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SPH4U1 Lesson 02 1. A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a speed \( v=0.80 \mathrm{c} \) relative to it. Assume that the three planets do not move relative to each other during the trip. The distance between Mars and Venus is \( 1.20 \times 10^{11} \mathrm{~m} \), as measured by a person on earth. (a) What does the Martian measure for the distance between Mars and Venus? (b) What is the time of the trip (in seconds) as measured by the Martian?

Ask by Washington Peterson. in Canada
Mar 09,2025

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**(a)** The Martian measures the distance between Mars and Venus as \( 7.20 \times 10^{10} \) meters. **(b)** The time of the trip, as measured by the Martian, is 300 seconds.

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To solve this problem, we’ll employ the concepts of special relativity, particularly length contraction and time dilation. (a) The Martian measures the distance between Mars and Venus using the length contraction formula: \[ L' = L \sqrt{1 - \frac{v^2}{c^2}} \] Where: - \( L \) is the distance measured by someone on Earth (\( 1.20 \times 10^{11} \, \text{m} \)), - \( v \) is the velocity of the spaceship (\( 0.80 \, c \)), - \( c \) is the speed of light. First, let's calculate \( \frac{v^2}{c^2} \): \[ \frac{v^2}{c^2} = (0.80)^2 = 0.64 \] Now we substitute this into the length contraction formula: \[ L' = 1.20 \times 10^{11} \sqrt{1 - 0.64} = 1.20 \times 10^{11} \sqrt{0.36} = 1.20 \times 10^{11} \times 0.6 = 7.2 \times 10^{10} \, \text{m} \] Thus, the Martian measures the distance between Mars and Venus as \( 7.2 \times 10^{10} \, \text{m} \). (b) To find the time of the trip as measured by the Martian, we can use the time dilation formula, but first, we need to find the proper time as perceived from Earth: \[ t_{Earth} = \frac{L}{v} = \frac{1.20 \times 10^{11} \, \text{m}}{0.80 \times c} \] Using \( c \approx 3.0 \times 10^8 \, \text{m/s} \): \[ t_{Earth} = \frac{1.20 \times 10^{11}}{0.80 \times 3.0 \times 10^8} = \frac{1.20 \times 10^{11}}{2.4 \times 10^8} \approx 500 \, \text{s} \] Now, we must calculate the time experienced by the Martian: \[ t' = t_{Earth} \sqrt{1 - \frac{v^2}{c^2}} = 500 \, \text{s} \times 0.6 = 300 \, \text{s} \] Therefore, the time of the trip as measured by the Martian is \( 300 \, \text{s} \).

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