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Space shuttle releases a sattelite into a circular orbit

  1. Oct 12, 2007 #1
    The space shuttle releases a sattelite into a circular orbit 680 km above the Earth. How fast must the shuttle be moving (relative to the Earth's center) when the release occurs?


    I know that G (m*m(earth)/r^2) = m (v^2/r)...with r = r(earth) + h. In this equation, r=6380+680=7060 km. There also a manipulation of the given equations in the examples for velocity, sqr(G*m(earth)/r))

    I am probably thinking along the wrong lines, but shouldn't the shuttle be moving at the same speed needed to keep the satellite in orbit?

    So that is my approach to the problem thus far - trying to find the velocity of the satellite and equating it to the velocity of the shuttle upon release. To do that, I've used the sqr(G*m(earth)/r)) equation, which gives me:

    sqr((6.67*10^-11)(5.98*10^24)/(7060))=237690 m/s. This isn't even close to the given answer in the book, 7520 m/s.
     
  2. jcsd
  3. Oct 12, 2007 #2


    What should the units be for R...I am getting an answer very close to what your book says.

    Casey
     
    Last edited: Oct 12, 2007
  4. Oct 12, 2007 #3
    Ah yes, it should be m instead of km. Often times I get caught up in the problem and lose focus in critical things such as units. Thanks for your help.
     
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