Question

A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of \(0.60 c\) and \(0.80 c .\) What is the relative speed of the two ships as measured by a passenger on either one of the spaceships?

Short answer

Answer: The relative speed of the two ships as measured by a passenger on either one of the spaceships is approximately \(0.9459c\).

Step-by-step solution

Identify the given velocities

The given velocities of the two spaceships are \(0.60c\) and \(0.80c.\) We will use these values in the formula for relative velocity in special relativity.

Use the relativistic velocity addition formula

To find the relative velocity of the spaceships, use the formula \(v_{rel} = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}},\) where \(v_1\) and \(v_2\) are the velocities of the spaceships. Substitute the given values: \(v_{rel} = \frac{0.60c + 0.80c}{1 + \frac{(0.60c) (0.80c)}{c^2}}.\)

Simplify the expression

Now, simplify the expression: \(v_{rel} = \frac{1.40c}{1 + \frac{0.48c^2}{c^2}}.\)

Cancel common terms

Cancel out the \(c^2\) terms in the denominator: \(v_{rel} = \frac{1.40c}{1 + 0.48}.\)

Calculate the relative velocity

Finally, calculate the relative velocity by dividing the numerator by the denominator: \(v_{rel} = \frac{1.40c}{1.48} = 0.9459c.\) The relative speed of the two ships as measured by a passenger on either one of the spaceships is approximately \(0.9459c.\)