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Orbital mechanics problem

  1. Sep 14, 2007 #1
    This is not a homework problem. I want to learn how to perform this kind of “approach” calculation on my own. So I’m not interested in someone solving this particular problem, with these specific values, for me. What I want to know is what equations to use, and how to apply them, so that in the future I can figure these things out on my own.

    The basic idea:
    I have a “carrier craft” that launches and recovers smaller “deployable craft”.
    The “carrier craft” is more or less spindle shaped, with a coaxial “launch/recovery ring” that spins.
    The “launch/recovery ring” is approximately 2621 feet (799m) in diameter, and spins at approximately 1.5 revolutions per minute, in order to produce about one gee of centripetal acceleration. There are many launch/recovery “hooks” evenly spaced around the outside of the launch/recovery ring.

    Docking a deployable craft with the carrier craft, IN THE ABSENCE OF A GRAVITY WELL, is fairly straight forward. All you have to do is to approach on a course tangent to the spinning launch/recovery ring, at a speed equal to the edge speed of the spinning ring (in this case aprox. 140mph or 62.5m/s), and time the approach so that the deployable craft and launch/recovery hook arrive at the tangency point at the same moment.

    However, this whole idea becomes considerably more complicated if the carrier craft is in orbit of a steep gravity well like a planet.

    To do that, I suspect something similar to the following would be required.

    The deployable craft would first have to enter an approach orbit that’s either a higher or lower orbit than the one occupied by the carrier craft, with both orbits laying in the same plane. (For example’s sake the carrier is orbiting Earth at an altitude of 500mi or 804672m, while the approach orbit of the deployable craft is, say 15mi or 24140m higher?).

    At the right moment the deployable craft would perform a retro burn that would start it falling toward the lower orbit occupied by the carrier craft. The delta-V of the burn would have to be calculated taking into account the mass of the deployable craft, including current fuel and stores loads, and in keeping with the requirement that it’s going to have to wind up crossing the orbit of the carrier craft “on tangent” (to the spinning launch/recovery ring), and “at speed” (i.e. a tangential approach speed equal to the edge speed of the spinning ring, 140mph or 62.5m/s).

    I also suspect that in the final moments of the approach, visually, the approach would appear to be a straight line. From the pilot of the deployable craft’s point of view, the launch/recovery ring would appear to be rolling toward him and about to roll over his craft.

    Additionally, since I doubt that a retro-burn could be performed perfectly enough to do the job with just a single burn, I expect that the pilot would require the assistance of essentially a “landing signals officer” and final approach instrumentation, to guide him in making small, but constant “tweaks” to the approach, using maneuvering thrusters, right up until his craft mates with its assigned launch/recover hook.

    In other words, the final moments of the approach would probably get “hairy”, but perhaps no more “hairy” than those immediately preceding attempts to hit the three wire during a landing on a carrier deck at night.

    In the event of a “wave off”, a blown approach that can’t be safely recovered from, the pilot would burn, off the tangent line, and away from an uncontrolled collision with the launch/recovery ring, and be instructed to perform a new burn to put his craft back into its approach orbit, and the whole process would have to be repeated.

    How would I calculate the required delta-v and timing of the retro-burn that would deliver the deployable craft from its (in this case 15mi higher) approach orbit, into a trajectory 90 degrees to the orbit of the carrier (tangent to the spinning ring), at a closure rate of 140mph (62.5m/s)?
  2. jcsd
  3. Sep 15, 2007 #2


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    First I'd turn the recovery craft so that the recoverery ring rotates in the same plane as the craft's orbit. This simplifies things greatly.

    As far as the actual burns go, the simpliest, if not most direct, method goes like this:

    Perform a burn that lifts your craft into a transfer orbit with a higher apogee of the proper value (to calculated below).

    Do a second burn that puts you into a circular orbit at the same altitude as the new apogee.

    Wait until you reach the correct point in your new orbit and do a braking burn which puts you into another tranfer orbit, one with a perigee equal to the altitude of the orbit of the recoverery craft. cone right, this gives you the correct relative speed to the recovery craft as you pass.

    The equations.

    To determine the altitude for the new orbit for the deployable craft:

    given that

    [itex]V_r[/itex] is the relative velocity the two craft must have to dock.
    [itex]M[/itex] is the mass of the planet being orbited
    [itex]R_1[/itex] is the radius of the recovery's craft's orbit. (as measured from the center of the planet)
    [itex]R_2[/itex] is the radius of the initial orbit of the deployable craft.
    [itex]G[/itex] is the gravitational constant.
    [itex]m[/itex] is the mass of the deployable of the craft.

    Then we use the conservation of energy to find the new orbit's radius.

    First we find the orbital velocity of the recovery craft by

    [tex]v_2 = \sqrt{\frac{GM}{R_1}}[/tex]
    The total energy of the deployable craft is as it passes the recovery craft is found by

    [tex] E_t =\frac{2(V_1+V_2)m}{2}- \frac{GMm}{R_1}[/tex]

    It is also found by

    [tex]E_t = -\frac{GMm}{2a}[/tex]

    a is the semimajor axis of the second transfer orbit.

    we equate the two equations and solve for a
    We find the apogee of this orbit by:

    [tex]R_{a1} = 2a-R_1[/tex]

    This is the same as the apogee for the first transfer orbit.

    Now you can find the delta v needed for the first burn.

    [tex]\Delta v1 = \sqrt{\frac{Gm}{R_2}} \left [ \sqrt{\frac{2R_{a1}}{R_2+R_{a1}}}-1 \right ][/tex]

    The second delta v burn needed to move into a circular orbit is:

    [tex]\Delta v2 = \sqrt{\frac{GM}{R_{a1}}} - \sqrt{\frac{2GMR_2}{R_{a1}(R_2+R_{a1})}}[/tex]

    The delta v needed for the last burn is:

    [tex]\Delta v3 = \sqrt{\frac{2GMR_{a1}}{R_1(R_1+R_{a1})}}[/tex]
  4. Sep 18, 2007 #3
    Huge Thanks!

    This is EXACTLY what I was looking for. And I absolutely love the idea of turning the recovery craft so that the launch/recovery ring spins in the same plane as its orbit, (or, for that matter, any orbital attitude where the spin axis of the launch/recovery ring is perpendicular to a line tangent to the "recovery" orbit). I can see where that simplifies the heck out of things.

    I can't wait to try plugging some numbers into this to see how things change when the recovery craft is in orbit of things other than Earth, like the moon (a smaller body), or Jupiter (a much bigger one), Earth's L5 or L4 LaGrange points (i.e. orbiting the Sun), etc.

    Absolutely fantastic.

    Thank you Janus!
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