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Newton's second law and how to calculate the force on the mo

  1. Jan 10, 2016 #1
    I, earlier were studying Newton's second law F=ma, or rather the free fall of objects, F=mg.
    A thought occurred to me, how do I apply this formula to the moon with the Gravity of 1.622m/^2 and came up with F=m(g/6), however, g/6 = 1.63m/s^2

    Normally I'd just use F=ma, but assuming people one day would live there, they'd use g = 1.622m/s^2, which I think many would find confusing. So I tried to make a generalized formula. Is there any way to improve upon the formula?
     
  2. jcsd
  3. Jan 10, 2016 #2

    Student100

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    What do you mean "apply it to the moon"? Do you mean objects falling on the moon, or the moon-Earth system?
     
  4. Jan 10, 2016 #3
    Objects falling towards the moon*
     
  5. Jan 10, 2016 #4

    Student100

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    Simply redefine g. That's all that needs to be done.
     
  6. Jan 10, 2016 #5
    Yeah, but what I was asking is there any way to use earths gravity as a template without actually redefining g, which now I think about, I did not specify, I got the formula f=m(g/6). But is there a better way of doing it? My formula skills aren't that great :/
     
  7. Jan 10, 2016 #6

    Student100

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    Why not use Newton's law of gravitation in that case?

    $$F=\frac{Gm_1m_2}{r^2}$$

    Although, g in F=mg is just a dummy variable, it doesn't have an intrinsic meaning until we give it one, so I'm not really understanding where you're going.
     
  8. Jan 10, 2016 #7
    We are going through that in friday, I have a basic concept of how to use that but not indept unfortunately.
     
  9. Jan 10, 2016 #8

    Student100

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    It doesn't use Earth as a template either, (not sure why you'd want to), it uses the nature of the system and the gravitational constant.
     
  10. Jan 12, 2016 #9
    Free fall acceleration on Earth is just ##g=G\frac{M_{Earth}}{R^2_{Earth}}##, here ##M_{Earth}## is Earth mass, and ##R_{Earth}## - its radius. For Moon you get ##g=G\frac{M_{Moon}}{R^2_{Moon}}##. Now if you want to express free fall acceleration for Moon in terms of Earth g you just need to take relation ##g_{Moon}/g_{Earth}##. So you will get ##g_{Moon}=\frac{M_{Moon}}{M_{Earth}}\frac{R^2_{Earth}}{R^2_{Moon}}g_{Earth}##
     
    Last edited: Jan 12, 2016
  11. Jan 12, 2016 #10

    sophiecentaur

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    It's interesting that the formula for the gravitational acceleration at or above the surface is true, even if the density of the planet varies with depth (i.e a massive iron core or shell) It's just the total mass and the radius that count. It makes life a lot simpler :smile:.
     
  12. Jan 13, 2016 #11
    Yeah I know, I had the same thought when I was thinking about it ^•^
     
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