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 $$ 1.20 imes 10^{12} \mathrm{~m} $$, 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?

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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 $$ 1.20 	imes 10^{12} \mathrm{~m} $$, 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?

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1. Determine the Lorentz factor:
   
   $$
   \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
   $$
   
   Given $$ v = 0.80\,c $$, we have:
   
   $$
   \gamma = \frac{1}{\sqrt{1-(0.80)^2}} = \frac{1}{\sqrt{1-0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} \approx 1.67
   $$
   
2. (a) Find the Martian’s measured distance (length contraction):

The proper distance measured from Earth is $$ L_0 = 1.20 \times 10^{12}\, \text{m} $$. In the moving frame (Martian’s frame), the distance is contracted:
   
   $$
   L = \frac{L_0}{\gamma} = \frac{1.20 \times 10^{12}\, \text{m}}{1.67} \approx 7.20 \times 10^{11}\, \text{m}
   $$
   
3. (b) Calculate the time of the trip as measured by the Martian:
   
   In the Martian’s frame, the trip time is given by the contracted distance divided by the speed:
   
   $$
   t' = \frac{L}{v} = \frac{7.20 \times 10^{11}\, \text{m}}{0.80\,c}
   $$
   
   Using $$ c = 3.0 \times 10^8\, \text{m/s} $$:
   
   $$
   t' = \frac{7.20 \times 10^{11}\, \text{m}}{0.80 \times 3.0 \times 10^8\, \text{m/s}} = \frac{7.20 \times 10^{11}}{2.4 \times 10^8} \approx 3000\, \text{s}
   $$
   
Thus, the Martian measures the distance between Mars and Venus as $$ 7.20 \times 10^{11}\, \text{m} $$ and the trip takes approximately $$ 3000\, \text{s} $$.
The Martian measures the distance between Mars and Venus as approximately $$ 7.20 \times 10^{11}$$ meters and the trip takes about 3000 seconds.

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