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Thread: Moon Orbit

  1. #1
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    Moon Orbit

    I have tried to compare between the Earth/Moon to Sun/Moon gravity force as follow:

    Earth Mass = M⊕ = (5.9722±0.0006)×1024 kg
    Moon Mass 7.342×1022 kg (0.012300 of Earth's) = 0.0123 M⊕
    Sun Mass = M☉ = 1.98855×1030 kg (333,000 × Earth) = 333,000 M⊕

    Radius (moon-earth) = 384399 km = 0.00257 AU = 2.57 x 10 ^ -3 AU

    Radius (earth-sun) = 149,597,887 km = 1 AU‏

    As the Moon is located close to Earth, than we can assume that this value also represents the Radius (moon-sun)
    Therefore,
    Radius (moon-sun) = Radius (earth-sun) = 149,597,887 km = 1 AU‏


    Gravity Force:

    F = G x M1 x M2 / R ^ 2

    F (Earth Moon) = G x M(moon) x M⊕ / Radius (moon-earth) ^ 2 =

    G x 0.0123 M⊕ x M⊕ / (0.00257 AU) ^ 2 = G x M⊕ ^ 2 x 0.0123 / (6.6 10 ^-6 x AU ^ 2) =

    G x M⊕ ^ 2 x AU ^ -2 x 1,863.63

    F (Sun Moon) = G x M⊕ x M☉ / Radius (moon-sun) ^ 2 =

    G x M⊕ x 333,000 x M⊕ / (1 AU) ^ 2 =

    G x M⊕ ^ 2 x AU ^ -2 x 333,000

    Therefore, with regards to the Moon;

    The direct Gravity force from the sun is stronger then the gravity force from the Earth:

    F (Sun Moon) / F (Earth Moon) = 330,000 / 1,863.63 = 177

    So, why our dear moon had decided to orbit the Earth, while it can get a 177 stronger gravity force directly from the Sun?

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    Quote Originally Posted by Dave Lee View Post
    ... So, why our dear moon had decided to orbit the Earth, while it can get a 177 stronger gravity force directly from the Sun?
    If you drew a graph of where the Moon moves in a year, it would certainly look like an ellipse around the Sun.
    Forming opinions as we speak

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    While it is true that the Sun "pulls" on the Moon much more strongly than the Earth does, it accelerates the Earth and the Moon in unison. It is the difference in the gravitational action that would try to separate the Earth and the Moon, not the absolute value of that action. This difference is only a small fraction of the Earth-Moon action. It is enough to severely perturb the orbit but not rip it asunder.

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    If we plot the moon's orbit as the Earth-moon system moves around the sun, we see that the moon's orbit is everywhere concave towards the sun. It's effectively in orbit around the sun, while continuously perturbed by the Earth. That's what makes the theory of the moon's orbit so complicated compared to most other planetary moons.

    Grant Hutchison

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    Thank you all

    Sorry, I have just noticed that I had a mistake in my calculation.
    It should be as follow:

    Gravity Force:

    F = G x M1 x M2 / R ^ 2

    F (Earth Moon) = G x M(moon) x M⊕ / Radius (moon-earth) ^ 2 =

    G x 0.0123 M⊕ x M⊕ / (0.00257 AU) ^ 2 = G x M⊕ ^ 2 x 0.0123 / (6.6 10 ^-6 x AU ^ 2) =

    G x M⊕ ^ 2 x AU ^ -2 x 1,863.63

    F (Sun Moon) = G x M moon x M☉ / Radius (moon-sun) ^ 2 =

    G x 0.0123 M⊕ x 333,000 x M⊕ / (1 AU) ^ 2 =

    G x M⊕ ^ 2 x AU ^ -2 x 0.0123 x 333,000 =

    G x M⊕ ^ 2 x AU ^ -2 x 4905.9

    Therefore, with regards to the Moon;

    The direct Gravity force from the sun is stronger then the gravity force from the Earth:

    F (Sun Moon) / F (Earth Moon) = 4905.9 / 1,863.63 = 2.17

    Hence, the ratio in gravity force is 2.17.

