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Newtons equation for gravitational force

  1. Jul 9, 2007 #1
    The question: assuming that the moon circles the earth in an orbit of radius d show thatthe periodic time T of the moon may be expressed as: T=2pi d sqrt d/r sqrt g where r is the radius of earth and g is acceleration due to gravity on earths surface


    i took up physics 2 to 3 months ago on my own and im definetly missing something here
    its eay to get Tsqrd=4pi Rcubed/GM but iv only readas faras chapter 12 and i cant think how to figure this out



    can get T=2pi r sqrtr/sqrt GMbut not much more :confused:
     
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  3. Jul 9, 2007 #2

    Kurdt

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    This is fairly simple using Newtons equation for gravitational force and some circular motion equation for force involving period. Can you attempt to show the steps you've gone through already?

    I'll write in latex the equation you're trying to find.

    [tex] T=2\pi \frac{d\sqrt{d}}{r\sqrt{g}} [/tex]

    Equations you'll need:

    [tex] F = md \frac{4\pi^2}{T^2} [/tex]

    [tex] F = G \frac{Mm}{d^2} [/tex]

    [tex] g = G \frac{M}{r^2} [/tex]
     
    Last edited: Jul 9, 2007
  4. Jul 9, 2007 #3
    how is force equal to mass by distance by 4 pi squared all over period squared.i should prob know :{ i think i may be more lost than i thought. thanks for the equations i'l be back if i make progress! il read the chapterinside oput again.
     
  5. Jul 9, 2007 #4
    pardon my typeing errors. physics is melting my brain. moreso! lol.
     
  6. Jul 9, 2007 #5

    Kurdt

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    The first equation I posted is used when dealing with circular motion. I do not know if you've studied circular motion yourself.

    There is a page here with various links to concise tutorials on the subject.

    http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html
     
  7. Jul 9, 2007 #6

    Dick

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    I think the only equations you really want to remember are i) F=ma, ii) F=GMm/r^2, and iii) a=v^2/r (for circular motion). Kurdt's first equation is just F=ma combined with a=v^2/r (and v expressed as 2*pi*r/T). Since you say you already have
    T=2*pi*r*sqrt(r)/sqrt(GM), then you are almost there. Just look at F=mg=GMm/r^2 (Kurdt's last equation and combination of i) with ii)). Solve for GM and put that into your expression for T. Be careful, there are two different radii around (the radius of the orbit d and the radius of the earth r) - be sure you use the correct one in each expression - because I haven't.
     
  8. Jul 9, 2007 #7
    i dont understand how earths radius and gravity come into it. centripetal acceleration is equal to the force of gravitational attraction. the equations say its at the height or distance of the moons orbit d.
    m*v squared/r =GMm/d squared. d is the distance of moons orbit from centre of earth.
    the acceleration of grav or force of grav on earth is completely different to that of moon in orbit with radius d as opposed to earths radius r??
    am i raving?
     
  9. Jul 9, 2007 #8

    Dick

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    You aren't raving, but you are looking in the wrong direction. You are correct that g at the earths surface has nothing to do with the orbit of the moon. But for some reason in the equation you are being asked to show they have substituted an expression for sqrt(GM) involving the earth's radius and g. Find an equation to solve for GM involving those two variables.
     
  10. Jul 9, 2007 #9
    thanks! :}

    guys ye dont know how much ye've actually just helped me and now that i go over wat ye said it seems very clear! i think iv got it ;} right so it goes:
    m*vsrd/d=GMm/dsqrd where d is moon orbital radius so vsqrd is GM/d.
    then time T=2pid/v. make t a squared, sub in v squared end up with:
    Tsqrd=4pi sqrd *dcubed/GM therefor Tsqrd*GM=4pi sqrd*dcubed and magically mathematically divide by the earth radius squared (rsqrd) so that the T sqrd*GM/rsqrd becomes T sqrd*g. divide across get roots and end up with desired mad answer
    T=2pid*sqrtd/r*sqrtg.
    now i can sleep! you guys are brill:!!) wen i figure out this physics stuff i promise il try to work out that latex thing that makes the maths look nicer
     
  11. Jul 9, 2007 #10

    Kurdt

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    I'm not personally very fond of questions like this that want you to make seemingly ridiculous substitutions. I'd personally prefer to leave it in terms of constants like G and the mass of the Earth. I suppose it helps you with your maths anyway.
     
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