Chapter 26, Problem 12P

(a)

To determine

The time observed from Earth.

Answer to Problem 12P

  $\text{11}\text{.0}\text{yr}$ .

Explanation of Solution

Given:

Rest length is $10.6ly$

Speed is $0.960c$

Formula used:

It is known that,

  $t=\frac{L_0}{v}$

Where,

  $L_0$ is rest length

  $t$ is time

  $v$ is velocity

Calculation:

Substitute the given values in the formula: $t_{earth}=\frac{L_0}{v}$

  $t_{earth}=\frac{\left(10.6ly\right)}{0.960c}=11.0yr$

(b)

To determine

Time taken observed from spaceship.

Answer to Problem 12P

  $3.09yr$ .

Explanation of Solution

Given:

Time is $\text{11}\text{.0}\text{yr}$

Speed is $0.960c$

Formula used:

It is known that,

  $\Delta t=\frac{\Delta t_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

Where,

  $t_0$ is rest time

  $t$ is time

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

Substitute the given values in the formula $\Delta t=\frac{\Delta t_0}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}$

  $11.0yr=\frac{\Delta t_0}{{\left[1-{\left(0.960\right)}^2\right]}^{\frac{1}{2}}}$

  $\Delta t_0=3.09yr$

(c)

To determine

The distance traveled.

Answer to Problem 12P

  $2.97ly$ .

Explanation of Solution

Given:

Rest length is $10.6ly$

Speed is $0.960c$

Formula used:

It is known that,

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

Where,

  $L_0$ is rest length

  $L$ is length (contracted)

  $v$ is velocity

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

Calculation:

To the spacecraft observer, distance to the star is contracted:

Substitute the given values in the formula $L=L_0{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}$

  $L=\left(10.6lightyear\right){\left[1-{\left(0.960\right)}^2\right]}^{\frac{1}{2}}$

  $L=2.97lightyear$

(d)

To determine

The speed computed by the occupants.

Answer to Problem 12P

The answer is $0.960c$ , as expected.

Explanation of Solution

Given:

The dilated time on the spacecraft is $3.09yr$

To the spacecraft observer, the distance to the star is contracted: $2.97ly$

Formula used:

It is known that,

  $v=\frac{L}{\Delta t}$

Where,

  $L$ is length (contracted)

  $v$ is velocity

  $t$ is time

Calculation:

Substitute the given values in the formula $v=\frac{L}{\Delta t}$

  $v=\frac{2.97ly}{3.09yr}=0.960c$