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How do proportional relationships derive physics equations?

  1. Feb 5, 2016 #1
    in particular f=Gm1m2/r^2?sorry if my question sounds very irrelevant.if f is proportional to m1m2 it implies f=some constant times m1m2.okay.at the same time f is inversely proportional to r^2 .so force = some other constant times 1/r^2.okay.but in most places i see that what is done is they take take the two proportional relationships with respect to f and say "THEY ARE COMBINED" into one which is f is proportional to m1m2 times 1/r^2 or f is proportional m1m2/r^2.im just not getting how those two proportional relationships "combine" to give that one relationship from which obviously the law of gravitation comes.Can you please explain why exactly two proportional relationships can be combined like that?also why doesnt it apply same way in f proportional to m and f proportional to a to give f proportional to ma and then f= k ma. why just f= ma? i really want to understand this.THANKS FOR YOUR TIME.
     
  2. jcsd
  3. Feb 5, 2016 #2
    So the only other alternative would be F = (some factor) x (Gm1m2/r^2). That factor could mean too things:
    • it's a universal constant in which case it's just a matter of units of G;
    • it's a quantity that depends on other physical properties such as time, substance, etc.
    We haven't found any evidence of the latter. G really seems to be universal and constant. So I guess the answer to your question is you need three propositions:
    • the force is proportional to m1;
    • the force is proportional to m2;
    • the force is proportional to 1/r^2;
    • the force isn't proportional to anything else.
    Therefore F is proportional to m1m2/r^2 or, in other words, F = Gm1m2/r^2
     
  4. Feb 5, 2016 #3

    HallsofIvy

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    If I know that "f is proportional to h" then I know that, as long as everything else is held fixed, f is equal to a constant times h: f= Ch. Of course, a lot may be hidden in that "C"- things that could be allowed to vary.
    If I also know that "f is inversely proportional to [itex]r^2[/itex]",then I know that, everything except r being fixed, [itex]f= \frac{K}{r^2}[/itex]. The two together are possible only if there is a "[itex]\frac{1}{r^2}[/itex]" "hidden" in that "C" or, conversely, "h" "hidden" in that "K". Replacing the "C" in f= Ch with [itex]\frac{K}{r^2}[/itex],we have [itex]f= \frac{Kh}{r^2}[/itex]. Replacing the "K" in [itex]f= \frac{K}{r^2}[/itex] with Ch, we have [itex]f= \frac{Ch}{r^2}[/itex], exactly the same formula, differing only in what we call the "constant of proportionality.
     
  5. Feb 5, 2016 #4

    jbriggs444

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    f = k ma is correct.

    It is just that we have chosen our units of force, mass and acceleration so that k = 1 in those units. If you were to use pounds force, pounds mass and feet per second squared then the corresponding k would be approximately 1/32.17 [32.17 is a standard acceleration of gravity, "g", expressed in feet per second squared].
     
  6. Feb 6, 2016 #5
    hmm.very thankful to everyone who answered but i think where im actually getting stuck is nothing high level.im just not getting why a proportional to b and a proportional to c implies a proportional to bc .i get it that if b changes by factor x then a changes by factor x and at the same time if c changes by y then that a.x would change by y resulting a net change a into a.x.y and also i mean yeah when you see a proprtional to bc it is evident that b,c getting changed by x,y makes a change by xy.but i want some more derived proof or something i guess.
     
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