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Homework Help: Graviation With Two Bodies / 1 Dimension

  1. Jan 7, 2015 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find the point at which FG attracting to the moon = FG attracting to the earth...

    5.98e24 kg = Earth Mass
    7.35e22 kg = Moon Mass
    3.84e8 m = Distance from Earth to Moon


    a = 3.84e8 - b
    b = 3.84e8 - a
    a+b = 3.84e8

    2. Relevant equations

    g = G m / r2

    (6.67e-11 * 5.98e24)/a² = (6.67e-11 * 7.35e22)/b²

    3. The attempt at a solution

    I wolfram'd this but it didn't give a step by step solution because I don't have a paid acct

    [(6.67e-11 * 5.98e24)/(3.84e8 - b)^2 = (6.67e-11 * 7.35e22)/(3.84e8 - a)^2, 3.84e8 = a+b]

    a = 3.45677e8
    b = 3.83233e7

    This isn't for an assignment but I'm really curious as to how to do the calculus with two unknown variables. Unfortunately, I don't have calculus till next semester... If you could show your solution so I can study and figure out how you did it, I would be greatly appreciative.

    Thanks for your advice and help!
    Last edited: Jan 7, 2015
  2. jcsd
  3. Jan 7, 2015 #2

    Doug Huffman

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  4. Jan 7, 2015 #3
    Care to go into detail? Is this a Lagrangian point I've rediscovered or something? I'm more interested in the calculus used to find the variables a and b, then what it is known as, if you could be so kind... Thank you!
  5. Jan 7, 2015 #4


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    No, it's nothing to do with Lagrangian points, and it doesn't require calculus.
    It's better to keep all the algebra symbolic until the final step. You have ##\frac E{a^2} = \frac M{(D-a)^2}##. Multiply out and gather the a's together.
  6. Jan 7, 2015 #5
    (G * n) / (3.84e8 - b)² = (G * m) / (3.84e8 - a)²

    Can you do the algebra so I can see it step by step?
  7. Jan 7, 2015 #6


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    As I posted, it's better to keep everything symbolic. Don't plug in any numbers yet. This is an important stylistic matter. Keeping everything symbolic has many benefits. It makes it quicker to write out, easier to spot mistakes, easier for others to follow, avoids loss of precision, quicker to correct when you find a mistake...
    No, we don't do that. I said to multiply out the expression I posted. Do you understand what that means?
  8. Jan 7, 2015 #7
    No... I find it hard to understand because I'm a hands-on learner, I can follow along to a finished solution and research along the way for things I don't know, but I can't find an example of this to reverse engineer and this is harder than what I know how to do on my own without help.
  9. Jan 7, 2015 #8

    Ray Vickson

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    Can you see why the equation
    [tex] \frac{E}{a^2} = \frac{M}{(D-a)^2} [/tex]
    is equivalent to (i.e., essentially the same as) the equation
    [tex] \frac{a^2}{E} = \frac{(D-a)^2}{M}? [/tex]
    If you cannot see it I would suggest you need to enhance your algebra background; if you can see it, can you then see what the next steps would be in the problem of finding the value of ##a##? Again, if you cannot see these next steps you need to enhance your algebra background. If you can see it, then carry out the steps to see what you get.
  10. Jan 7, 2015 #9
    I know A/B = C/D becomes D/B = C/A

    (G * n) / (d - b)² = (G * m) / (d - a)² == (d - a)² / (d - b)² = (G * n) / (G * m)

    But I can't find it this way I don't think because of the 2 variables ... sorry, Its not that I'm not trying or just want a short cut, this isnt even homework I'm just curious.
  11. Jan 7, 2015 #10
    I just don't know where you're getting

    [tex] \frac{E}{a^2} = \frac{M}{(D-a)^2} [/tex]
    is equivalent to
    [tex] \frac{a^2}{E} = \frac{(D-a)^2}{M}? [/tex]
  12. Jan 7, 2015 #11

    Ray Vickson

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    Can you not see that
    [tex] \frac{1}{2} = \frac{7}{14}[/tex]
    says the same thing as
    [tex] \frac{2}{1} = \frac{14}{7}?[/tex]
    That's all there is to it.
  13. Jan 7, 2015 #12


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    The general rule for algebraic equations is that you can do exactly the same operation to both sides of the equation. Ray inverted both sides, i.e. replaced x/y by y/x.

    There are a couple of places where you have to be careful.
    You must not divide both sides by zero. Division by zero is illegal anyway, but you might do it without realising:
    4(x-y) = 2(x-y)
    Dividing both sides by x-y:
    The problem is that x-y might be zero, so whenever you divide both sides by something you have to say "either (that something) is zero or..". In this case, the solution is x-y=0 so x=y.

    Another situation is where an operation can be ambiguous, like taking a square root.

    I had advised you to 'multiply the expression out'. Whenever you have fractions in the equation one option is to multiply both sides of the equation by the denominator. So x/y = z becomes x = y*z. Whether you do that or use Ray's inversion method is a matter of taste. They will lead to the same answer.
  14. Jan 7, 2015 #13


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    It seems contradictory for you to claim you're a hands-on learner, yet you are asking someone else to do the problem for you. You'll learn more if you take a stab at the problem yourself and ask questions when you get stuck.

    You started with two equations:
    $$\frac{G M_\text{Earth}}{a^2} = \frac{G M_\text{Moon}}{b^2}$$ and
    $$a + b = D.$$ To attack this problem, the idea is to write (only) one variable in terms of the other, so if you want to eliminate ##b##, you can solve the second equation for ##b## to get ##b=D-a## and then replace every instance of ##b## in the first equation by ##D-a##, giving
    $$\frac{G M_\text{Earth}}{a^2} = \frac{G M_\text{Moon}}{(D-a)^2}.$$ Now you have an equation with just one variable, ##a##, to solve for. This is the equation haruspex is asking you to solve. The mistake you made was replacing ##a## by ##D-b##, which is okay to do but not useful as it gets you back to an equation with two variables.

    The next step I would take is to cancel any common factors on both sides of the equation, and then perhaps use Roy's hint or whatever method you feel comfortable with to get ##a## in the numerator on both sides.
  15. Jan 7, 2015 #14


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    Good point.
    That sounds to me like playing a piano piece by copying the finger movements of someone else playing the same piece. You don't actually learn to pay the piano that way. At some point you have to discern the principles and try playing straight from the score.
  16. Jan 7, 2015 #15
    But the terms in your equation, what is E for ?
    Thanks, this is exactly what I need to hear to get going.
  17. Jan 7, 2015 #16
    Thank you for everyone's help, and sorry for sounding grumpy and lazy. I really appreciate your input and reference and working with me.
  18. Jan 7, 2015 #17


    Staff: Mentor

    Well, of course it's easier to look at someone else's work than to grind it all out yourself, but as already pointed out, you don't learn much by merely observing. Also, as already hinted at, our philosopy here at Physics Forums is to make the poster do the bulk of the work, and provide guidance along the way when he or she takes a wrong turn. A major part of learning is taking those wrong turns, and eliminating the steps that produce those dead ends.
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