anonymous
  • anonymous
hi. we know gravitational force F is prop. to m1m2, it is also inversely prop. to d^2, we then combined these two proportions and wrote that F is prop. to m1m2/d^2. How can we do that? does that mean if x is prob to a, and x is prop to b, then x is prop to ab? im confused.
MIT 8.01 Physics I Classical Mechanics, Fall 1999
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anonymous
  • anonymous
hi. we know gravitational force F is prop. to m1m2, it is also inversely prop. to d^2, we then combined these two proportions and wrote that F is prop. to m1m2/d^2. How can we do that? does that mean if x is prob to a, and x is prop to b, then x is prop to ab? im confused.
MIT 8.01 Physics I Classical Mechanics, Fall 1999
katieb
  • katieb
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anonymous
  • anonymous
yes. we can just multiply.
anonymous
  • anonymous
but i need a mathematical explanation.
anonymous
  • anonymous
Mathematically, i dont know how to prove it. But if u multiply it, then the resulting expression will satisfy the required conditions. which are, F will still be proportional to m1m2 and F will still be inversely proportional to R^2. So inorder to represent both using only one expression, we write it that way.

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anonymous
  • anonymous
but it doesnt work that way
anonymous
  • anonymous
y doesnt it? we need an expression that satisfies both conditions. and it does. whats wrong?
anonymous
  • anonymous
it doesnt because that would lead to the confusion that i posted about, x prop to a and xx prop to b doesnt mean x prop to ab....
anonymous
  • anonymous
it does. x prop ab cos if u now double ab, it could be taken as doubling a and keeping b the same or vice versa. and the result would be a doubling of x.
anonymous
  • anonymous
lets say x is prop to a, then x=k1a similarly, x=k2b if we multiply them, then x^2=k1k2ab or x^2 is prop to ab (k1k2 is constant) so u see, x^2 will be prop to ab, not x itself.
anonymous
  • anonymous
Why do u multiply? you just need an expression that satisfies both conditions. F is never equal to k*m1*m2 It is ONLY PROPORTIONAL to m1*m2. The espression for force is not derived in the way u mentioned. newton just said that force would increase when m1*m2 increases, and would decrease proportional to r^2 So we now need an expression that can satisfy both these conditions. and F = Gm1m2/r^2 Thats all. it DID NOT come from F = k1*m1m2 and F = k2/r^2
anonymous
  • anonymous
yes it did, it came from F=Gm1m2/r^2 where G is the constant of proportionality. i guess there are a few ideas of proportionality u haven't come across yet.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Quote: "i guess there are a few ideas of proportionality u haven't come across yet." Come on, @tanvirms! What @Saikam explained is perfectly clear; no need to sneer! Instead of complicating things with equalities, proportionality constants and so on, get back to common sense and think again what the word "proportionality" means. If X is proportional to a, b², 1/c ,1/d² , cos \(\alpha\) and 1/ln r at the same time, is means that there is ONE proportionality constant k so that: \(\large X=k\;\LARGE \frac{ab^2\cos\alpha}{cd^2\ln r}\)
anonymous
  • anonymous
but Vincent, shouldn't there be a mathematical explanation? again, why isn't x proportional to a+b? or a-b? why only ab?
anonymous
  • anonymous
i perfectly agree to your given equation....but we can only write that when the proportionality constant 'k' is same for every proportion, isn't it? what if the constant is different for ever proportion?
anonymous
  • anonymous
Your Q is valid Its a matter of an experiment. Initially people found that F grav. is proportional to masses of the objects and also to R(-2) and the when they varied both mass and distance they found the eqn. holds too. the above process neednt hold for everything on earth
anonymous
  • anonymous
thanks people! i emailed Dr.Math and got a very satisfying answer.... @srikrishnapriya thats what i wanted to know :D

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