spaceships for ground observer is found by length contraction formula.

Determination of trip time in rest frame of messenger

(a)

The speed of the messenger in rest frame is given as:

$\mathrm{v}=\frac{\mathrm{u}-{\mathrm{v}}^{\text{'}}}{1-\frac{{\mathrm{uv}}^{\text{'}}}{{\mathrm{c}}^2}}$

Substitute all the values in the above equation.

$\begin{array}{l}v=\frac{0.800c-0.950c}{1-\frac{\left(0.800c\right)\left(0.950c\right)}{c^2}}\\ v=-0.625c\end{array}$

The length of armada in rest frame of messenger is given as:

$L=L_0\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Substitute all the values in the above equation.

$\begin{array}{l}L=\left(1\text{\hspace{0.17em}ly}\right)\sqrt{1-{\left(\frac{-0.625c}{c}\right)}^2}\\ L=0.781\text{\hspace{0.17em}ly}\end{array}$

The trip time in the rest frame of messenger is given as:

$t=\frac{L}{\left|v\right|}$

Substitute all the values in above equation.

$\begin{array}{l}t=\frac{0.781\text{\hspace{0.17em}ly}}{|-0.625c|}\\ t=\frac{\left(0.781\text{\hspace{0.17em}ly}\right)\left(\frac{1c\text{\hspace{0.17em}y}}{1\text{\hspace{0.17em}ly}}\right)}{0.625c}\\ t=1.25\text{\hspace{0.17em}y}\end{array}$

Therefore, the trip time in the rest frame of messenger is $1.25\text{\hspace{0.17em}y}$.

Determination of trip time in rest frame of armada

(b)

The trip time in the rest frame of armada is given as:

$t_a=\frac{L_0}{\left|v\right|}$

Substitute all the values in above equation.

$\begin{array}{l}{t}_a=\frac{1\text{\hspace{0.17em}ly}}{|-0.625c|}\\ t_a=\frac{\left(1\text{\hspace{0.17em}ly}\right)\left(\frac{1c\text{\hspace{0.17em}y}}{1\text{\hspace{0.17em}ly}}\right)}{0.625c}\\ t_a=1.60\text{\hspace{0.17em}y}\end{array}$

Therefore, the trip time in the rest frame of armada is $1.60\text{\hspace{0.17em}y}$.

Determination of trip time for observer in ground frame S

(c)

The length of armada for observer in ground frame S is given as:

$L_1=L_0\sqrt{1-{\left(\frac{u}{c}\right)}^2}$

Substitute all the values in the above equation.

$\begin{array}{l}{L}_1=\left(1\text{\hspace{0.17em}ly}\right)\sqrt{1-{\left(\frac{0.800c}{c}\right)}^2}\\ L_1=0.6\text{\hspace{0.17em}ly}\end{array}$

The trip time for observer in ground frame S is given as:

$t_g=\frac{L_1}{v^{\text{'}}-u}$

Substitute all the values in above equation.

$\begin{array}{l}{t}_g=\frac{0.6\text{\hspace{0.17em}ly}}{0.950c-0.800c}\\ t_g=\frac{\left(0.6\text{\hspace{0.17em}ly}\right)\left(\frac{1c\text{\hspace{0.17em}y}}{1\text{\hspace{0.17em}ly}}\right)}{0.150c}\\ t_g=4\text{\hspace{0.17em}y}\end{array}$

Therefore, the trip time for observer in ground frame S is$4\text{\hspace{0.17em}y}$ .