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A rock, the earth, an ellipse and the joys of angular momentum

  1. Oct 31, 2004 #1
    I understand that this is a bit cheeky but I've found a problem on one of my mechanics problem sheets that is giving me a headache. As much as I could just ignore it I'd rather try and gain some more understanding of the vague world of rotation and orbit

    Even at this early stage I'm bewildered and my searchings and own ponderings still leave me confused. If anyone can quite highlight how L can progress anywhere beyond the obvious r x v (where x is the cross product, and v is the tangental velocity of the rock at it's nearest point to Earth) I'd be forever in their debt

    For fun, here is the rest which flows from this first section

    I'm not asking people to do the question for me (although feel free ;) ), any general direction would be pretty damn useful

    Thanks in Advance
  2. jcsd
  3. Oct 31, 2004 #2
    To answer your first part, we have

    angular momentum L = mvr

    energy E = 1/2(mv^2) - GMm/r

    rearrange energy equation until you have v = something....

    substitute that into L = mvr for v

    multiply m into the squareroot and thats your answer.
  4. Oct 31, 2004 #3


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    Start with the Vis Viva equation.

    V^2 = GM (2/R - 1/a)

    The kinetic energy is

    K = (1/2) m V^2

    K = (GMm/2)(2/R - 1/a)

    The gravitational potential energy is

    U = -GMm/R

    The total energy is

    E = K + U

    E = (GMm/2)(2/R - 1/a) - GMm/R

    E = GMm { (1/2)(2/R - 1/a) - 1/R }

    E = -GMm/(2a)

    Notice that the expression for the total orbital energy, just above, is the answer to Part 2 of your homework problem.

    Solve for the semimajor axis.

    a = -GMm/(2E)

    The angular momentum L is

    L = R x mV

    The magnitude of which is

    L = RmV sin Q

    Where Q is the angle from R to V. Since at periapsis this angle is pi/2 radians, the angular momentum in your homework problem reduces to

    L = RmV

    Refer again to the Vis Viva equation.

    L = Rm sqrt{ GM (2/R - 1/a) }

    Substitute the expression for (a) found above.

    L = Rm sqrt{ GM (2/R + 2E/GMm) }

    Rearrange terms.

    L = Rm sqrt{ 2EGM/GMm + 2GM/R }


    L = Rm sqrt{ 2E/m + 2GM/R }

    Move the orbiting object's mass inside radical.

    L = R sqrt{ 2Em + 2GMm^2/R }

    Factor out the m.

    L = R sqrt{ 2m (E + Gm/R) }

    It looks to me as if Part 3 and Part 4 are really one question, and it is a trick question. An orbit with total energy equal to zero is, by definition, barely unbound having exactly the escape speed for its geocentric distance.

    Jerry Abbott
    Last edited: Oct 31, 2004
  5. Nov 3, 2004 #4
    Ah, ta muchly guys. That clears everything up
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