Chapter 26, Problem 45P

(a)

To determine

The speed of second spaceship relative to Earth when it moves in direction of first spaceship.

Answer to Problem 45P

  $0.98c$

Explanation of Solution

Given info:

Speed of first spaceship with respect to the earth is $v=+0.71c$ .

Speed of second spaceship is $u\text{'}=+0.87c$

Formula used:

  $u=\frac{\left(v+u\text{'}\right)}{1+\frac{vu\text{'}}{c^2}}$

Where,

  $m$ is relativistic mass

  $v$ is velocity (with the respect to the earth)

  $u\text{'}$ is a second speed (or speed respect to first speed)

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

Note: If $u\text{'}$ is in the opposite direction from $v$ then $u\text{'}$ must have a minus sign in the above mentioned expression.

Calculation:

Take positive direction in the direction of the first spaceship.

In the reference frame of the earth, first spaceship is moving at $+0.71c$ and the second spaceship is moving at $+0.87c$ relative to the first.

So, speed of the second spaceship relative to the first is

  $u=\frac{\left(0.71c+0.87c\right)}{1+\left(0.71\right)\left(0.87\right)}=0.98c$

(b)

To determine

The speed of second spaceship relative to the Earth when it moves directly backward towards Earth.

Answer to Problem 45P

The answer is $-0.42c$

Explanation of Solution

Given info:

Speed with respect to the earth is $v=+0.71c$ .

Speed of second spacecraft is $u\text{'}=-0.87c$ .

Formula used:

  $u=\frac{\left(v+u\text{'}\right)}{1+\frac{vu\text{'}}{c^2}}$

Where,

  $m$ is relativistic mass

  $v$ is velocity (with the respect to the earth)

  $u\text{'}$ is a second speed (or speed respect to first speed)

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

Note: If $u\text{'}$ is in the opposite direction from $v$ then $u\text{'}$ must have a minus sign in the above mentioned expression.

Calculation:

Take positive direction in the direction of the first spaceship.

In the reference frame of the earth, first spaceship is moving at $+0.71c$ and the second spaceship is moving at $-0.87c$ relative to the first.

So, speed of the second spaceship relative to the Earth is

  $u=\frac{\left(0.71c+\left(-0.87c\right)\right)}{1+\left(0.71\right)\left(-0.87\right)}=-0.42c$