Figure 37-37 shows three situations in which a starship passes Earth (the dot) and then makes a round trip that brings it back past Earth, each at the given Lorentz factor. As measured in the rest frame of Earth, the round-trip distances are as follows: trip 1, 2D; trip 2, 4D; trip 3, 6D. Neglecting any time needed for accelerations and in terms of D and c, find the travel times of (a) trip 1, (b) trip 2, and (c) trip 3 as measured from the rest frame of Earth. Next, find the travel times of (d) trip 1, (e) trip 2, and (f) trip 3 as measured from the rest frame of the starship. (Hint: For a large Lorentz factor, the relative speed is almost c.)

The distance for trip 1 is$d_1=2D$

The Lorentz factor for trip 1 is ${\gamma }_1=10$

The distance for trip 2 is$d_2=4D$

The Lorentz factor for trip 2 is ${\gamma }_2=24$

The distance for trip 3 is $d_3=6D$

The Lorentz factor for trip 3 is ${\gamma }_3=30$

The speed of star ship is $v=c$

The Lorentz factor is the factor which is introduced in classical formulas is particle moves with speed of light. The length contraction and time dilation is calculated by using Lorentz factor.

The travel time for trip 1 in rest frame of earth is given as:

$t_1=\frac{d_1}{v}$

Substitute all the values in the above equation.

$t_1=\frac{2D}{c}$

Therefore, the travel time for trip 1 in rest frame of earth is $\frac{2D}{c}$.

The travel time for trip 2 in rest frame of earth is given as:

$t_2=\frac{d_2}{v}$

Substitute all the values in the above equation.

$t_2=\frac{4D}{c}$

Therefore, the travel time for trip 2 in rest frame of earth is $\frac{4D}{c}$.

The travel time for trip 3 in rest frame of earth is given as:

$t_3=\frac{d_3}{v}$

Substitute all the values in the above equation.

$t_3=\frac{6D}{c}$

Therefore, the travel time for trip 3 in rest frame of earth is $\frac{6D}{c}$.

The travel time for trip 1 in rest frame of starship is given as:

$T_1=\frac{d_1}{{\gamma }_{1}v}$

Substitute all the values in the above equation.

$$\begin{gathered} T_1=\frac{\left(2D\right)}{\left(10\right)\left(c\right)} \\ {T}_1=\frac{D}{5c} \end{gathered}$$

Therefore, the travel time for trip 1 in rest frame of starship is $\frac{D}{5c}$.

The travel time for trip 2 in rest frame of starship is given as:

$T_2=\frac{d_2}{{\gamma }_{2}v}$

Substitute all the values in the above equation.

$$\begin{gathered} T_2=\frac{\left(4D\right)}{\left(24\right)\left(c\right)} \\ {T}_2=\frac{D}{6c} \end{gathered}$$

Therefore, the travel time for trip 2 in rest frame of starship is $\frac{D}{6c}$.

The travel time for trip 3 in rest frame of starship is given as:

$T_3=\frac{d_3}{{\gamma }_{3}v}\pm$

Substitute all the values in the above equation.

$$\begin{gathered} T_3=\frac{\left(6D\right)}{\left(30\right)\left(c\right)} \\ {T}_3=\frac{D}{5c} \end{gathered}$$

Therefore, the travel time for trip 3 in rest frame of starship is $\frac{D}{5c}$.