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Angular momentum of a satellite

  1. May 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A satellite is in a circular orbit of radius R from the planet's center of mass around a planet of mass M.

    The angular momentum of the satellite in its orbit is:
    I. directly proportional to R.
    II. directly proportional to the square root of R
    III. directly proportional to the square root of M.

    The correct answer apparently is II and III.
    2. Relevant equations


    3. The attempt at a solution
    OK, I thought that since the satellite's angular momentum is mvR * sin(90), then its angular momentum would be directly proportional to its mass, velocity, and/or even the radius orbit.

    I was thus the only thing that made sense.

    Why is the correct answer II and III? Thanks so much in advance for the help!
     
  2. jcsd
  3. May 8, 2016 #2
    angular momentum is moment of momentum , in a way you may have correctly thought of the dependence ;
    but is there a dependence of v on R?
    may be v is proportional to R^ -1/2 ! then the ang. momentum can be said to be dependent on R^1/2
     
    Last edited: May 8, 2016
  4. May 8, 2016 #3

    PeroK

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    This is a slightly nasty question because, in general, angular momentum is directly proportional to ##r## and ##v##. So, if the question were:

    An object is moving past a planet at a distance ##R##, its angular momentum is ...? Then "proportional to R" would be correct.

    But, in a circular orbit, ##v## depends on ##R## and the mass of the planet and the gravitational constant. So, it was necessary to express angular momentum in terms of all the known variables first.

    The question is a bit tricky in my opinion. Slightly disengenuous, perhaps.
     
  5. May 8, 2016 #4
    If I have G * mass planet * mass sattelite / r^2 = mass sat. * v^2 /r

    Then I have v = sqrt( G * mass planet/ r).

    So the velocity is not directly proportional to the square root of r, but inversely proportional! Right?
     
  6. May 8, 2016 #5

    haruspex

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    Yes.
     
  7. May 9, 2016 #6
    your angular momentum is equal to m.v. r and if you say v to be proportional to 1/sqrt(r) ,

    then you get the ang. momentum as directly proportional to sqrt(r) ;
    i think that is what you need .
    regarding dependence on mass - if you replace v in terms of mass and r which are under sqrt sign ---one can say that angular momentum is proportional to sqrt(mass) also

    -

    so at least it corroborates the answer
     
  8. May 9, 2016 #7
    No, I think it would be inversely proportional???
     
  9. May 9, 2016 #8
    so , again i repeat for your consideration:
    L = m.v.r ,
    v = constant / sqrt(r) ; substitute in L (ang. momentum)
    L= m. constant / sqrt(r) . r = m . constant. sqrt(r) so L is proportional to sqrt(r)
    as r = sqrt(r). sqrt(r)
     
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