Problem

$\bullet$ You measure the length of a futuristic …

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Problem 9 Easy Difficulty

A spacecraft flies away from the earth with a speed of
$4.80 \times 10^{6} \mathrm{m} / \mathrm{s}$ relative to the earth and then returns at the
same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting
point 365 days $(1$ year) later, as measured by the clock that
remained on earth. What is the difference in the elapsed times
on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the smaller elapsed time?

Answer

clockin the spacecraft

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Physics 101 Mechanics

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Video Transcript

Okay, So in this problem, we have a spacecraft flying away from the earth with a speed that sport in here you equals four point eight times. Tend to the poorer six meters per second. This is speed this measure relative to the reference frame of the earth. Also in the reference frame of the earth, we have elapsed time Delta t off one year. We know that this ship come back after our year off. A time measure owned the year. Okay, so the first item off this problem, we have to answer What is the difference in time measured by a person of the year And what time off a person on this space shift spacecraft. And we need to answer this in hours. So first of all, let's put the time of the person off the Earth measure in hours. So we know that one years as 300 and 65 days and each day's each, they have 24 hours, So this is going to be equal to 8000 760 hours. Okay, so this is the time a person on the Earth measure now, what is the time that the person owned a spacecraft measure. We know that the time of the person on the spacecraft is dilated in relation. So the person on the earth So we can say that the person on the earth it's equal. D'oh! Time off The person of the spacecraft dilated off gamma. So since we want to calculate Delta T s, we can see that doubt t s is equal doubt the tea divided by gamma. And we know that this needs to be called out t square root off one miners years crime divided by C Square. And we just need to calculate this now and we know what is the speed of the light. Immediate per second is beautiful I to just three times 10 to the power off eight meters per second. So let's calculate first, the time of the person of the spacecraft is going to be 8000 moments. The time of the burst on the spacecraft is going to be 1000 and seven, 760. That multiplies the square root off one miners, four point eight time stand to the power off! Six All this square divided by three times 10 to the power off eight. All this is quiet also in the square root. So after calculating all this, we're gonna have a time off eight 1000 in 700 and 58 point 88 hours. So the difference in the laps of times, No doubt, de miners Delta T s is just one point 12 hours, and that's the first answer off this problem. The second question is just which one of the times, Which one of the clocks and measures the smaller time. Wow, we already calculate. This is just second item I can be. The time is that the clock is the clock off the person on the spacecraft, and that's all Thanks for watching.

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