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Kinetic and potential energy (satellite low orbit)

  1. Dec 15, 2015 #1
    • Thread moved from the technical physics forums, so no HH Template is shown.
    Hello, I have this problem statement : "Due to air drag, the radius of a satellite’s circular orbit decreases from r to r - △r, where the positive quantity △r is much less than r. The mass of the satellite is m. Show that the increase in orbital speed is △v = +(△r/2)[(GM/r^3)^1/2]; that the change in kinetic energy is △K = + (GMm/2r^2); that the change in gravitational potential energy is △U = -2△K= - (GMm/r^2)△r; and that the amount of work done by the force of air drag is W = - (GMm/2r^2)△r."
    Well, I did the derivative for the velocity equation and I got v' = -(1/2)[(GM/r^3)^1/2], but another way to solve it could be using ""r - △r = (r - △r/r)r = (1-△r/r)r"", my problem is that I don't know how to apply that to the velocity equation.

    Doing the derivative of the velocity I get the given results in the problem statement for the kinetic and potential energy, but if instead of doing the derivative...which other method could I use to get the same results?

    △K = + (GMm/2r^2)
    △U = -2△K= - (GMm/r^2)△r
     
  2. jcsd
  3. Dec 15, 2015 #2

    mfb

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    I don't understand that approach, and the units do not match, so the formula cannot be right (probably just a typo however). What is the goal of that transformation?
     
  4. Dec 15, 2015 #3
    Instead of use r as the unknown radius, use r - △r because the statement given says:
    Due to air drag, the radius of a satellite’s circular orbit decreases from r to r - △r, where the positive quantity △r is much less than r.

    But maybe I'm wrong and the only way to demonstrate that the variation of velocity is:
    △v = +(△r/2)[(GM/r^3)^1/2]
    Is by using derivation.

    Ps: this problem is from the book of Sears Zemansky University Physics Vol. I
     
  5. Dec 15, 2015 #4

    mfb

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    You can calculate the velocity at radius ##r-\Delta r## and at radius ##r## independently, of course, but approximating this to first order is very similar to taking the derivative.
     
  6. Dec 15, 2015 #5
    Okay, thank you :)!
    Another question, for the kinetic and potential energy...should I derivate the equations too?
     
  7. Dec 15, 2015 #6

    mfb

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    Something like a derivative of some equation is probably useful. You don't need it, but it helps.
     
  8. Dec 15, 2015 #7
    I'm so grateful for your help, thank you!
     
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