    However, I still don't understand why the moon orbits around the Earth (while their center of mass orbits around the Sun), instead of going directly around the Sun.
    In other words: Why the Moon needs the Earth?

    In order to make it even more difficult, let me introduce the following scenario:

    Let's assume that group of scientists are located at a nearby solar system.

    They have full knowledge about Newton and Kepler laws.
    They can see the Sun; They can also see the Moon orbital path.

    So, they see the following image:

    https://www.quora.com/How-would-I-ch...around-the-sun

    However, Unfortunately, they can't see the Earth.

    So, they try to understand the source for that in/out orbital path of the moon.

    Based on their calculation, the ratio between the radiuses of the Sun/Moon to that in/out path is:

    Radius (earth-sun) / Radius (earth-sun)

    Radius (moon-earth) = 384399 km = 0.00257 AU

    Radius (earth-sun) = 149,597,887 km = 1 AU‏

    Hence, the ration is:

    1: 0.00257.

    They claim that this ratio is in the Error bar of their instruments.
    Therefore, they have decided that this in/out path of the moon has no real effect of the Moon/Sun orbit.

    They have also found that the Sun gravity is stronger by at least two times from any sort of virtual object which the moon might orbit.

    Hence, they have concluded that there is no way that the moon orbits around some virtual object.

    So, how can we convince them that the moon orbits around this virtual object which we call "Earth"?
    Last edited by Dave Lee; 2018-Jan-26 at 10:36 PM.

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    How can they see the moon if they can't see the Earth?
    If the Earth were improbably black, then they'd decide (within the measurement error you've just invented) that the moon orbits the sun. Which would be a perfectly reasonable approximation.
    But the Earth isn't black, so we don't need to convince them.

    Grant Hutchison

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    Dave's hypothetical scenario is analogous to what Bessel experienced while observing the motion of Sirius. He could see the wavy path and inferred a massive but dark companion, perhaps a burned out remnant of a star that was no longer shining. In a thought exercise we could replace the Earth with a black hole of equal mass. In the real thing such an idea would be absurd.

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    Quote Originally Posted by Hornblower View Post
    Dave's hypothetical scenario is analogous to what Bessel experienced while observing the motion of Sirius. He could see the wavy path and inferred a massive but dark companion, perhaps a burned out remnant of a star that was no longer shining. In a thought exercise we could replace the Earth with a black hole of equal mass. In the real thing such an idea would be absurd.
    Thanks for the support.

    Quote Originally Posted by grant hutchison View Post
    How can they see the moon if they can't see the Earth?
    But the Earth isn't black, so we don't need to convince them.

    Grant Hutchison
    Our Blue might be their black.
    So, what we clearly see from our location might be invisible from a different location or different technology.
    Based on our current technology, we can detect many black holes, but we can't see all the black objects in the space.
    So, could it be that there are many black objects (even near our solar system) which we just can't see with our current technology?


    Quote Originally Posted by grant hutchison View Post
    If the Earth were improbably black, then they'd decide (within the measurement error you've just invented) that the moon orbits the sun. Which would be a perfectly reasonable approximation.

    Grant Hutchison
    No, that is a severe mistake.
    They shouldn't ignore that in/out movement of the moon!
    They shouldn't assume that it is in the error bar of their technology.
    If they don't see the Earth, it doesn't mean that the Earth doesn't exsist.
    They should understand that even if the gravity force of the Sun is stronger than the local gravity force of the invisible object (which we call - earth), That earth has the power to hold the Moon in an orbital path around it.
    Let's try to understand what might be the outcome of the decision to ignore the earth as a main orbital center for the moon.
    let's assume that they can detect the mass of the Moon, but they need to extract the mass of the Sun.

    As the moon mass is:

    Moon Mass = 7.342×1022 kg = 0.012300 of Earth's = 0.0123 M⊕

    So, instead of using the real total orbital mass of Earth + Moon (or even just earth mass), they have only used the Moon mass.
    The outcome is that their calculated Sun mass is less by about 0.012300 from its real value.
    Therfore, instead of getting the real mass value of:
    Sun Mass = 333,000 M⊕
    They have found that the calculated Sun mass is only:
    Calculated Sun mass = 333,000 M⊕ * 0.012300 = 4,096 M⊕

    So, don't you agree that it is a severe mistake to ignore this small in/out movement variation in the orbital path?
    Last edited by Dave Lee; 2018-Jan-27 at 10:15 AM.

