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Putting a satellite into orbit

  1. Feb 9, 2014 #1
    1. The problem statement, all variables and given/known data
    It is required to put a satellite of mass m into orbit with apogee 2.5 times the radius of the planet of mass M. The satellite is to be launched from the surface with speed v0 at an angle of 30° to the local vertical.

    Use conservation of energy and angular momentum to show that

    [itex]v_0^2 = \frac{5GM}{4R}[/itex]

    assuming that the planet is spherical, not rotating and atmospheric effects can be ignored.

    2. Relevant equations

    [itex]L = mr^2\dot{\theta}[/itex]

    [itex]V = -\frac{GMm}{r}[/itex]

    where r is the distance from the centre of the planet to the satellite's position at time t.

    3. The attempt at a solution
    At t = 0, the angular momentum is L = mRv0sin(30) = (1/2)mRv0. Therefore:

    [itex]\dot{\theta} = \frac{L}{mr^2} = \frac{Rv_0}{2r^2}[/itex]

    Total energy at time t is:

    [itex]E = \frac{1}{2}mv^2 - \frac{GMm}{r}[/itex]

    [itex]E = \frac{1}{2}mr^2\dot{\theta}^2 - \frac{GMm}{r}[/itex]

    [itex]E = \frac{mR^2v_0^2}{8r^2} -\frac{GMm}{r}[/itex]

    I don't really know what to do with these equations. At the apogee, dr/dt = 0, so I could differentiate the E equation, but then I just get 0 = 0. Also I notice that E should be the same for all r as well as t, but clearly E(r) is not constant, so is the energy equation wrong?

    Any help would be greatly appreciated.
  2. jcsd
  3. Feb 9, 2014 #2
    The energy equation is wrong, because you neglected the radial velocity.
  4. Feb 9, 2014 #3
    Well, don't I feel silly now.

    So including the radial velocity gives a term with dr/dt. I set the energy equation equal to the initial energy at t=0 {i.e. (1/2)mv02 - GMm/R}. Then at the apogee, r=2.5R and dr/dt = 0, solve for v_0 which gives the required result.

    Thank you for pointing out my silly error.
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