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Easy Newtonian Gravitation viewed from a rusty perspective.

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data

    (1) Fsm + Fem = Mm x Ams
    (2) Ams = Ame + Aes
    (3) Res ~= Rms

    Show that (4) Fem ~= Mm x Ame

    'Fsm' is the force between the sun and moon, 'Fem' the force between earth and moon, etc.
    'Mm' is the mass of the moon, 'Ams' the accelleration of the moon round the sun, 'Aes' the accelleration of the earth in the frame of the sun, etc.
    'Res' is the distance between the earth and sun, etc.
    And '~=' is a rubbish looking 'approximately equal to' sign.

    2. Relevant equations
    Oh... see above.

    3. The attempt at a solution
    Right, well, trust me, I'm not trying to cheat, I do have a physics degree gained in 2001 (a mere 2.1), but am finding myself more rusty/stupid than I realised. I'm going through my undergraduate physics book slowly, refreshing my memory and destroying my ego simultaneously.

    I'm not going to detail my working (although written from my future self's proof reading perspective, it would have been easier), but effectively my attempt to solve this has centred around the given statement (3) that the distances between the earth and sun, and the moon and sun, can be assumed to be approximately equal. The only relevance I can see for this statement is to enable you to make the assumption that Aes and Ams are also approximately equal (the mass of the earth and moon being irrelevant obviously and the only variable being R). However, once I've made this assumption I'm stuck with a problem. Looking at (2), Ame now may as well ~=0. Which is completely the opposite of what I want. Clearly, by assuming Res and Rms are equal, I'm supposed to take Ams and Aes out of the picture. But I don't know how!

    I've tried using (1), which is an addition of the forces around the moon, resulting in it's circular path around the Sun, constructing a similar expression for the earth and rearranging... but no luck.

    Any help will be appreciated. My book only gives the answers to problems with a numerical answer, so I'm never going to find out otherwise!

  2. jcsd
  3. Jan 22, 2009 #2


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    Homework Helper

    Welcome to PF.

    First of all have a free " ≅ ".

    I think it is a fairly simple vector problem.

    Plug the identity of 2) into 1).

    What they are saying is that using Res ≅ Rem,

    I think you can infer that using the relationship A = V2/R

    that Fsm ≅ MmAes
  4. Feb 12, 2009 #3
    So, plugging 2) into 1) gives:

    Fsm + Fem = (Mm x Ame) + (Mm x Aes)

    Then eliminating Fem = Mm x Ame

    gives Fsm = Mm x Aes

    If you assume Aes ≅ Ams (given A = V2/R and that Res ≅ Rem)

    All good? I feel like that was too easy.
  5. Feb 12, 2009 #4


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    Homework Helper

    Sometimes the math is easy, but the understanding harder to grasp.

    (This makes for a lot of sloped foreheads.)

    Given that they give you Res ≅ Rem... it looks to me any way like it just falls out the bottom after cranking not even half a turn.
  6. Feb 13, 2009 #5
    Awesome. Thanks for that! I truly am rusty. However, leaving it a couple of weeks and looking at some electromag for a while made it a lot easier with fresh eyes. Your help was much appreciated!
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