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    Quote Originally Posted by Dave Lee View Post
    They claim that this ratio is in the Error bar of their instruments.
    Therefore, they have decided that this in/out path of the moon has no real effect of the Moon/Sun orbit.
    If it is within the error bars, then they are correct; they cannot detect the presence of the Earth.

    When their measurements get more accurate, then they will be able to say that there is some "dark matter" that affects the path of the Moon. They would naturally assume this was a planet that they couldn't see.

    They may decide to name this unknown planet after their sea god. (That is a subtle reference to the fact that this is pretty much how Neptune was first detected.)

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    Quote Originally Posted by Dave Lee View Post
    No, that is a severe mistake.
    They shouldn't ignore that in/out movement of the moon!
    If it is within the error bars, then there is no in-out movement. (That is what error bars mean.)

    They shouldn't assume that it is in the error bar of their technology.
    But you said it was. And there is no assumption involved. They know what the error range is on their measurements. Either the movement is larger than the errors in measurement or it isn't.

    Let's try to understand what might be the outcome of the decision to ignore the earth as a main orbital center for the moon.
    Why should we? Why would anyone decide that? You seem to be making arbitrary assumptions to try an make a point. But I don't understand what the point is.

  11. #11
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    Quote Originally Posted by Dave Lee View Post
    So, why our dear moon had decided to orbit the Earth, while it can get a 177 stronger gravity force directly from the Sun?
    Rather than force, look at the gravitational acceleration, and compare the gravitational acceleration of the sun on the moon to that of the sun on Earth.

    You'll find that they're basically identical (the difference in force is exactly canceled by the difference in mass), they are in near-identical orbits around the sun. And when you subtract out the portion of the influence from the sun which is common to both, you'll find that they are also orbiting each other, just as when you go up in scale instead, you'll find that the whole solar system is orbiting through the galaxy, with the galaxy's gravitational influence being essentially identical on everything in the solar system.

    If you were to plug Earth, Luna, and the sun into a gravitational simulation and delete one of them: if you deleted either Earth or Luna, the other would continue in its orbit around the sun with almost no change. If you deleted the sun, Earth and Luna would continue to orbit each other with almost no change. There is no physical meaning to "main orbital center" in reality, it depends on what you're looking at.

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    Quote Originally Posted by cjameshuff View Post
    There is no physical meaning to "main orbital center" in reality, it depends on what you're looking at.
    What do you mean by that?

    I assume that if we want to be accurate we have to say that the Earth and the Moon orbit around their common center of mass, while this center of mass orbits around the Sun.

    However, just as a general idea, can we claim that:
    The Earth is the "main orbital center" to the Moon?
    Or, the Sun is the "main orbital center" to the Earth (or Earth-moon center of mass)?
    Or, the SMBH is the "main orbital center" to the Sun?
    How shall we call it? Please correct it as needed.
    Last edited by Dave Lee; 2018-Jan-27 at 03:35 PM.

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    It appears that we have two distinct issues here. One is not understanding how the Earth and Moon can be gravitationally bound where the Sun's gravity is so strong, and the other is about the ability to observe the wavy component of the Moon's heliocentric motion in the hypothetical case of an invisible Earth-mass body. Let's deal with the second one by returning to Sirius.

    Observers in the 17th century could see that Sirius had moved from Ptolemy's position, but their astrometry was too primitive to detect anything but a straight line of motion. In the 19th century Bessel had what we could call the instrument of their dreams, a very stable refracting telescope, and he was able to see the wavy component from which he inferred a dark but massive companion. A few decades later Alvan Clark, with a much larger telescope, saw the almost dark companion. Successive generations of observers in our dark-Earth thought exercise could do likewise.

