If 'a' is proportional to 'b' for some known range, then does it mean it would always

  • #1
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Hi

I'm not a math student. I can understand general basic math.

While solving many problems we take relations "a directly proportional to b" for granted and don't care to ponder what if it doesn't always hold all the way. Let me try to explain it a bit what I'm trying. If 'a' is proportional to 'b' for some known range, then does it mean it would always hold? I think gravitational force on some object is proportional to the distance from earth. But once the object has landed on earth, then this relation won't hold. Gravitational force is inversely proportional to the distance when being calculated from the surface of earth toward the center.


Is there any difference in saying "a directly proportional to b" or "a proportional to b"?

Please help me with this. Thank you for guidance and time.
 

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  • #2
tiny-tim
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hi jackson6612! :smile:
While solving many problems we take relations "a directly proportional to b" for granted and don't care to ponder what if it doesn't always hold all the way. Let me try to explain it a bit what I'm trying. If 'a' is proportional to 'b' for some known range, then does it mean it would always hold?
If it's exactly proportional, then yes.

But you can't be sure it's exact …

for example, gravitational https://www.physicsforums.com/library.php?do=view_item&itemid=269" is proportional to height (PE ~ mgh) near the Earth's surface, but is 1/distance a long way from the Earth

(btw, gravitational force is inversely proportional to the square of the distance)
Is there any difference in saying "a directly proportional to b" or "a proportional to b"?
they mean the same thing …

i think people use "directly proportional" as the opposite of "inversely proportional", but it's not necessary
 
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  • #3
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Hi Tim

Thank a lot for the help.

If it's exactly proportional, then yes.
I don't understand what you really meant by "exactly". Could you please elaborate a little? I think you meant if it holds all the way but it cannot always be guaranteed. Could you give me some more scientific and non-scientific examples?

gravitational potential energy is proportional to height (PE ~ mgh) near the Earth's surface, but is 1/distance a long way from the Earth
You mean as an object is taken too far away from earth, then its earthly PE gets reversed and becomes inversely proportional to distance. Right?

Thank you for all the guidance and time.
 
  • #4
tiny-tim
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By "exactly", I meant that when you make measurements you can never make them exactly. :wink:

From the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=269"

Derivation of mgh:

[tex]\Delta (PE)\ =\ \Delta(-mMG/r)[/tex]

[tex]=\ \frac{-mMG}{r_{earth}\,+\,H\,+\,h}\ -\ \frac{-mMG}{r_{earth}\,+\,H}[/tex]

which is approximately:

[tex]\frac{-mMG(H\,-H\,-\,h)}{r_{earth}^2}\ =\ \frac{mMGh}{r_{earth}^2}\ =\ mgh[/tex]
 
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  • #5
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You mean as an object is taken too far away from earth, then its earthly PE gets reversed and becomes inversely proportional to distance. Right?
From the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=269"

Derivation of mgh:

[tex]\Delta (PE)\ =\ \Delta(-mMG/r)[/tex]

[tex]=\ \frac{-mMG}{r_{earth}\,+\,H\,+\,h}\ -\ \frac{-mMG}{r_{earth}\,+\,H}[/tex]

which is approximately:

[tex]\frac{-mMG(H\,-H\,-\,h)}{r_{earth}^2}\ =\ \frac{mMGh}{r_{earth}^2}\ =\ mgh[/tex]
Tim, I do believe my follow-up question had a simply answer with not too much mathematics!:confused: So, could you please confirm what I said about PE? Thanks.
 
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  • #7
tiny-tim
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You mean as an object is taken too far away from earth, then its earthly PE gets reversed and becomes inversely proportional to distance. Right?
wrong … the "earthly PE" is a good approximation for the full (remote) version :wink:
 
  • #8
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Hi Tim

I'm not a math or science student, I'm quite a layman. Please keep your replies simple and explanatory. Thanks a lot.

By "exactly", I meant that when you make measurements you can never make them exactly.
When "a" is measured to one decimal place and so is "b". We find that when "a" is doubled, so does "b". That would mean that at least to one decimal place we are pretty sure that "a directly proportional to b" holds. Right?

gravitational potential energy is proportional to height (PE ~ mgh) near the Earth's surface, but is 1/distance a long way from the Earth
What does "a/distance" mean? Aren't you saying that when an object is not near the earth's surface, PE becomes inversely proportional to distance? Please guide me.

Thanks.
 
  • #9
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Hi

Is there any difference in saying "a directly proportional to b" or "a proportional to b"?

Please help me with this. Thank you for guidance and time.
To answer that question:

if a is directly proportional to b or proportional to b makes no difference:

If a changes then b changes to maintain the same proportion

a/b = c, where c is the constant of the proportion
or you could say a=c*b

So for this to be true, if I change a, I must change b to keep the equality. I can't change c because it is a constant.
 
