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Two spaceships \( A \) and B are exploring a new planet. Relative to this planet, spaceship A has a speed of 0.60 c , and spaceship B has a speed of 0.80 c . What is the ratio \( \mathrm{D} / \mathrm{Da}_{3} \) of the values for the planet's diameter that each spaceship measures in a direction that is parallel to its motion?

Ask by Goodwin Martinez. in Canada
Mar 09,2025

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The ratio of the planet's diameter measured by spaceship A to that measured by spaceship B is 4:3.

Solución

1. The proper diameter of the planet (i.e., in its rest frame) is \( D \). 2. Due to length contraction, an observer moving relative to the planet will measure the diameter (parallel to the direction of motion) as \[ D_{\text{measured}} = D \sqrt{1 - \frac{v^2}{c^2}}. \] 3. For spaceship \( A \) with speed \( 0.60\, c \), the measured diameter is \[ D_A = D \sqrt{1 - (0.60)^2} = D \sqrt{1 - 0.36} = D \sqrt{0.64} = 0.8\, D. \] 4. For spaceship \( B \) with speed \( 0.80\, c \), the measured diameter is \[ D_B = D \sqrt{1 - (0.80)^2} = D \sqrt{1 - 0.64} = D \sqrt{0.36} = 0.6\, D. \] 5. The ratio of the two measured diameters is \[ \frac{D_A}{D_B} = \frac{0.8\, D}{0.6\, D} = \frac{0.8}{0.6} = \frac{4}{3}. \] Thus, the ratio \(\frac{D}{\mathrm{Da}_3}\) of the planet's diameter that spaceship \( A \) measures to that which spaceship \( B \) measures is \(\frac{4}{3}\).

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To find the ratio \( D / D_A \) of the planet's diameter measured by the two spaceships \( A \) and \( B \) in a direction parallel to their motion, we will use the concept of length contraction from the theory of special relativity. The length contraction formula states that the length measured in a moving frame (denoted as \( L' \)) is related to the proper length (denoted as \( L \)) by: \[ L' = L \sqrt{1 - \frac{v^2}{c^2}} \] where \( v \) is the speed of the moving observer and \( c \) is the speed of light. 1. For spaceship \( A \) with speed \( v_A = 0.60c \): - The contracted diameter of the planet as observed by \( A \) is given by the formula: \[ D_A = D \sqrt{1 - \left(0.60\right)^2} \] \[ D_A = D \sqrt{1 - 0.36} = D \sqrt{0.64} = D \cdot 0.8 \] 2. For spaceship \( B \) with speed \( v_B = 0.80c \): - The contracted diameter of the planet as observed by \( B \) is given by the formula: \[ D_B = D \sqrt{1 - \left(0.80\right)^2} \] \[ D_B = D \sqrt{1 - 0.64} = D \sqrt{0.36} = D \cdot 0.6 \] 3. Now, we need to find the ratio \( \frac{D}{D_A} \) and \( \frac{D}{D_B} \): - From \( D_A = D \cdot 0.8 \): \[ \frac{D}{D_A} = \frac{D}{D \cdot 0.8} = \frac{1}{0.8} = 1.25 \] - From \( D_B = D \cdot 0.6 \): \[ \frac{D}{D_B} = \frac{D}{D \cdot 0.6} = \frac{1}{0.6} \approx 1.6667 \] 4. Finally, we are looking for the ratio \( \frac{D_A}{D_B} \): \[ \frac{D_A}{D_B} = \frac{0.8}{0.6} = \frac{4}{3} \approx 1.3333 \] Thus, the ratio of the values for the planet’s diameter that each spaceship measures in a direction that is parallel to its motion is \( \frac{D_A}{D_B} = \frac{4}{3} \).

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