Chapter 26, Problem 44P

(a)

To determine

Speed of the first spaceship relative to the second spaceship.

Answer to Problem 44P

  $0.80c$

Explanation of Solution

Given:

Speed with respect to the earth is $v=0.50c$

Speed of second spacecraft is $u\text{'}=0.50c$

Formula used:

It is known that

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

Where,

  $m$ is relativistic mass

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

  $u\text{'}$ is a speed (Or speed respect to the first speed)

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

Calculation:

In the reference frame of the second spaceship, the Earth is moving at $0.50c$ and the first spaceship is moving at $0.50c$ relative to the Earth.

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

  $u=\frac{\left(0.50c+0.50c\right)}{1+\left(0.50\right)\left(0.50\right)}=0.80c$

(b)

To determine

The speed of the second spaceship relative to the first spaceship.

Answer to Problem 44P

  $-0.80c$

Explanation of Solution

Given:

Speed with respect to the earth is $v=-0.50c$

Speed of second spacecraft is $u\text{'}=-0.50c$

Formula used:

It is known that

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

Where,

  $m$ is relativistic mass

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

  $u\text{'}$ is a speed (Or speed respect to the first speed)

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

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

Calculation:

In the reference frame of the first spaceship, the Earth is moving at $-0.50c$ and the second spaceship is moving at $-0.50c$ relative to the Earth.

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

  $u=\frac{\left(-0.50c+\left(-0.50c\right)\right)}{1+\left(-0.50\right)\left(-0.50\right)}=-0.80c$