SPHAU1 Lesson 02 Modern Phys A Martian leaves Mars in a spaceshipt that is heading to Venus. On the way, the spacest passes earth with a speed \( v=0.80 \mathrm{c} \) relative to it. Assume that the three planets do not move relative to each pther during the trip. The distance between Mars and Venus is \( 1.20 \times 10^{12} \mathrm{~m} \), as neasured by a person on earth. reference (a) What does the Nratian-measure for the distance between Mars and Venus? Detern (b) What is the time of the trip (in seconds) as measured by the Martian?

To solve this problem, we will use the concepts of special relativity, specifically time dilation and length contraction. ### (a) Distance measured by the Martian The length contraction formula is given by: \[ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} \] Where: - \( L \) is the contracted length (as measured by the Martian) - \( L_0 \) is the proper length (the distance measured by someone at rest relative to both ends of the measurement, which in this case is \( 1.20 \times 10^{12} \, \mathrm{m} \)) - \( v \) is the speed of the spaceship, \( 0.80 \, c \) - \( c \) is the speed of light First, we calculate \( \frac{v^2}{c^2} \): \[ \frac{(0.80c)^2}{c^2} = 0.64 \] Now we substitute \( L_0 \): \[ L = 1.20 \times 10^{12} \sqrt{1 - 0.64} = 1.20 \times 10^{12} \sqrt{0.36} \] \[ = 1.20 \times 10^{12} \times 0.6 = 0.72 \times 10^{12} = 7.2 \times 10^{11} \, \mathrm{m} \] So the distance between Mars and Venus as measured by the Martian is \( 7.2 \times 10^{11} \, \mathrm{m} \). ### (b) Time of the trip as measured by the Martian To find the time of the trip as measured by the Martian, we can use the formula: \[ t' = \frac{L}{v} \] Where: - \( L \) is the contracted length found in part (a), \( 7.2 \times 10^{11} \, \mathrm{m} \) - \( v \) is the speed of the spaceship \( 0.80 \, c = 0.80 \times 3.00 \times 10^8 \, \mathrm{m/s} = 2.40 \times 10^8 \, \mathrm{m/s} \) Substituting the values: \[ t' = \frac{7.2 \times 10^{11}}{2.40 \times 10^8} \approx 3.00 \times 10^3 \, \mathrm{s} \] So the time of the trip as measured by the Martian is approximately \( 3000 \, \mathrm{s} \), or \( 50 \, \mathrm{minutes} \).