What is the separation of two masses when they have their average speed in orbit relative to their center of mass?

Present your answer as a ratio with the orbit's semimajor axis, and as a function of the orbit's eccentricity.

Update:

Mike, I have the utmost contempt for the kind of evasion that you answer exemplifies. If you don't know the answer, or how to find the answer, then either be silent or else be honest enough to say that you don't know. I am not a cheating student. I'm a 60-year-old retired physicist, and I already know the answer to the question I've asked.

The other answerer, Ronald 7, is merely confused. There is an exact and calculable answer, which I will provide hereafter in an answer of my own.

Update 2:

The distance between two masses when they have their average speed in their orbit around their mutual center of mass is found as follows:

circumference of an ellipse,

C = 4a ∫(0,π/2) √(1−e²sin²θ) dθ

period in an elliptical orbit

P = 2π√[a³/(GM)]

speed in orbit as a function of distance

v = √[GM(2/r − 1/a)]

average speed in orbit

vₐ = C/P

4a ∫(0,π/2) √(1−e²sin²θ) dθ / 2π√[a³/(GM)] = √[GM(2/r−1/a)]

r = 2 { [4/(π²a)] [ ∫(0,π/2) √(1−e²sin²θ) dθ ]² + 1/a }⁻¹ 

Update 3:

r = 2 { [4/(π²a)] [ ∫(0,π/2) √(1−e²sin²θ) dθ ]² + 1/a }⁻¹

The true anomaly in the orbit at which the object has its average speed in orbit, where r is the value calculated just above, is

θ = ± arccos{ [ (a/r)(1−e²) − 1 ] / e }

Note that

if e=0, then r=a & θ is indefinite

if e→0⁺, then r=a & θ→±π/2

if e>0, then r>a & θ is toward the apoapsis from ±π/2

if e→1⁻, then r→1.423199122a & θ→π

3 Answers

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  • 4 months ago
    Favourite answer

    You might as well be asking how long is a piece of string ?

    The answers is Twice the Length of 50% of it

    (2Xx1/2X)

    Attachment image
  • Long cylinders, initially rotating around their physical axis of symmetry (a line running down the length through the middle) are rotationally unstable versus small deviations in the internal distribution of mass. They start to precess, and then they tumble end-over-end, just as the Explorer 1 satellite did.

    Short, squat cylinders are, on the other hand, rotationally stable around their physical axis of symmetry.

    Where's the break-even point in frustum length between cylinders that rotate stably around their physical axis of symmetry, and those that are unstable? Rotating objects tend to seek the rotational axis having the greatest moment of inertia, which is usually the shortest axis.

    The moments of inertia for a hollow cylinder of inner radius Rᵢ , outer radius Rₒ , and length h are

    Rotation around the physical axis of symmetry (desired)

    I∥ = (M/2)(Rᵢ²+Rₒ²)

    Rotation perpendicular to the physical axis of symmetry (to be avoided)

    I⟂ = (M/12) [ 3(Rₒ²+Rᵢ²) + h² ]

    Set them equal to each other.

    (M/2)(Rᵢ²+Rₒ²) = (M/12) [ 3(Rₒ²+Rᵢ²) + h² ]

    6Rᵢ² + 6Rₒ² = 3Rₒ² + 3Rᵢ² + h²

    3Rᵢ² + 3Rₒ² = h²

    Assuming that the thickness of the frustum wall (Rₒ−Rᵢ) is negligible compared with the diameter of the cylinder,

    6R² ≈ h²

    h = √6 R = 2.449489743 R

    When the length, h, of the cylinder is less than √6 R, the cylinder is rotationally stable versus small asymmetries in mass distribution around the frustum. But when h is greater than √6 R, the cylinder is rotationally unstable.

    Wasn't that simple?

  • Bob
    Lv 7
    4 months ago

    In other words, do your homework for you and format it so you can just copy/paste it.

    Sorry, if you had asked as if you really wanted to learn something I would have helped.

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