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Elliptical orbits problem

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data

    It is required to put a satellite into an orbit with apogee of 5R/2, where R is the radius of the planet. The satellite is to be launched from the surface with a speed vo at 300 to the local vertical. If M is the mass of the planet show that v0=5GM/R (use conservation of energy and angular momentum). Assume that the planet is not rotating and that effects due to the planetary atmosphere can be ignored.

    2. Relevant equations

    E= 1/2mv2 +J2/(2mr2)- GMm/r

    J=mvr

    l/r= 1+ecos(x)

    3. The attempt at a solution

    i have attempted the question more than once, the following working is the attempt which i thought was closest to the answer.

    rearranging the first equation gives

    2E/m+2GM/r -J2/(m2r2)=0

    at rmax, r=5R/2, dr/dt=0, x=pi

    therefore
    rmax= -GMm/2E [ 1+ sqrt(1+2EL2/G2m3M2)]

    i dont know where to go from here.

    2nd attempt:

    v=v0sin30 (using launch angle)

    therefore E= mv02/4- GMm/r

    E= -2.integral[Fdr] + 2GMm/r

    therefore
    2GM/r= v02/4- GM/r

    v02= 12GM/r and r=5R/2

    therefore
    v02= 24GM/5R
     
  2. jcsd
  3. Apr 20, 2009 #2

    Cyosis

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    Homework Helper

    I don't quite follow what you're doing exactly, especially your first attempt at a solution. As for the second attempt. The kinetic energy of an object is related to its velocity not a component of its velocity. That said are you sure you listed the value we should get for v0 correctly?. First of all the units don't match, although this is fixed by squaring it. Secondly I get a factor 5/4 instead of 5.

    Either way you should make good use of the hint.

    Both energy and angular momentum is conserved.
    So E(launch)=E(apogee). What is the total energy at the launch, potential and kinetic?
    What is the total energy when the satellite reaches apogee?

    Same questions for angular momentum, however keep in mind that you have to take the angle into account here. What is the angle between r and p at apogee?
     
  4. Apr 21, 2009 #3
    this is correct, sorry i typed it wrong the question was supposed to say
    v02= 5GM/4R.

    i've attempted the question as you suggested but my answer is a factor of 2/3 outs maybe you could spot where ive gone wrong as i cant see any mistakes.

    E(launch)= 1/2 mv02+ 3mMG/5R

    E(apogee)= J2/2mr2 - GMm/r

    J(launch)=J(apogee)= mvorsin30= mv0r/2

    therefore E(apogee)= mv02/8 -2GMm/5R

    equating the two values for E gives

    3/8v02= GM/5R

    which gives v02=8Gm/15R.
     
  5. Apr 21, 2009 #4

    Cyosis

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    The first mistake is:

    E(launch)= 1/2 mv0^2+ 3mMG/5R

    During launch the distance between the satellite and the center of the planet is not (5/3)R, but is...?

    The second mistake comes into life from being a bit sloppy with notation.

    E(apogee)= J^2/2mr^2 - GMm/r

    This expression is correct, but you are given r at apogee so enter it in the equation right here this will prevent you from making the mistake you're about to make.

    J(launch)=J(apogee)= mvorsin30= mv0r/2

    Same problem here. You know the r for J(launch) so you can express J(apogee) in terms of that.

    The mistake that follows from this is going to this step:

    E(apogee)= mv0^2/8 -2GMm/5R

    You say that J^2/2mr^2=mv0^2/8, but this is not true. You used J^2=(mv0r/2)^2/(2mr^2). This results in the radii cancelling, but these radii are not the same!

    What you should have is: J^2=(mv0 r(launch)/2)^2/(2mr(apogee)^2). Would you have entered the appropriate radii in E(apogee) and J(launch) from the start you wouldn't have made this mistake.

    I hope it's clear now, if not just ask again.
     
    Last edited: Apr 21, 2009
  6. Apr 21, 2009 #5
    Thank you, that was really helpful.
     
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