    For the first issue, let's make an analogy with a tugboat towing a big ship with a lightweight towline which it could break easily if the towline was attached to an immovable object. If the ship was dead in the water, the tug would break the towline trying to move it. If the ship's engines are working hard enough that both vessels would make the same speed without the towline, there will be no tension on the towline. If the big ship reduces power just slightly, the towline can make up the difference without breaking. Here the tug's thrust corresponds to the Sun's gravitational pull on the Moon, the ship's thrust to the Sun's pull on the Earth, and the towline to the Earth-Moon mutual gravitation toward one another. This corresponds to having the Moon in conjunction with the Sun.

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    The term is barycentre.

    The Earth-Moon barycentre is inside the Earth, about 400km from its centre.

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    Quote Originally Posted by Hornblower View Post
    It appears that we have two distinct issues here. One is not understanding how the Earth and Moon can be gravitationally bound where the Sun's gravity is so strong, and the other is about the ability to observe the wavy component of the Moon's heliocentric motion in the hypothetical case of an invisible Earth-mass body. Let's deal with the second one by returning to Sirius.

    Observers in the 17th century could see that Sirius had moved from Ptolemy's position, but their astrometry was too primitive to detect anything but a straight line of motion. In the 19th century Bessel had what we could call the instrument of their dreams, a very stable refracting telescope, and he was able to see the wavy component from which he inferred a dark but massive companion. A few decades later Alvan Clark, with a much larger telescope, saw the almost dark companion. Successive generations of observers in our dark-Earth thought exercise could do likewise.

    For the first issue, let's make an analogy with a tugboat towing a big ship with a lightweight towline which it could break easily if the towline was attached to an immovable object. If the ship was dead in the water, the tug would break the towline trying to move it. If the ship's engines are working hard enough that both vessels would make the same speed without the towline, there will be no tension on the towline. If the big ship reduces power just slightly, the towline can make up the difference without breaking. Here the tug's thrust corresponds to the Sun's gravitational pull on the Moon, the ship's thrust to the Sun's pull on the Earth, and the towline to the Earth-Moon mutual gravitation toward one another. This corresponds to having the Moon in conjunction with the Sun.
    You are just on the main points!

    Let me concentrate on the first issue:
    "how the Earth and Moon can be gravitationally bound where the Sun's gravity is so strong"

    So, you gave a good explanation by using a tugboat.
    However, let's go one step back.
    How could it be that the moon had selected the Earth as its tugboat or Braycenter while it gets much higher gravity force from the Sun?
    Actually, it comes to the first point of setting the moon in its current orbital path.

    https://www.space.com/19275-moon-formation.html

    "There are three theories as to how our planet's satellite could have been created: the giant impact hypothesis, the co-formation theory and the capture theory".

    So, we have to start from the early days, when the moon (or any sort of other object which had set that giant impact) was free from Earth gravitation.
    At that time all the objects were directly under the Sun gravitation.
    However, now we know that the Sun gravity is stronger than the Earth gravity.
    So, if the early moon was under the sun gravitation, why it will prefer to go after the Earth, which offers much lower gravity power.
    I think that we must look again at those theories and try to understand why any object in the early solar system, which was under a direct sun gravity force, will chose to abandon this strong force and join to a planet with much lower gravity force.

    It seems to me that we are missing key issue in our understanding how the solar system (or especially - the moon - Earth system) had been established
    Last edited by Dave Lee; 2018-Jan-27 at 05:32 PM.

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    Quote Originally Posted by Dave Lee View Post
    It seems to me that we are missing key issue in our understanding how the solar system (or especially - the moon - Earth system) had been established
    It certainly seems to me that you are missing key issues in understanding.
    The moon hasn't made a choice about orbiting the Earth rather than the sun. It orbits both, and is continuously affected by the gravity of both.

    Grant Hutchison

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    The Moon did not "choose" to "abandon" anything. In the prevailing collision theory, the glob of material that became the Moon was bludgeoned by the force of the collision into a configuration where it could remain gravitationally bound, while several times as much stuff went flying off into space, never to return.

