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If an observer on earth sees two space ships approaching each other and she measures the speed of ship 1 wrt (with respect to) herself to be \( 0.8 c \) to the right and ship 2 to be 0.7 c to the left. An observer in ship 1 also sees the approach of ship 2. What velocity does he see ship 2 moving at wrt his ship 1?

Ask by Stuart Burns. in Canada
Mar 09,2025

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Answer

The observer in ship 1 sees ship 2 moving at approximately \( 0.227c \) to the right.

Solution

To find the velocity of ship 2 as observed from ship 1, we can use the relativistic velocity addition formula. The formula is given by: \[ u' = \frac{u + v}{1 + \frac{uv}{c^2}} \] where: - \( u' \) is the velocity of ship 2 as observed from ship 1, - \( u \) is the velocity of ship 2 as observed from Earth (which is \( -0.7c \) since it is moving to the left), - \( v \) is the velocity of ship 1 as observed from Earth (which is \( 0.8c \)), - \( c \) is the speed of light. Now, substituting the values into the formula: \[ u' = \frac{-0.7c + 0.8c}{1 + \frac{(-0.7c)(0.8c)}{c^2}} \] Let's simplify this step by step. 1. Calculate the numerator: \[ -0.7c + 0.8c = 0.1c \] 2. Calculate the denominator: \[ 1 + \frac{(-0.7c)(0.8c)}{c^2} = 1 - 0.56 = 0.44 \] Now we can substitute these results back into the equation for \( u' \): \[ u' = \frac{0.1c}{0.44} \] Now, let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{0.1}{0.44}\) - step1: Convert the expressions: \(\frac{\frac{1}{10}}{\frac{11}{25}}\) - step2: Multiply by the reciprocal: \(\frac{1}{10}\times \frac{25}{11}\) - step3: Reduce the numbers: \(\frac{1}{2}\times \frac{5}{11}\) - step4: Multiply the fractions: \(\frac{5}{2\times 11}\) - step5: Multiply: \(\frac{5}{22}\) The calculation gives us: \[ u' = \frac{0.1}{0.44} \approx 0.22727c \] Thus, the observer in ship 1 sees ship 2 moving at approximately \( 0.227c \) towards the right.

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To find the velocity of ship 2 as observed from ship 1, we can use the relativistic velocity addition formula: \[ v' = \frac{u + v}{1 + \frac{uv}{c^2}} \] where: - \( v' \) is the velocity of ship 2 relative to ship 1, - \( u = 0.7c \) (velocity of ship 2 relative to earth, to the left, we'll take left as negative, so \( u = -0.7c \)), - \( v = 0.8c \) (velocity of ship 1 relative to earth, to the right). Plugging these values into the formula: \[ v' = \frac{0.8c + (-0.7c)}{1 + \frac{(0.8c)(-0.7c)}{c^2}} \] Calculating the numerator: \[ 0.8c - 0.7c = 0.1c \] Calculating the denominator: \[ 1 + \frac{(0.8)(-0.7)c^2}{c^2} = 1 - 0.56 = 0.44 \] Now plugging these into the equation: \[ v' = \frac{0.1c}{0.44} \approx 0.2273c \] Therefore, the observer in ship 1 sees ship 2 moving towards him at a speed of approximately \( 0.2273c \).

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