A spacecrat approaching the earth launches an exploration vehicle. After the launch, an observer on earth sees the spececrat approaching at a speed of 0.50 c and the exploration vehicle approaching at a speed of \( 0.70 c \). what is the speed of the exploration vehide relative to the spacestip?

We use the relativistic velocity addition formula. Let \[ u = 0.70\,c \quad \text{(speed of the exploration vehicle in the Earth frame)} \] and \[ v = 0.50\,c \quad \text{(speed of the spacecraft in the Earth frame)} \] The velocity of the exploration vehicle relative to the spacecraft is given by: \[ u' = \frac{u - v}{1 - \frac{uv}{c^2}} \] Step 1: Compute the numerator: \[ u - v = 0.70\,c - 0.50\,c = 0.20\,c \] Step 2: Compute the denominator: \[ 1 - \frac{uv}{c^2} = 1 - \frac{(0.70\,c)(0.50\,c)}{c^2} = 1 - 0.35 = 0.65 \] Step 3: Combine the results: \[ u' = \frac{0.20\,c}{0.65} \approx 0.3077\,c \] Thus, the exploration vehicle is moving at approximately: \[ u' \approx 0.31\,c \quad \text{relative to the spacecraft.} \]