Satellite ME is orbit around the Moon and Earth. Satellites E is around Earth. I understand the low orbit satellites orbit at higher speeds. Can Sat ME match the speed of any of Satellites E at any altitude to be in close proximity for a half orbit?

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Since you add the clarification of "Can Sat ME match the speed of any of Satellites E at any altitude to be in close proximity for a half orbit?", I assume that by speed, which is a scalar quantity (no direction information), you actually mean velocity, which is a vector quantity, having both magnitude and direction. Two satellites going at the same speed but in directions 90° apart haven't performed a rendezvous!

So my assumption is that you're asking if two satellites at pretty much the same relative position for an extended period (half an orbit), but one staying in LEO and the other going out to near-lunar distances, can have velocity vectors that are fairly closely parallel and equal. Is that an accurate restatement of your question?

If so, the answer is no. Orbital mechanics says that if two objects are in the same position, with parallel and equal velocity vectors, then their orbits will be the same.

The ME and E example is a good one to demonstrate just how far from matched those velocity vectors are. A satellite in a 400 km (altitude) circular orbit at Earth has a speed of ~7.67 km/s. An object in a "free-return" Earth-moon orbit (which can't do a lunar flyby on every orbit!) with a 400 km perigee has a perigee speed of ~10.74 km/s. So even if you line up their velocity directions, there's more than 3 km/s of difference in their speeds.

Summary: There is no orbit that goes out to near-lunar distances that would even roughly rendezvous with an object in LEO.

  • $\begingroup$ Thank you for making it where I can visualize it. So at best you would only get 2 near misses 1 coming and 1 going? What if you had a pole that is .5km long could the close proximity be extended? $\endgroup$ – Muze Oct 9 '18 at 18:31
  • $\begingroup$ If by near misses you mean the lunar trajectory object wizzing by the other object at breck-neck speeds, sure. Unless you learn how to drop your apogee very quickly to match the other object at the intersection. $\endgroup$ – Magic Octopus Urn Oct 9 '18 at 18:46
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    $\begingroup$ Take a look at how fast that system with the 1/2 km pole would have to be rotating to get the pole's tip to match speeds, even instantaneously, with the other satellite. The slowest possible case would be where the velocity vectors are parallel at the closest approach point, so the orbits have equal inclinations. As mentioned above, at 400 km that difference is ~3 km/s (1/2 km difference in altitude won't make much difference in speed at all). So a pole rotating with a radius of 1/2 km has to have the tip going 3 km/s: ~1 revolution per second! (Assume some counterweight has the whole... $\endgroup$ – Tom Spilker Oct 9 '18 at 23:55
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    $\begingroup$ ...pole rotating around one extreme end, not around its center). If it rotates around its center, it's even worse. The rendezvous time would be a tiny fraction of a second. And the centrifugal acceleration at the tip of the pole would be ~18,000 m/s^2, nearly 2,000 g's! If the pole rotates around its center, the radius of gyration is only 250 m but the tip still has to go 3 km/s, so the rotation rate goes to ~1.91 revs per sec, and the centrifugal acceleration goes up to 36,000 m/s^2, nearly 4,000 g's. $\endgroup$ – Tom Spilker Oct 10 '18 at 0:06
  • $\begingroup$ @TomSpilker I'm still working on centrifugal force hah! Going off of this and what you said earlier about the femur in space :). Awesome explanation though. I'm at chapter 7/13 in my orbital mechanics textbook and, somehow, they've not mentioned the rotation of the Earth yet. $\endgroup$ – Magic Octopus Urn Oct 10 '18 at 1:07

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