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Equation Rearrangement inv. Point of Gravitational Equilibrium btw Earth and Moon

  1. Jan 2, 2013 #1
    1. The problem statement, all variables and given/known data

    The mass of the Moon is 7.35*10^22kg. At some point between the Earth and the Moon, the force of gravitational attraction cancels out. Calculate where this will occur, relative to Earth.

    Me = 5.98*10^24 kg
    Mm = 7.35*10^22 kg
    D = 3.84*10^8 m
    g = 6.67*10^-11 Nm^2/kg^2
    **Mx(object) = 1 kg

    2. Relevant equations

    r(moon) + r(earth) = D
    r(earth) = D-r(moon)

    FeG = FmG
    G(MeMx)/r^2 = G(MmMx)/(D-r)^2 *I had some trouble deciding where to put the (D-r)


    3. The attempt at a solution
    Here's where I had difficulty...I simplified the equation into a quadratic that yields two answers:

    r^2(Me-Mm) - 2DMe*r +MeD^2 = 0

    Plugging in the numbers, I get:
    r = 4.319*10^8 m or r = 3.46*10^8 m (this appears to be the correct answer).

    I researched this question and found the equation:

    r = (D[Me-(MeMm)^1/2]) / (Me-Mm)

    I was wondering how to isolate r and simplify the equation in order to get it to look like the one above. I am having trouble understanding where the 'Mm' term in the '(MeMm)^1/2' came from; also, does the '^1/2' come from taking the square root or does it have something to do with the '2' in the initial equation being taken out? Aside, what is a suitable explanation regarding the existence of the fallacious answer in the initial unsimplified quadratic?

    Thank you very much, I really appreciate the help.
     
  2. jcsd
  3. Jan 2, 2013 #2

    symbolipoint

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    Shown symbolically only,

    D is distance between centers of earth and moon;
    re is NOT radius but distance from 1 kg mass to center of earth;
    rm is NOT radius but distance from 1 kg mass to center of moon;
    D=re+rm
    Mx=1, the 1 kg mass between earth and moon;

    Forces between the mass and either large body are equal will be equal at some re and rm


    [itex]\[
    \frac{{GM_e M_x }}{{r_e^2 }} = \frac{{GM_m M_x }}{{r_m^2 }},\quad \quad D = r_e + r_m {\rm }
    \]
    [/itex]
    Solve for either re or rm .
     
  4. Jan 2, 2013 #3

    ehild

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    Your quadratic equation gives two distances from the Earth where the forces are equal: one distance shorter, the other distance longer that the distance between Moon and Earth. There are tow points really where both the Earth and the Moon attracts a body with equal forces. But you were asked about the point in between, so choose the smaller root of your quadratic equation.

    ehild
     
  5. Jan 2, 2013 #4
    While researching, I came across the concept of "Lagrange Points." While the answer, which I labeled correct, of the quadratic conforms to one of these (i.e. the point between the Earth and the Moon) the other answer seems to correspond to the point beyond the moon. Is this correct? Simple if true I guess, I should have seen it.

    Also, I have been out of the math game for quite some time...just coming back to it after several years. The main problem I'm having is changing the quadratic into the isolated variable equation I presented. The book I saw it in does not do this. I've tried taking derivatives as well as various algebraic manipulations (more than a few of which I'm sure are incorrect) but am unable to figure out how it was done. Could someone perhaps point me toward a specific online source (or even a printed one I can order on amazon) that will detail such operations or point me in the right direction (with specific attention to those areas I mention...i.e. the (MeMm)^1/2 section).

    Thank you so much for your explanation ehild, it was just enough to point me in the right direction while not spoiling the benefit of thinking out the problem for myself. Thank you.
     
  6. Jan 2, 2013 #5

    haruspex

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    No, Lagrange points are different. You found the two places, on a straight line through the two centres, where the magnitudes of the forces are equal. At one of those points the forces cancel, but at the other they add.
    At a Lagrange point the forces do not cancel. Instead, the (vector) sum of the forces is the same as would be experienced if the small satellite were to replace the smaller of the two other bodies. That is, it is a point where the satellite can orbit the larger of the other two with the same period as the smaller. Given any two bodies in orbit around each other, there are five Lagrange points: three on the line through their centres (one between them, one each beyond them), one 60 degrees ahead in orbit and one 60 degrees behind. At the Lagrange point between Earth and moon, the net gravity is towards the Earth; the gravity cancellation point would be closer to the moon.

    Returning to your question, it seems to be just a matter of solving quadratic equations. I suggest you Google that.
     
  7. Jan 2, 2013 #6

    ehild

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    Your equation "FeG = FmG" means forces equal in magnitude, and says nothing about the direction. And there are two points along the line connecting Earth and Moon, where the forces are of the same magnitude. Between Earth and Moon, these forces are opposite, so they cancel. A the point beyond the Moon, they point in the same direction, toward the Earth.
    The URL below explains the quadratic formula: The solution of a quadratic equation. You can use the formula also with symbols.

    The general form of the quadratic equation is ax2+bx+c=0
    Your equation is : r^2(Me-Mm) - 2DMe*r +MeD^2 = 0
    a=Me-Mm b=-2DMe c=MeD2.

    The quadratic formula

    [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    is applied to your equation, giving

    [tex]r=\frac{2DM_e\pm\sqrt{(2DM_e)^2-4(M_e-M_m)M_eD^2}}{2(M_e-M_m)}[/tex]

    Expand the parentheses and simplify the expression under the square root:

    [tex]\sqrt{(2DM_e)^2-4(M_e-M_m)M_eD^2}=\sqrt{4DM_e^2-4DM_e^2+4M_mM_eD^2}=\sqrt{4D^2M_eM_m}[/tex]
    You can factorize the square root:
    [tex]\sqrt{4D^2M_eM_m}=\sqrt{4D^2}\sqrt{M_eM_m}=2D\sqrt{M_eM_m}[/tex]

    I think you can proceed from here.

    ehild
     
  8. Jan 3, 2013 #7
    Ha...it never donned on me to put it into a quadratic and then simplify it. Thank's a lot...I look forward to playing with it and manipulating it around.

    You're truly great--I really appreciate it. Thanks again ehild... ****stars
     
  9. Jan 3, 2013 #8

    ehild

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    You are welcome. And thank you for the compliment. :smile:

    ehild
     
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