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How does the moon pull on the earth?

  1. May 7, 2014 #1
    I hope I posted this in the right section. Here are two questions I have.

    1) If I pull on a very heavy object, that object's mass resists my attempt to accelerate it. This is known as inertia. When that object resists my attempt to accelerate it, it essentially pulls back on me with an equal force, correct?

    2) Earth's gravity pulls on the moon. That pull accelerates the moon. The moon's mass resists that acceleration (inertia). When the moon's mass resists that acceleration with its inertia, does that cause the moon to pull back on the earth with an equal force?
     
    Last edited: May 7, 2014
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  3. May 7, 2014 #2

    Nugatory

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    yes, and if you're pushing instead of pulling it will push back.

    yes. The earth is much more massive than the moon, so the moon's pull on the earth doesn't move the earth as much the earth's pull on the moon moves the moon - that's why we can draw pictures of the moon orbiting a stationary earth. But in fact the earth and the moon are both orbiting around their common center of gravity.
     
  4. May 7, 2014 #3
    So the moon pulls on the earth with an equal force from its inertia? I thought the moon pulled on the earth with an equal force from its gravity?
     
    Last edited: May 7, 2014
  5. May 7, 2014 #4

    Nugatory

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    The earth pulls on the moon and the moon pulls on the earth; the magnitude of both forces is given by Newton's ##F=GM_eM_m/r^2## where ##M_e## is the mass of the earth, ##M_m## is the mass of the moon, and ##r## is the distance between them. The mass (and hence inertia) of both bodies appears in this equation and cannot be separated, so it's best to think of the force between them as coming from both masses.
     
  6. May 7, 2014 #5
    Yes I am familiar with that equation. Thanks for your answer.

    Imagine two masses, m1 and m2. If m1 accelerates m2, m2's inertia resists the acceleration. This resistance to acceleration by m2 pulls back on m1 with an equal magnitude. When it pulls back with an equal magnitude, it becomes a force that is acting on m1. So now m2 is pulling on m1 with an equal force. if m2 is now pulling on m1 with an equal force, that equal force is now acting as though it is m2's gravity isn't it? There is an equal force between m1 and m2 at this point. That is what we define as gravity isn't it?
     
  7. May 8, 2014 #6

    UltrafastPED

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    Newton's Third Law of Motion.


    Newton's Universal Law of Gravitation describes the forces - and allows us to calculate them (via Newton's Second Law of Motion); the Third Law of Motion still applies to the result, as all force laws must be consistent with it.
     
  8. May 8, 2014 #7

    A.T.

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    The cause-effect-chain you are building here is not what Newtons 3rd says. Both forces act simultaneously. You cannot say which causes which. The common terminology of "action & reaction" is very misleading here.
     
  9. May 8, 2014 #8

    NascentOxygen

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    Any object resists, even the tiniest. If it has mass, it has inertia.
     
  10. May 8, 2014 #9

    sophiecentaur

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    I don't see the point of introducing the term "inertia" into this explanation. It is a rather wooly term (you won't find it defined in many text books) and doesn't come with a special Unit or a value, when you are dealing with simple classical mechanics.
    You can discuss this sort of situation with
    1)Gravitational Force = GM1M2/d2

    2)Force = Mass times Acceleration
    and

    3)Centripetal Force = Mv2/r
    which don't involve Inertia, yet they give you the answer to what will happen in Orbits - and in most other situations we have to deal with.

    Add in Newton's 3rd law and you can predict what will happen with two masses orbiting round each other. The cm of the Earth Moon system is way below the surface of the Earth and the effect of the orbit of each body around this point is to give the Earth a monthly 'wobble', as the Moon 'goes round' it.

    [Edit: But the terms 'inertia' is absolutely fine - inspired, even - for your name! :smile:]
     
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