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Law of Gravitation

  1. Sep 3, 2004 #1

    cepheid

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    I just have a question about relationships between physical quantities. Is it true that if:

    [tex] z \propto x \ \textrm{and}\ \ z \propto y [/tex]

    then [itex] z [/itex] will always be in the form:

    [tex] z = Cxy [/tex]

    Where [itex] C [/itex] is a constant?

    What prompted me to ask was Newton's Universal Law of Gravitation. In the textbook I have, it wastes no time arriving at the typical equation by simply stating that Newton determined that the gravitational force between two masses was directly proportional to the product of those masses and inversely proportional to the square of the distance between them. So my question could be restated: In the expression:

    [tex] F = G\frac{m_{1}m_{2}}{r^2} [/tex]

    Is the fact that F is proportional to the product of the masses a necessary consequence of the mathematics? I.e., does the math demand that it be so in order to satisfy the condition that the gravitational force is directly proportional to (varies linearly with) m1 and also varies linearly with m2?

    Or is there an alternative (non-physical) way of formulating the expression that still satisfies the stated condition. For instance, would:

    [tex] F \propto m_1 + m_2 [/tex]

    fit the bill mathematically? It seems to me that when you consider the relationship between F and each one of the masses, all other things being equal, it is still linear.
     
  2. jcsd
  3. Sep 3, 2004 #2
    Actually, if z is proportional to x, then z = kx, where k is the constant of proportionality, and likewise, if z is also proportional to y, then z = ry, where r is again the constant of proportionality. So, we can right:

    kx = ry, but I don't see how to get z = Cxy, unless you want to know what z^2 is, then you have z^2 = krxy or Cxy, where C = kr
     
  4. Sep 3, 2004 #3

    Tide

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    If [itex]z^2 = Cxy[/itex] then z would vary as [itex]\sqrt x[/itex] and as [itex]\sqrt y[/itex]. Obviously, that's inconstent with the original assertions.
     
  5. Sep 5, 2004 #4

    cepheid

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    Umm...(as Tide pointed out)...no. z cannot be expressed as a function of only one of either x or y. If z if proportional to both x and y, then any expression for z must include both variables on which z is dependent i.e. it must involve some combination of the two. The question I was asking was, in the case of the gravitational force law, this combination takes the form of a product. Is there any other mathematically (if not physically) valid form e.g. like the sum, as I asked in my first post, that still satisfies the condition that F is proportional to m1 and F is also proportional to m2?
     
  6. Sep 5, 2004 #5

    robphy

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    It seems to me...
    If [itex] F \propto m_1 [/itex], then doubling [itex]m_1[/itex] will double [itex]F [/itex].
    However, if [itex] F \propto (m_1 + m_2) [/itex], doubling [itex]m_1[/itex] will not double [itex]F[/itex] unless [itex]m_2=k m_1[/itex].
     
  7. Sep 5, 2004 #6

    HallsofIvy

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    If [tex] z \propto x[/tex] then, all other things held constant, z= Cx. If [tex]z \propto y[/tex] then, all other things held constant, z= Dy. In the second equation, x is one of the things being held constant: If y is held constant and x varies then I must still have z= D y except that, now, y is constant and D is varying: z= Cx= (Ex)y.

    There is an unfortunate ambiguity in the term "linear". In very basic math, y= ax+ b is called "linear" because its graph is a line. In more advanced math, the only "linear" functions are of the form y= ax.

    As robphy pointed out, If z= Exy and you double x, z= E(2x)y= 2(Exy)= 2z. If you double y, z= Ex(2y)= 2(Exy)= 2z. On the other hand, if z= x+ y and you double x,
    z= 2x+ y which is NOT 2z.
     
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