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[spaceflight mechanics] given r and e vector, calculate inclination?

  1. Mar 12, 2009 #1
    this problem deal with classical orbit parameters. a bold letter indicates a vector. ex:r is a vector while r is a scalar.

    1. The problem statement, all variables and given/known data
    given that, relative to the geocentric equatorial frame,
    r=-6634.2i-1261.8j-5230.9k[km]
    the eccentricity vector is
    e=-0.40907i-0/48751j-0.63640k
    and the satellite is flying towards perigee, calculate the inclination of the orbit.

    2. Relevant equations
    eq1: e=1/u*(vXh-r/r
    eq2: h=rXv

    formular for inclination is i=arccos(h.k/h)

    3. The attempt at a solution
    i managed to find vXh using eq1,
    since r.(vXh=h.(rXv=h.h=h^2
    so i find h, but idk how can i find the vector h. anyone have a better approach?
     
  2. jcsd
  3. Mar 13, 2009 #2

    D H

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    Staff Emeritus
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    You don't need to find h per se. You just need to find some vector that is parallel or anti-parallel to h. You have two vectors here; what operation can be performed on them to yield another vector? That bit of information about the satellite flying toward perigee is important.
     
  4. May 9, 2009 #3
    Forgive my bump, and this is likely too late for the OP, but it may serve other people well in the future. Here's how I did it:

    Start by solving for your true anomaly. The direction that the satellite is flying (towards perigee) will give you a clue as to which quadrant theta is in. I came up with 330 degrees.

    With the given R vector, you can calculate a radius, and then use the fundamental orbit equation to figure out your angular momentum h.

    Now calculate the radial velocity of the object using

    Code (Text):
    Vr = (mu/h)*e*sind(theta)
    where sind is just sin in degrees. The perpendicular velocity is just (mu/r). Now you can square both of the terms, add them, and then square root them to get a speed.

    Code (Text):
    V = ((mu*E) - ((v^2) - (mu)/r)*R)/(-r*vr)
    Using this formula above, solve for the V vector. It's a rearranged form of the equation you provided above (eq1). You can now find your H vector by crossing V with R. Pick off the third element of the vector, divide it by h, take the inverse cosine, and there's your answer.

    A wrote a quick little .m file in MATLAB to do this. Copy/paste and run to see the inclination. Note that is all done assuming a 2 body system:

    Code (Text):
    R = [-6634.2, -1261.8, -5230.9];

    E = [-.40907, -.48751, -.63640];

    mu = 398600;

    r = norm(R);

    e = norm(E);

    theta = 360 - acosd((dot(R,E))/(e*r));

    h = sqrt(r*mu*(1 + e*cosd(theta)));

    vr = (mu/h)*e*sind(theta);

    vperp = (mu/r);

    v = sqrt((vr^2) + (vperp^2));

    V = ((mu*E) - ((v^2) - (mu)/r)*R)/(-r*vr);

    H = cross(R,V);

    disp('The inclination of the orbit is (in degrees):')
    i = acos((H(3))/h)*(180/pi)
     
  5. May 9, 2009 #4

    D H

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    Since this is an oldish post, giving out answers is not really breaking the rules. It will after all help people solve similar problems in the future.

    That said, you sure solved this the hard way. All that is needed is a vector parallel to (or anti-parallel to) the specific orbital momentum vector, [itex]\boldsymbol h \equiv \boldsymbol r \times \boldsymbol v[/itex]. One such vector is [itex]\boldsymbol r \times \boldsymbol e[/itex].

    To see that this is the case, it's easiest to work in cylindrical curvilinear coordinates:

    [tex]\aligned
    \hat{\boldsymbol r} &\equiv \frac{\boldsymbol r}{||\boldsymbol r||} \\
    \hat{\boldsymbol h} &\equiv \frac{\boldsymbol h}{||\boldsymbol h||} \\
    \hat{\boldsymbol \theta} &\equiv \hat{\boldsymbol h}\times \hat{\boldsymbol r}
    \endaligned[/tex]

    Note that (rhat, thetahat, hhat) forms a right-handed system. With this,

    [tex]\aligned
    \boldsymbol r &= r \hat{\boldsymbol r} \\
    \boldsymbol v &= \dot r \hat{\boldsymbol r} + r\dot{\theta} \hat{\boldsymbol \theta} \\
    \boldsymbol h &= \boldsymbol r \times \boldsymbol v
    = r^2\dot{\theta} \hat{\boldsymbol h} = h \hat{\boldsymbol h}
    \endaligned[/tex]

    The eccentricity vector is

    [tex]\aligned
    \boldsymbol e &= \frac{\boldsymbol v \times \boldsymbol h}{\mu} - \hat{\boldsymbol r} \\
    &= \left(\frac{h^2}{\mu r}-1\right)\hat{\boldsymbol r}
    - \frac{\dot r h}{\mu} \hat{\boldsymbol \theta}
    \endaligned[/tex]

    Now compute r×e:

    [tex]\boldsymbol r \times \boldsymbol e
    = -\,\frac{\dot r h}{\mu} \hat{\boldsymbol h}[/tex]

    When the vehicle is flying toward apogee, rdot is positive so r×e is anti-parallel to the specific angular momentum vector during this first half of the orbit. When the vehicle is flying toward perigee (as is the case in this problem), rdot is negative, making r×e parallel to the specific angular momentum vector during the latter half of the orbit.

    In this particular problem,

    [tex]\hat{\boldsymbol h} = \frac{\boldsymbol r \times \boldsymbol e}{||\boldsymbol r \times \boldsymbol e||} =
    \bmatrix -0.45451 \\ -0.54168 \\ 0.70711 \endbmatrix[/tex]

    As the inclination is the inverse cosine of the z-component of the hhat unit vector, the vehicle is in an orbit with inclination of 45o.
     
  6. May 9, 2009 #5
    Not trying to pick a fight or anything, but I don't think my method is necessarily the 'hard way'. Took about 2 minutes to do in MATLAB.
     
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