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    Quote Originally Posted by Dave Lee View Post
    So, we have to start from the early days, when the moon (or any sort of other object which had set that giant impact) was free from Earth gravitation. [...]
    Dave Lee,

    This is the Q&A forum. It’s beginning to sound like you’re lecturing, rather than asking questions about a subject you are not clear on. Have a care. If you wish to advocate something, please do it in the appropriate forum.
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    Quote Originally Posted by Dave Lee View Post
    So, we have to start from the early days, when the moon (or any sort of other object which had set that giant impact) was free from Earth gravitation.
    At that time all the objects were directly under the Sun gravitation.
    However, now we know that the Sun gravity is stronger than the Earth gravity.
    So, if the early moon was under the sun gravitation, why it will prefer to go after the Earth, which offers much lower gravity power.
    How do you think the Earth itself formed?

    It formed because a denser-than-average volume of dust and gas that was orbiting the sun coalesced into a solid, gravitationally-bound object. If your idea of solar system gravity held any water, nothing in the solar system could have formed larger than a mote.

    The Earth did form, as did all the other planets (including, not trivially, Mercury), and it formed from the coalescence of motes into rocks into boulders into planetoids into planets - all in the Sun's orbit.




    Gravity falls as the square of the distance. The Earth's pull on the Moon is much stronger than the Sun's pull on it.

    You know the rubber sheet analogy?

    While the Earth's well is not as deep as the Sun's, if you are close enough to the Earth, you can still be pretty far down the well, on a highly-sloped section) - which is the same as saying you are in a strong gravitational influence.
    The the Sun's well is very deep, we are very far from it. Out here where the Earth is, it's almost flat .


    Look here:



    The above diagram of Earth-Moon could very well be embedded in the Sun's well - but it's hard to tell - the Sun's well is just too flat to even notice the slope.


    i.e.the Moon is much more influenced by the Earth than by the Sun.
    Last edited by DaveC426913; 2018-Jan-27 at 08:27 PM.

  20. #20
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    Here's another way of looking at it with all three bodies:



    That point L1 is Lagrange Point One. It is defined as the point where the Sun's and the Earth's gravity are balanced. You're at the top of a hill, and you could slide down either way.

    The Moon is much farther down Earth's side of the hill than the Sun's.

    (Here it is in 3D, by Yours Truly.)
    Attached Images Attached Images
    Last edited by DaveC426913; 2018-Jan-27 at 08:47 PM.

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    I made this diagram of the Moon's orbit around the Sun. It is to
    accurate scale, but is still crude because the sizes of the Earth and
    Moon and the Moon's orbit around the Earth are tiny compared to
    the size of the pixels and the size of of the Earth's and Moon's orbit
    around the Sun.

    http://www.freemars.org/jeff2/MoonOrb1.png

    I used no smoothing on the pixels. If anything looks even slightly
    fuzzy, it is because your browser is adjusting the size of the image
    to try to fit it on your screen.

    -- Jeff, in Minneapolis
    http://www.FreeMars.org/jeff/

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    Here's a simulation of the Earth and Moon orbiting the Sun. I made Earth black and told it to trace no path, so you can't see it. So the path you see is only from the Moon. I can barely tell there is something strange about its shape.

    This sim will run in your browser:
    http://orbitsimulator.com/gravitySim...lackEarth.html

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    Quote Originally Posted by Jeff Root View Post
    .... your browser is adjusting the size of the image
    to try to fit it on your screen.
    Who sends 1/16th of a picture!!

    Do you know how long I sat here, like a feckless dingwad, looking at that short arc of Earth's orbit - patiently waiting for the other 15/16ths of the picture to load??

    ("Hm. It seems to have stopped rendering after half the screen. I'll refresh. Huh, it froze in exactly the same place. I'll give it more time...")
    Last edited by DaveC426913; 2018-Jan-28 at 04:25 AM.

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    I'd like to have a monitor that could display the whole orbit!

    I did see a demonstration of a computer monitor once, back in
    the 1980's, that would have been able to display the entire orbit.
    Square screen -- aspect ratio of 1:1. But it was monochrome.
    Orange and black. And of course it was a CRT. Like my diagram,
    it used no antialiasing to smooth out jaggies, because it had so
    many pixels that the designers felt it wasn't needed.

