Chapter 20, Problem 3SP

Suppose an astronaut travels to a distant star and returns to Earth. Except for brief intervals of time when he is accelerating or decelerating, his spaceship travels at the incredible speed of v = 0.995c relative to the Earth. The star is 46 light-years away. (A light-year is the distance light travels in 1 year.) a. Show that the factor γ for this velocity is approximately equal to 10.

a.    How long does the trip to the star and back take as seen by an observer on Earth?

b.    How long does the trip take as measured by the astronaut?

c.    What is the distance traveled as measured by the astronaut?

d.    If the astronaut left a twin brother at home on Earth while he made this trip, how much younger is the astronaut than his twin when he returns?

To determine

To show that the factor $\gamma$ for $0.995c$ velocity of spaceship is $10$.

Answer to Problem 3SP

It is shown that the factor $\gamma$ for $0.995c$ velocity of spaceship is $10$.

Explanation of Solution

Given info: The velocity of the spaceship is $0.995c$. The star is $46\text{lightyears}$ away from earth.

Write the expression to find the $\gamma$ factor.

$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Here,

$v$ is the velocity spaceship

$c$ is the speed of light

Substitute $0.995c$ for $v$ in the above equation.

$\begin{array}{c}\gamma =\frac{1}{\sqrt{1-\frac{{\left(0.995c\right)}^2}{c^2}}}\\ =10\end{array}$

Conclusion:

Therefore, it is shown that the factor $\gamma$ for $0.995c$ velocity of spaceship is $10$.

To determine

The time taken for the travel to the star and back to earth as seen by an observer on earth.

Answer to Problem 3SP

The time taken for the travel to the star and back to earth as seen by an observer on earth is $92.46\text{ yr}$.

Explanation of Solution

Write the expression to find the distance travelled in meter.

$\begin{array}{c}d=46\text{lightyears}+46\text{lightyears}\\ =92\text{lightyears}\\ \text{=8}\text{.7}\times {\text{10}}^{17}\text{ m}\end{array}$

Write the expression to find the time taken by the spaceship to travel back and forth.

$t=\frac{d}{v}$

Here,

$d$ is the distance travelled in meter

$v$ is the velocity of the spaceship

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $\text{8}\text{.7}\times {\text{10}}^{17}\text{ m}$ for $d$ and $0.995c$ for $v$ in the above equation.

$\begin{array}{c}t=\frac{\text{8}\text{.7}\times {\text{10}}^{17}\text{ m}}{\left(3\times {10}^8\text{m}/\text{s}\right)\left(0.995\right)}\\ =2914572864\text{ s}\\ \text{=92}\text{.46 yr}\end{array}$

Conclusion:

Therefore, the time taken for the travel to the star and back to earth as seen by an observer on earth is $92.46\text{ yr}$.

To determine

The time taken for the travel to the star and back to earth as seen by the astronaut.

Answer to Problem 3SP

The time taken for the travel to the star and back to earth as seen by the astronaut is $9.246\text{ yr}$.

Explanation of Solution

Write the expression to find the distance travelled in meter.

$\begin{array}{c}d=46\text{lightyears}+46\text{lightyears}\\ =92\text{lightyears}\\ \text{=8}\text{.7}\times {\text{10}}^{17}\text{ m}\end{array}$

Write the expression to find the time taken by the spaceship to travel back and forth as seen by the astronaut.

$t=\frac{d}{v}\gamma$

Here,

$d$ is the distance travelled in meter

$v$ is the velocity of the spaceship

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$, $\text{8}\text{.7}\times {\text{10}}^{17}\text{ m}$ for $d$, $10$ for $\gamma$ and $0.995c$ for $v$ in the above equation.

$\begin{array}{c}t=\frac{\text{8}\text{.7}\times {\text{10}}^{17}\text{ m}}{\left(3\times {10}^8\text{m}/\text{s}\right)\left(0.995\right)}\left(10\right)\\ =29145728640\text{ s}\\ \text{=9}\text{.246 yr}\end{array}$

Conclusion:

Therefore, the time taken for the travel to the star and back to earth as seen by the astronaut is $9.246\text{ yr}$.

To determine

The distance travelled as measured by the astronaut.

Answer to Problem 3SP

The distance travelled as measured by the astronaut is $9.19\text{ ly}$.

Explanation of Solution

Write the expression to find the distance as measured by the astronaut.

$d=vt$

Here,

$t$ is the time measured by the astronaut

Substitute $0.995c$ for $v$,$3\times {10}^8\text{m}/\text{s}$ for $c$ and $9.246\text{ yr}$ for $t$ in the above equation.

$\begin{array}{c}d=\left(0.995\right)\left(3\times {10}^8\text{m}/\text{s}\right)\left(9.246\text{ yr}\right)\\ =\left(0.995\right)\left(3\times {10}^8\text{m}/\text{s}\right)\left(9.246\text{ yr}\right)\\ =9.19\text{ly}\end{array}$

Conclusion:

Therefore, the distance travelled as measured by the astronaut is $9.19\text{ ly}$.

To determine

The age difference between the astronaut and his twin after he returns from his space journey.

Answer to Problem 3SP

The age difference between the astronaut and his twin after he returns from his space journey is $83.2\text{ y}$.

Explanation of Solution

Write the expression to find the age difference between the astronaut and his twin.

$t=t_e-t_s$

Here,

$t_e$ is the time of travel observed by observer on earth

$t_s$ is the time of travel observed by the astronaut in the spaceship

Substitute $9.246\text{ y}$ for $t_s$ and $92.46\text{ y}$ for $t_e$ in the above equation.

$\begin{array}{c}t=92.46\text{ y}-9.246\text{ y}\\ \text{=}83.2\text{ y}\end{array}$

Conclusion:

Therefore, the age difference between the astronaut and his twin after he returns from his space journey is $83.2\text{ y}$.