Question

Suppose that as you travel away from Earth in a spaceship, you observe another ship pass you heading in the same direction and measure its speed to be $0.50 c .$ As you look back at Earth, you measure Earth's speed relative to you to be \(0.90 c .\) What is the speed of the ship that passed you according to Earth observers?

Short answer

Answer: The speed of the ship that passed you according to Earth observers is approximately \(0.966c\).

Step-by-step solution

List the known values

We know the following speeds: - Speed of the spaceship relative to Earth: \(u = 0.90c\) - Speed of the other ship relative to you: \(v = 0.50c\) - Speed of light: \(c = 1\) (in natural units)

Substitute the values into the relativistic velocity addition formula

Now, we can substitute the values into the formula: $$ w = \frac{0.90c + 0.50c}{1 + \frac{(0.90c)(0.50c)}{c^2}} $$

Simplify the equation

To simplify the equation, we can cancel out \(c\) since it's equal to \(1\) in natural units: $$ w = \frac{0.90 + 0.50}{1 + (0.90)(0.50)} $$

Calculate the result

Now, we can perform the calculations: $$ w = \frac{1.40}{1 + 0.45} = \frac{1.40}{1.45} $$

Convert the result to the speed of light

To convert the result back to the speed of light, we can multiply by \(c\) (in this case, \(c=1\), so no need to multiply): $$ w = \frac{1.40}{1.45}c $$

Final answer

The speed of the ship that passed you according to Earth observers is: $$ w \approx 0.966c $$