Chapter 21, Problem 30Q

(a)

To determine

The time interval of the trip as measured by an observer on Earth, given that an astronaut flies at a speed, that is, 80% of the speed of light, to a distant star, and the time taken for one side of the trip is 15 years, as measured by the astronaut.

Answer to Problem 30Q

Solution:

$25\text{ years}$

Explanation of Solution

Given data:

An astronaut flies at a speed that is 80% of the speed of light, to a distant star, and the time taken for one side of the trip is 15 years, as measured by the astronaut.

Formula used:

Write the expression for Lorentz transformation of time

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

Here, $T$ is the time interval measured by an astronaut (moving relative to the phenomenon), $T_o$ is time interval measured by an observer on Earth (not moving relative to the phenomenon), $v$ is the speed of the moving observer, relative to the phenomenon, and $c$ is the speed of light.

Explanation:

Consider spaceship observer moving relative to the phenomenon with time interval $T_o$ and clock on earth not moving relative to the phenomenon with time interval $T$.

Let the speed of light, $c$, be equal to $3\times {10}^8\text{ m/s}$.

Refer to the expression for Lorentz transformation of time

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

Substitute $15\text{ years}$ for $T_o$ and $0.8c$ for $v$,

$\begin{array}{c}T=\frac{\left(15\text{ years}\right)}{\sqrt{1-{\left(\frac{0.8c}{c}\right)}^2}}\\ =\frac{15}{\sqrt{1-{\left(0.8\right)}^2}}\\ =25\text{ years}\end{array}$

Conclusion:

Hence, the time interval of the trip, as measured by an observer on Earth, is $25\text{ years}$.

(b)

To determine

The distance from Earth to star as measured by observer on earth and as measured by astronaut. Given that an astronaut flies at speed 80% of speed of light to a distant star and the time taken in one-way trip is 15 years as measured by astronaut.

Answer to Problem 30Q

Solution:

$d=12\text{ light year}$; $d^{\prime }=20\text{ light year}$

Explanation of Solution

Given data:

An astronaut flies at speed 80% of speed of light to a distant star and the time taken in one-way trip is 15 years as measured by astronaut.

Formula used:

Write the expression of speed, time, and distance.

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

Here, $v$ is the speed, $d$ is distance travelled, and $t$ is the time.

Explanation:

Let speed of light be $c$ equals to $3\times {10}^8\text{ m/s}$. Also $d^{\prime }$ and $d$ be the distance measured by observer on earth and that measured by astronaut.

As an astronaut flies at speed 80% of speed of light to a distant star takes 15 years in one-way trip.

Calculate distance measured by astronaut.

Refer to the expression of speed, time, and distance.

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

Substitute $0.8c$ for $v$ and $15\text{ years }$ for $t$,

$\begin{array}{c}\left(0.8c\right)=\frac{d}{\left(15\text{ years }\right)}\\ d=\left(0.8c\right)\left(15\text{ years }\right)\left(\frac{365\text{ d}}{1\text{ years}}\right)\left(\frac{24\text{ h}}{1\text{ d}}\right)\left(\frac{60\text{ min}}{1\text{ h}}\right)\left(\frac{60\text{ s}}{1\text{ min}}\right)\end{array}$

Substitute $3\times {10}^8\text{ m/s}$ for $c$,

$\begin{array}{c}d=\left[0.8\left(3\times {10}^8\text{ m/s}\right)\right]\left(15\text{ years }\right)\left(\frac{365\text{ d}}{1\text{ years}}\right)\left(\frac{24\text{ h}}{1\text{ d}}\right)\left(\frac{60\text{ min}}{1\text{ h}}\right)\left(\frac{60\text{ s}}{1\text{ min}}\right)\\ =1.135\times {10}^{14}\text{ km}\left(\frac{\text{1 light year}}{9.461\times {10}^{12}\text{ km}}\right)\\ =12\text{ light year}\end{array}$

Calculate distance measured by observer on earth.

Refer to the expression of speed, time, and distance.

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

Substitute $0.8c$ for $v$ and $25\text{ years }$ for $t$,

$\begin{array}{c}\left(0.8c\right)=\frac{d^{\prime }}{\left(15\text{ years }\right)}\\ d^{\prime }=\left(0.8c\right)\left(25\text{ years }\right)\left(\frac{365\text{ d}}{1\text{ years}}\right)\left(\frac{24\text{ h}}{1\text{ d}}\right)\left(\frac{60\text{ min}}{1\text{ h}}\right)\left(\frac{60\text{ s}}{1\text{ min}}\right)\end{array}$

Substitute $3\times {10}^8\text{ m/s}$ for $c$,

$\begin{array}{c}{d}^{\prime }=\left[0.8\left(3\times {10}^8\text{ m/s}\right)\right]\left(25\text{ years }\right)\left(\frac{365\text{ d}}{1\text{ years}}\right)\left(\frac{24\text{ h}}{1\text{ d}}\right)\left(\frac{60\text{ min}}{1\text{ h}}\right)\left(\frac{60\text{ s}}{1\text{ min}}\right)\\ =1.892\times {10}^{14}\text{ km}\left(\frac{\text{1 light year}}{9.461\times {10}^{12}\text{ km}}\right)\\ =20\text{ light year}\end{array}$

Conclusion:

Hence, distance from earth to star measured by astronaut is $12\text{ light year}$ and the measured by observer on earth is $20\text{ light year}$.