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Thread: gravity

  1. #31
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    Let's keep this thread on topic, based on the OP please.
    No golden ratio discussions.
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  2. #32
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    Quote Originally Posted by Ken G View Post
    Also, it is a common misconception that the Langrange point is where the gravity of the two objects cancel. That's not the case, to get the Lagrange point you go into a reference frame that is corotating with the orbit of the two bodies (which are assumed to be in circular orbits). Then the Lagrange points are where the gravity of the two objects add up to cancel the centrifugal force in that rotating frame, so that's effectively three forces that add up to zero, not two.
    .
    That makes sense and I was curious what size vector the c.f. is relative to the other two. Solving for a 1 kg mass where the two grav. forces act equally upon it, it is just shy of the actual L1 by a factor of...5.79! Wow. If my napkin work is correct, that's surprising, huh?

    iphone
    Last edited by George; 2018-Feb-01 at 09:55 PM.
    We know time flies, we just can't see its wings.

  3. #33
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    Quote Originally Posted by George View Post
    That makes sense and I was curious what size vector the c.f. is relative to the other two. Solving for a 1 kg mass where the two grav. forces act equally upon it, it is just shy of the actual L1 by a factor of...5.79! Wow. If my napkin work is correct, that's surprising, huh?

    iphone
    My bold. Not at all surprising. The booby trap here is thinking in terms of a static solution, when in fact we have orbital motion to consider. The Sun's force at L1 is vastly stronger, but the Earth's component creates just enough of an offset that a spacecraft there can orbit the Sun at that shorter radius in one year. As Ken G has shown above, this is a situation where the rotating frame of reference with its fictitious centrifugal force is a handy device for calculation.

  4. #34
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    Quote Originally Posted by George View Post
    That makes sense and I was curious what size vector the c.f. is relative to the other two. Solving for a 1 kg mass where the two grav. forces act equally upon it, it is just shy of the actual L1 by a factor of...5.79! Wow. If my napkin work is correct, that's surprising, huh?
    What did you use for the mass ratio between the two bodies? If the bodies are equal mass, the L1 is at the barycenter and there's no centrifugal contribution. If one body is way more massive than the other, the balance is all between the big gravity and the centrifugal force. (The mass of the test particle is of no consequence.)

  5. #35
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    I suppose the fact that the c.f. wasn't addressed until Ken's post led me to assume it was more of a tweak than the dominant vector. So much so that erroneous math won't be too surprising, huh? [I stumbled in obtaining an exact solution with these simple three vectors; the algebra got tricky for me, though I'm on the run using iPhone & iPad.]
    We know time flies, we just can't see its wings.

  6. #36
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    Yeah, it's no tweak, it's key.

  7. #37
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    Quote Originally Posted by Ken G View Post
    What did you use for the mass ratio between the two bodies?
    Sun & Earth, thus I can steal the actual L1 and compare it to my calculated point where the grav force balance for any given mass, giving the difference factor of 5.79 (using 1.5M km for L1).

    If the bodies are equal mass, the L1 is at the barycenter and there's no centrifugal contribution.
    . Good point, pun unintended.
    (The mass of the test particle is of no consequence.)
    Right since this mass cancels in the equations, but I mentioned it to separate it from the earlier point about gravitational forces being equal, which is true but different for our test mass.
    We know time flies, we just can't see its wings.

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