Question

Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of spaceship 1 relative to spaceship 2?

Answer 1

The speed of spaceship 1 relative to spaceship 2 is 0.

Step-by-step explanation:

The relative velocity between two spaceships can be found by using the equation for relative velocity:

Vrel = (V1 - V2) / (1 - (V1 * V2 / c2))

In this case, spaceship 1 is moving with a velocity of 0.60c and spaceship 2 is also moving with a velocity of 0.60c. Plugging in these values into the equation, we get:

Vrel = (0.60c - 0.60c) / (1 - (0.60c * 0.60c / c2))

Vrel = 0

Therefore, the speed of spaceship 1 relative to spaceship 2 is 0.

Answer 2

The relative speed of 1 relative to 2 is 0.88c

Step-by-step explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

`v_(ab)=(v_(a)+v_(b))/(1+(v_(a)v_(b))/(c^(2)))`

Since it is given that

`v_(a)=0.6c\\\\v_(b)=0.6c`

Applying values in the formula we get

`v_(ab)=(0.6c+0.6c)/(1+((0.6c)^(2))/(c^(2)))\\\\v_(ab)=0.88c`