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Homework Help: Agravic point between planets

  1. Jan 20, 2005 #1
    Ok, this extra credit assignment is driving crazy. I keep trying to put it away and work on something important, but it can't get it out of my mind.

    The full question is:
    On the agonic line between Earth and Mars lies the agravic point with respect to the gravitational fields of the two planets. Determine the location of that point as a function of the locations of both Earth and Mars.

    I assume they are talking about the point between the planets where the force of gravity of the earth equals the force of gravity of mars.

    I've tried a few different things to get a formula that would tell me how far from earth the point would be if the distance between them was known. It keeps leading me to a quadratic with no real solution if the distance between them is large.

    Even if I can figure out what I'm doing wrong there, I'm not totally sure it's what the question is asking. What do they mean by "as a function of the locations of both planets"? Shouldn't I be free to set up my own coordinate system with Earth at the origin?

    Here's what I have so far:

    G = gravitational constant
    Me = mass of Earth
    Mm = mass of Mars
    x = distance between the center of the Earth and the agonic point (I think this should end up being the dependant variable)
    d = distance between the Earth and Mars (I think this should be the independent variable)

    Since I want to find the point where the gravitational field of the Earth is equal to the gravitational field of Mars, I set them equal to each other.

    [tex]\frac{G*Me}{x^2}=\frac{G*Mm}{(d-x)^2}[/tex]

    The G's cancel, after a little algebra, I end up with:

    [tex]d^2-2dx+(x^2-\frac{Mm}{Me}x^2)[/tex]

    If I try to solve this for x I end up with a negative under the square root sign if d is large.

    Can someone help me out with this? Am I even getting close?
     
  2. jcsd
  3. Jan 20, 2005 #2

    HallsofIvy

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    Science Advisor

    "If I try to solve this for x I end up with a negative under the square root sign if d is large."

    I don't see why you would. Using the quadratic formula, the "discriminant" (b2- 4ac) for this equation is
    [tex]4d^2- 4(1-\frac{M_m}){M_e}d^2}= 4d^2(\frac{M_m}{M_e})[/tex].

    That's positive as long as the masses are!
     
    Last edited by a moderator: Jan 20, 2005
  4. Jan 20, 2005 #3
    Thanks, I see it now. I had used B=-2 instead of B=-2d and then just kept making the same mistake over an over.
     
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