M4U1 Modern Physics A Martian leaves Mars in 5 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.

**Step 1. Calculate the time elapsed in the Earth frame** The distance between Mars and Venus as measured by an observer on Earth is \[ d = 1.20 \times 10^{11} \, \text{m}. \] The spaceship passes Earth with a speed \[ v = 0.80\,c, \] where \( c = 3.00 \times 10^8 \, \text{m/s} \). The time elapsed in the Earth frame is given by \[ t = \frac{d}{v} = \frac{1.20 \times 10^{11} \, \text{m}}{0.80 \, c}. \] Substitute the value of \( c \): \[ t = \frac{1.20 \times 10^{11} \, \text{m}}{0.80 \times 3.00 \times 10^8 \, \text{m/s}}. \] Now calculate the denominator: \[ 0.80 \times 3.00 \times 10^8 \, \text{m/s} = 2.40 \times 10^8 \, \text{m/s}. \] Thus, \[ t = \frac{1.20 \times 10^{11}}{2.40 \times 10^8} \, \text{s} = 500 \, \text{s}. \] --- **Step 2. Calculate the proper time elapsed in the spaceship (Martian) frame** Due to time dilation, the proper time \( \tau \) (time elapsed in the spaceship’s own frame) is related to the Earth frame time \( t \) by \[ \tau = \frac{t}{\gamma}, \] where the Lorentz factor \( \gamma \) is given by \[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}. \] Since \( v = 0.80\,c \), we have \[ \gamma = \frac{1}{\sqrt{1 - (0.80)^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} \approx 1.67. \] Now, plugging in the values: \[ \tau = \frac{500 \, \text{s}}{1.67} \approx 300 \, \text{s}. \] --- **Summary** - Time elapsed in the Earth frame: \[ t \approx 500 \, \text{s}. \] - Proper time elapsed in the spaceship (Martian) frame: \[ \tau \approx 300 \, \text{s}. \]