Chapter 15, Problem 37A

Before leaving the planet Hislaurels for a starship voyage, you pack a meterstick in your luggage. After the ship has settled down to a steady speed of 0.50c, you take the meterstick out of your bag.

a. How long will you measure the meterstick to be?

b. If the meterstick is moving parallel to an observer resting on Hislaurels, how long will the observer measure the meterstick?

To determine

The length of the meterstick measured by the person.

Answer to Problem 37A

The length of the meterstick is $1\text{m}$ as there is no relative motion between the meterstick and the person.

Explanation of Solution

Given:

Consider a person leaves the planet for a starship voyage.

The person has a meterstick in his luggage.

After settling down, the ship moves with steady speed of $0.50c$ .

Formula used:

The relativistic length contraction expression is,

  $L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Here, $L$ is the length of the meterstick measured by the observer, $L_0$ is the actual length of the meterstick, $v$ is the speed of the star ship, and $c$ is the speed of the light.

Calculation:

The ship moves with a steady speed of $0.50c$ .

Consider the person takes the meterstick of length $1\text{m}$ out of the luggage.There is no relative motion between the meterstick and the person.The person measures the length of the meterstick as $1\text{m}$ as there is no relative motion between the meterstick and the person.

Conclusion:

Thus, the person measures the length of the meterstick as $1\text{m}$ as there is no relative motion between the meterstick and the person.

To determine

The length of the meterstick measured by the observer

Answer to Problem 37A

The length of the meterstick measured by the observer is $L=0.87\text{m}$ .

Explanation of Solution

Given:

Consider a person leaves the planet for a star ship voyage.

The person has a meterstick in his luggage.

After settling down, the ship moves with steady speed of $0.50c$ .

The ships moves parallel to the observer resting in the planet.

Formula used:

The relativistic length contraction expression is,

  $L=L_0\sqrt{1-\frac{v^2}{c^2}}$

Here, $L$ is the length of the meterstick measured by the observer, $L_0$ is the actual length of the meterstick, $v$ is the speed of the star ship, and $c$ is the speed of the light.

Calculation:

The ship moves with a steady speed of $0.50c$ .

Consider the person takes the meterstick of length $1\text{m}$ out of the luggage.

The ship moves parallel to the observer resting in the planet.

The length of the meterstick measured by the observer is,

  $\begin{array}{l}L=L_0\sqrt{1-\frac{v^2}{c^2}}\\ L=1\text{m}\sqrt{1-\frac{{\left(0.5c\right)}^2}{c^2}}\\ L=1\text{m}\sqrt{1-\frac{0.25c^2}{c^2}}\\ L=1\text{m}\sqrt{1-0.25}\\ L=0.87\text{m}\end{array}$

Conclusion:

Thus, the length of the meterstick measured by the observer is $L=0.87\text{m}$ .