    -- Jeff, in Minneapolis
    http://www.FreeMars.org/jeff/

    "I find astronomy very interesting, but I wouldn't if I thought we
    were just going to sit here and look." -- "Van Rijn"

    "The other planets? Well, they just happen to be there, but the
    point of rockets is to explore them!" -- Kai Yeves

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    Quote Originally Posted by Jeff Root View Post
    I'd like to have a monitor that could display the whole orbit!

    I did see a demonstration of a computer monitor once, back in
    the 1980's, that would have been able to display the entire orbit.
    Square screen -- aspect ratio of 1:1. But it was monochrome.
    Orange and black. And of course it was a CRT. Like my diagram,
    it used no antialiasing to smooth out jaggies, because it had so
    many pixels that the designers felt it wasn't needed.

    -- Jeff, in Minneapolis
    So its meant to be only a tiny bit of the orbit? (I assumed I was having download problems, as well.) It would be interesting to see the whole orbit in one image - then you could zoom out to see the whole thing or zoom in to see the little slice you have posted.

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    Quote Originally Posted by DaveC426913 View Post
    How do you think the Earth itself formed?

    It formed because a denser-than-average volume of dust and gas that was orbiting the sun coalesced into a solid, gravitationally-bound object. If your idea of solar system gravity held any water, nothing in the solar system could have formed larger than a mote.

    The Earth did form, as did all the other planets (including, not trivially, Mercury), and it formed from the coalescence of motes into rocks into boulders into planetoids into planets - all in the Sun's orbit.
    Thanks for the explanation.

    So, you claim that only based on the Sun Gravity, noting would be formed:
    "If your idea of solar system gravity held any water, nothing in the solar system could have formed larger than a mote."
    What is the reason for that?
    Do you mean that the Sun gravity by itself is too weak for that task?

    However, you also claim that the formation of all the planets and moons in the solar system:
    "The Earth did form, as did all the other planets (including, not trivially, Mercury), and it formed from the coalescence of motes into rocks into boulders into planetoids into planets - all in the Sun's orbit."

    Is due to:

    "It formed because a denser-than-average volume of dust and gas that was orbiting the sun coalesced into a solid, gravitationally-bound object."

    In one hand you claim that "dust and gas that was orbiting the sun", so it seems that this matter is under the Sun gravity, however, in the next hand you claim: "It formed because a denser-than-average volume".
    So, what is the source of that power which had set that denser-than-average volume? if it isn't the Sun Gravity while it is orbiting the Sun, than what is it?

    Did I understand you correctly?

    Few more question:
    Why do we assume that the Earth and Moon had been formed separately and then they had need glued together by gravity?
    Why can't we assume that they both had been formed together at the same moment? So, from the first day of birth, they were already orbiting each other?
    Why we set a gape of about 2 Billion years between the Sun formation to Planets/Moons formation?
    Why the same power (or process) which had set the Sun formation, couldn't also set all planets with all their moons at the same spot and at the same time?
    Last edited by Dave Lee; 2018-Jan-28 at 11:19 AM.

  27. #27
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    Quote Originally Posted by DaveC426913 View Post
    Here's another way of looking at it with all three bodies:

    SNIP image

    That point L1 is Lagrange Point One. It is defined as the point where the Sun's and the Earth's gravity are balanced. You're at the top of a hill, and you could slide down either way.
    The Lagrange Point One is not where the sun's gravity and earth's gravity is balanced. The balance point is much closer to the earth than the moon is--that's the idea behind the OP's (improved) calculations. The OP is just trying to understand how orbits work, in the light of that "close-in balance point".
    The Moon is much farther down Earth's side of the hill than the Sun's.

    (Here it is in 3D, by Yours Truly.)

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    Dave Lee

    You seem to think that an object like the moon can only be gravitationally attracted to one object, but that is not correct. In fact all objects are attracted to all other objects; it just depends on the size and distance.

    The "motes" in the early solar system can be attracted to each other and to the sun at the same time.
    At night the stars put on a show for free (Carole King)

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  29. #29
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    Let me try a somewhat different approach to making sense of the gravitational binding of the Moon to the Earth. It is not enough to just look at the magnitudes of the forces. Since the three bodies are in free fall, we must look at how the Earth and the Moon move in response to the gravitational forces on them.

