If you enjoyed Rhett Allain’s post last Friday as much as we did (see “Using astronaut Mike Fossum’s YouTube video to measure ATV acceleration“), then – like the rest of us here at the ATV blog – you’re probably a bit of a space physics gearhead!
Therefore, you might enjoy reading (and answering!) a series of ‘Reboost Homework’ challenge questions that Rhett’s posted over at Wired’s dotphysics blog.
My favourite challenge question?
Assume the ISS has an acceleration of 0.03 m/s2 (this is the acceleration just due to the ATV reboost, not the gravitational acceleration). If an astronaut started in the Destiny module and let go, how long would it take to “fall” [as the astros are seen doing in the YT video] all the way to the back. You might want to look up this distance, but I think it is around 50 meters.
Interested? We’d like to invite all our ATV blog readers to answer any one (or all!) of Dr Allain’s ‘ATV Homework Questions’. Send us your answers, and we’ll publish the best (or, at least, correct) ones as assessed by the ATV Operations team. The answer judged most imaginative will win a super ATV prize from our mission goody closet. Submit your own work only, please. All judging final. No cash value. One submission per entrant. Due date Friday, 3 May, 12:00 CEST.
Astronaut selection requires three fundamental tenets: health, brains, and experience. You have to be able to pass the toughest medical in the world, so stay in shape and eat right. You have to demonstrate the ability to learn complex things, so an advanced technical degree is needed. And you have to demonstrate good decision-making when the consequences really matter.
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Discussion: 5 comments
This sent in today by Ante Medić, from Croatia. Thanks!
Well there is simple formula (x=v(0)t+1/2 *at^2), where x is our change in distance [or 50 m – the Station length, approximately] and since we can assume the astronaut starts with zero velocity (at least relative to ISS), then v(0) has zero value and then t [t stands for time] is also zero.
Then all that’s left is the acceleration of 3 cm/s^2 and when we toss all this into the equation we get a quadratic equation with two solutions: -57.7 and +57.7 seconds.
Of course, time can’t be negative, so +57.7 or almost one minute is the correct answer.. 🙂
From this we can also calculate that velocity of astronaut at the end will be 1.73 m/s or roughly 6 km/h
😀
Ante,
Yes – that is correct!
Rhett (via mail)
The time can be negative. In this case the astronaut would have to start with speed 1.73m/s and after 57.7 seconds you will find him at the other end with zero speed. Then he will start moving in the opposite direction and after another 57.7 seconds he would return to where he started, with speed 1.73m/s
Neglecting the air friction of course.
I know it can be but that was not the question or problem 🙂
Any way sorry for wrong interpretation I should not mention it! 🙂