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A spaceship makes the long trip from earth to the…

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Problem 13 Medium Difficulty

A rocket ship flies past the earth at 85.0$\%$ of the speed of
light. Inside, an astronaut who is undergoing a physical exami-
nation is having his height measured while he is lying down
parallel to the direction the rocket ship is moving. (a) If his
height is measured to be 2.00 $\mathrm{m}$ by his doctor inside the ship,
what height would a person watching this from earth measure
for his height? (b) If the earth-based person had measured
$2.00 \mathrm{m},$ what would the doctor in the spaceship have measured
for the astronaut's height? Is this a reasonable height? (c) Sup-
pose the astronaut in part (a) gets up after the examination and
stands with his body perpendicular to the direction of motion.
What would the doctor in the rocket and the observer on earth
measure for his height now?

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Video Transcript

okay. And this problem, we have a rocket ship traveling with a speech off 85 percent off this beautiful light. What's put in here? 80 five percent. This speed of light. This is just zero point 85 c. Okay, so we have here, um, spaceship. And here we have, uh, astronaut that lay down in the ground in the in parallel to the motion access off the flight. If we measured that, he's proper length. Let's put al zero is two meters. What should be the length measure by a person on the earth watching this astronaut laid out in this space shift. So let's calculate this in here. We know that since the astronaut is under relativistic effect, the the length that a person on the Earth should measure needs to be the lens off the astronaut contract by gamma. This is equals two at zero. That's pretending. Here, Al zero. Describe it off. One minus. You. Square C square. Okay, so let's calculate this. Here we have l equals two and multiplies this grand route off. One miners, zero point 85 square. Business to be equal. Two, 21 point. This is equal. That's put a red that this is one point 05 meters. That means that if the US around lake house in the Arlo to them motion access off the flight of person will see his length contract and ah, height off 1.5 meters. Okay, so let's make the opposite. Now. This is the first item. The second item. We need to calculate the Rio the proper length off the astronauts, knowing that the same situation the lens off the astronaut, measured by a personal dear should be two meters. So he's going to be the same equation as this one. But this time we want to calculate l zero. So we have in here. Zero equals gamma l. This is a close to El divided by square roots of one minus. You're square C square, all this in the square root. So let's calculate this. We have in here two divided by one minus little point 85 square in this crab route. So after calculating this, we're gonna get height off three point eight meters and this is not a reasonable height for a human. So the answer to the second item is No. This is not a reasonable height. And in the final item of this problem we have the same space shift. But this time the astronaut he's not lay down here is up, and we have to calculate his height in both their reference frame. Both the earth and space shift well, the lens off object moving with respect to the observer we know is shortening in the direction of motion. So since the astronaut is not in the direction of motion, it is no length contraction in the direction perpendicular to the emotion. So both jobs servers. We'll measure this same height for the astronaut because he's not in the barrel out of the direction of motion, his perpendicular to the direction of motion. So this is the final answer.

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Problem

A spaceship makes the long trip from earth to the…

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