HW 9 - Evan Klein

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University of Michigan *

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102

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Astronomy

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Jan 9, 2024

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1. Long Trips at Constant Acceleration: Earth Time. Suppose you stay on Earth and watch a spaceship leave on a long trip at a constant acceleration 1g (Note: little , 𝑔 is Earth’s gravitational acceleration . not big , universal( 𝑔 = 9. 8 𝑚 / ? 2) 𝐺 gravitational constant). a. At an acceleration of 1g, approximately how long will it take before you see the spaceship traveling away from Earth at half the speed of light? Explain. A = 9.8 m/s^2 9.8*(seconds) = 1.5*10^8 Seconds = 1.5*10^7 → 250000 Minutes → 4166 Hours → 173.6 Days b. Describe how you will see its speed change as it continues to accelerate. Will it keep gaining speed at a rate of 9. 8 𝑚 / ? 2 each second? Why or why not? As the spaceship continues to accelerate, its speed relative to Earth will keep increasing for some time. As the spaceship approaches the speed of light, the added speed for each additional second drops off. This occurs due to time dilation - from the perspective of a stationary observer, time on the accelerating spaceship appears to slow down c. Suppose the ship travels to a star 500 light-years away. From your perspective on Earth, approximately how long will this trip take? Explain. If the spaceship takes about 174 days to reach half the speed of light, then it would take 348 days to reach the full speed of light. For a star that is 500 light years away, the trip would then take just over 500 years because after 348 days the ship is traveling at the speed of light.
2. Long Trips at Constant Acceleration: Spaceship Time. Consider again the spaceship from the previous problem on a long trip with a constant acceleration of 1g. As long as the ship is gone from Earth for many years, the amount of time that passes on the spaceship during the trip turns out to be approximately: ??ℎ𝑖𝑝 = 2 𝑐 / 𝑔 ln ( 𝑔𝐷 / 𝑐 ^2) Use the formula to determine how much time will pass on the ship during its trip to a star that is 500 light-years away. Compare this to the amount of time that will pass on Earth. 2 𝑐 / 𝑔 ln ( 𝑔𝐷 / 𝑐 ^2) = 2 ln(d) where d is in light years 2 ln (500) = 12.4 years This is a fraction of the time that will pass on Earth which we have determined to be about 500 years. 500/12.4 = 40.3. The trip will appear about 40 times longer for people on Earth. Trip to the Center of the Galaxy. Use the same scenario as in Problem 3, but this time suppose the ship travels to the center of the Milky Way Galaxy, about 27,000 light years away. a. How much time will pass on the ship? Time (ship) = 2ln (27000) Time (ship) = 20.4 years b. Compare this to the amount of time that will pass on Earth. 27,000/20.4 = 1323.5. The trip will take 1323.5 times as long from the perspective of Earth than from the spaceship. 4. Trip to Another Galaxy. The Andromeda Galaxy is about 2.5 million light-years away. Suppose you had a spaceship that could constantly accelerate at 1g. a. Could you go to the Andromeda Galaxy and come back within your Lifetime? Time (ship) = 2ln(2.5*10^6) Time (ship) = 29.5 years b. Explain.
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