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Old 10-24-12, 06:42 PM   #1
Captain Blight
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Neil DeGrasse Tyson hasn't answered me

... On the Facebook, so naturally I thought I'd ask the foosters.

Is orbital velocity the same no matter which dirction the orbit is moving? I say a polar orbit, a retrograde orbit, and an orbit with the rotation of a body are all the same given equal altitudes. My friend, who is WRONG, says that a retrograde orbit needs to be twice the speed of a co-rotational orbit.

What says Foo? Asshat if you must, but if someone actually knows and can point me to the actual math, you'll have the satisfaction of knowing you've helped me win a 6er of delicious beer.
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Old 10-24-12, 07:23 PM   #2
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I think he is wrong. The formula V = square root of G*M/r seems straight forward. However, achieving retrograde orbit for an earth launched satellite takes much more energy to achieve that placing an earth launched satellite in prograde orbit. The end orbital velocity of both would be the same, assumed both satellites were placed into orbit equal distance from the center of the earth.
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Phobias are for irrational fears. Fear of junk ripping badgers is perfectly rational. Those things are nasty.

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Old 10-24-12, 08:17 PM   #3
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That's what I thought too, thanks for the math. I know planar changes are energy-expensive but if one were to (theoretically) launch in a co-rotational plane and over the course of several hundred orbits change direction over the pole, you could end up with a retrograde orbit. I suppose that it's like anything else, doing it slowly takes less energy than trying to do it fast.
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Old 10-24-12, 08:20 PM   #4
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My favorite Tyson video clip: http://www.youtube.com/watch?v=8B6jSfRuptY
A bit long, but the punchline (Cameron's response) is worth it.
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Old 10-24-12, 08:36 PM   #5
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But the earth is not a perfect sphere is it? I wonder if this has something to do with it.
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Old 10-24-12, 08:47 PM   #6
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Moot discussion. The answer is always 42.
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Old 10-24-12, 08:58 PM   #7
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But the earth is not a perfect sphere is it? I wonder if this has something to do with it.
jsharr posted the correct formula. Where the center of mass is located, is all that matters to the object orbiting it.
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