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Time for astronaut to reach the satellite

  1. Feb 19, 2014 #1
    So I found this problem in my 10th grade Physics workbook and since its test time tomorrow I must be able to work everything out. The problem I'm having here is just the calculations which give me a time short of one hour from the answer on the book. Anysways, please have a go:
    1. The problem statement, all variables and given/known data
    An astronaut is flying on a shuttle that's on the same orbit(400 km above the surface of Earth) as a satellite that he needs to repair. He is 50 km behind the satellite and in order to reach it, the astronaut lowers his orbiting radius by 2 km. What is the time needed for the astronaut to reach the satellite?
    It doesn't say anything about it being an aerial distance, so it must be taken to be circular.

    2. Relevant equations
    The equation for the Law of Universal Attraction must be used and also that of the centripetal force and acceleration in circular motion:
    Fc=mv^2/r and FG=(Gm1m2)/r^2

    3. The attempt at a solution
    I tried going around this problem calculating the relative linear speed of the astronaut to the satellite and then the time of approach would just be the ratio of the distance the astronaut had to travel relative to the satellite(50km) with the astronaut's relative linear velocity. I know I'm missing something here because my calculations(which in this problem are loathsome to be honest) come out ≈ 1.5 hours less than the answer on the book, which wouldn't surprise me if it was wrong because this book has plenty of mistakes in it.
    Formula and answer are found in the attachment.

    Attached Files:

    • t.docx
      File size:
      13.3 KB
    Last edited: Feb 19, 2014
  2. jcsd
  3. Feb 19, 2014 #2
    It does not look good.
    The expressions under square roots are accelerations (F/m). Taking the square root does not produce a velocity as you should have in order to get t from that formula.
  4. Feb 19, 2014 #3
    Oh I am so sorry the formula for t had a writing mistake in it. I had wrote the R^2 in the denominator inside the square roots mistakenly. Now I reuploaded the attachment to amend it. I had calculated for R not R^2 though, so it's all fine in that aspect.
  5. Feb 19, 2014 #4


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    Using your numbers in your formula I get 12.3 hours, which is close the book's answer. Not sure how you get 10.7 hours.
  6. Feb 19, 2014 #5
    OMG THANK YOU SO MUCH! For me, a physical problem ends when there is found a final formula that has everything known on the right side and the unknown on the left so as far as that goes I seem to be right. Getting a nice and clean formula in the end is the most exhilarating thing for me! The rest I think is computer-plug-and-chug. Thank you again for your time.
  7. Feb 19, 2014 #6


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    OK, Good work!
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