    Dave Lee’s corrected calculations are in good agreement with mine, indicating that the Sun’s gravity pulls on the Moon about twice as hard as does the Earth’s gravity. If we could magically immobilize the Earth with some sort of opposing nongravitational force and leave the Moon free to gravitate whichever way the vector sum of the gravitational forces is directed, then yes indeed the Sun would strip the Moon free of Earth’s grasp. However, in the real universe the Earth will fall toward the Sun nearly in unison with the Moon. Let’s look at the case of the new Moon position, where the Moon is about 1 part in 400 closer to the Sun. From the inverse square law we find that the Moon accelerates about 1 part in 200 more quickly than does the Earth. That shows the same sort of separation tendency as if the Sun were absent and something with about 1/200 of the solar gravitational force was tugging on the Moon but not on the Earth. Since the Earth’s gravity is about 100 times as strong as this hypothetical force, the two bodies remain gravitationally bound.

    Clear as mud? I know these compound motion cases can be tricky to visualize. If Dave Lee or anyone else reading this thread does not have a sufficient understanding of the fundamentals of inertial and gravitational mechanics to understand what I am saying, that OK. We all need to start somewhere. We are here to assist you in gaining more understanding. It can be awkward in a forum like this. It is easier in a face to face dialogue in real time in a classroom, where we can ask each other for clarifications of a point immediately and make sketches to visually enhance our presentations and responses.

  30. #30
    Join Date
    Sep 2015
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    Quote Originally Posted by Hornblower View Post
    Let me try a somewhat different approach to making sense of the gravitational binding of the Moon to the Earth. It is not enough to just look at the magnitudes of the forces. Since the three bodies are in free fall, we must look at how the Earth and the Moon move in response to the gravitational forces on them.

    Dave Lee’s corrected calculations are in good agreement with mine, indicating that the Sun’s gravity pulls on the Moon about twice as hard as does the Earth’s gravity. If we could magically immobilize the Earth with some sort of opposing nongravitational force and leave the Moon free to gravitate whichever way the vector sum of the gravitational forces is directed, then yes indeed the Sun would strip the Moon free of Earth’s grasp. However, in the real universe the Earth will fall toward the Sun nearly in unison with the Moon. Let’s look at the case of the new Moon position, where the Moon is about 1 part in 400 closer to the Sun. From the inverse square law we find that the Moon accelerates about 1 part in 200 more quickly than does the Earth. That shows the same sort of separation tendency as if the Sun were absent and something with about 1/200 of the solar gravitational force was tugging on the Moon but not on the Earth. Since the Earth’s gravity is about 100 times as strong as this hypothetical force, the two bodies remain gravitationally bound.

    Clear as mud? I know these compound motion cases can be tricky to visualize. If Dave Lee or anyone else reading this thread does not have a sufficient understanding of the fundamentals of inertial and gravitational mechanics to understand what I am saying, that OK. We all need to start somewhere. We are here to assist you in gaining more understanding. It can be awkward in a forum like this. It is easier in a face to face dialogue in real time in a classroom, where we can ask each other for clarifications of a point immediately and make sketches to visually enhance our presentations and responses.
    Thanks
    I do understand your point of view.
    Actually, I fully accept the idea that the Moon orbits around the Earth, although, it can get twice stronger gravity force from the Sun.
    It seems to me that even if the gravity force of the Sun will be 4 times stronger than the Earth, the Moon will still continues to orbit around the Earth.
    Based on my knowledge as Electronic Engineer, I will call it "Threshold".
    Which means, that the moon stay at that state because it was already position at that state (bonded by Earth gravity).
    However, based on the "Threshold" analogy, the moon won't set that gravity connection at the first stage without getting higher gravity force by the Earth.
    Do you agree with that analogy?
    If so, we have to ask: Why the Earth could offer higher gravity force at the first moment?
    Last edited by Dave Lee; 2018-Jan-28 at 08:28 PM.

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