  • #10
HallsofIvy
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Hi

I'm not a math student. I can understand general basic math.

While solving many problems we take relations "a directly proportional to b" for granted and don't care to ponder what if it doesn't always hold all the way. Let me try to explain it a bit what I'm trying. If 'a' is proportional to 'b' for some known range, then does it mean it would always hold? I think gravitational force on some object is proportional to the distance from earth.
No, that's not true. Gravitational force of the earth on a body is inversely proportional to the square of its distance to (the center of) the earth.

But once the object has landed on earth, then this relation won't hold. Gravitational force is inversely proportional to the distance when being calculated from the surface of earth toward the center.
No, that's also not true. The gravitational force is still inversely proportional to the distance from the center of the earth.

Now, that's because we can treat all of the mass of the earth as being concentrated at the center of the earth. Once the object goes beneath the surface of the earth, you can show that only the mass below your distance from the center is relevant. Since mass is proportional to the volume and so is proportional to the radius cubed, the gravitational force is proportional to [itex](1/r^2)(r^3)= r[/itex]. NOW the gravitational force is directly proportional to the distance to the center of the earth.

Is there any difference in saying "a directly proportional to b" or "a proportional to b"?
No, there isn't. While there are many different kinds of proportion that might require adjectives, just "proportional" is always interpreted as "directly proportional".

Please help me with this. Thank you for guidance and time.
 
  • #11
tiny-tim
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When "a" is measured to one decimal place and so is "b". We find that when "a" is doubled, so does "b". That would mean that at least to one decimal place we are pretty sure that "a directly proportional to b" holds. Right?
Yes. :smile:
What does "a/distance" mean? Aren't you saying that when an object is not near the earth's surface, PE becomes inversely proportional to distance?
The full version of PE measures the difference between the PE at radius R + h and the PE at infinity.

The near-Earth's-surface version of PE measures the difference between the PE at radius R + h and the PE at radius R.

In both these, R is the radius of the Earth, and h is the height above the Earth's surface.

The formulas look different because they're measured relative to different positions (infinity and the surface). If you add a constant to one of them, they are approximately the same (to see why, you really need to follow the maths).
 
  • #12
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I'm not a science or math student. So, please keep your replies simple. Thanks.

Halls: Thanks a lot.

Now, that's because we can treat all of the mass of the earth as being concentrated at the center of the earth. Once the object goes beneath the surface of the earth, you can show that only the mass below your distance from the center is relevant. Since mass is proportional to the volume and so is proportional to the radius cubed, the gravitational force is proportional to LaTeX Code: (1/r^2)(r^3)= r. NOW the gravitational force is directly proportional to the distance to the center of the earth.
I did not understand anything where you say LaTeX Code: (1/r^2)(r^3)= r, can you please explain it in simple terms - I mean the steps to derive this.

Tim: Thanks.

When "a" is measured to one decimal place and so is "b". We find that when "a" is doubled, so does "b". That would mean that at least to one decimal place we are pretty sure that "a directly proportional to b" holds. Right?
You okayed it. But suppose we know that at other decimal places beyond one decimal places relation doesn't hold. Then, how can we be sure that the relation will always hold for one decimal place to infinity?
 
  • #13
Office_Shredder
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I'm not a science or math student. So, please keep your replies simple. Thanks.

Halls: Thanks a lot.



I did not understand anything where you say LaTeX Code: (1/r^2)(r^3)= r, can you please explain it in simple terms - I mean the steps to derive this.
The gravitational force on an object is inversely proportional to the square of the distance between the object and the source of the gravitational force. I'm going to write this as F~1/r2 (~ is often used to mean proportional to). In fact we can go a step further, because it's also directly proportional to the mass of the source of the gravitational force.

F~M/r2. When outside of the earth, M is constant and the gravitational force is inversely proportional to the square of the distance from the center of the earth

Now suppose we're inside of the earth. The only part of the mass of the earth that counts is the part between you and the center of the earth (proof omitted). If p is the density of the earth (pretend it's constant for the moment), then the mass of the earth "underneath you" when you are a distance of r away from the center is [tex]M=\frac{4}{3} \pi p r^3[/tex].

So [tex]\frac{M}{r^2}=\frac{4/3 \pi p r^3}{r^2} = Kr[/tex] where K is a constant (4/3 p*pi to be specific). So the gravitational force when inside the earth is proportional to the distance from the center of the earth.

You end up getting a graph that looks like this
http://en.wikibooks.org/wiki/File:Grav_force_sphere.svg

The horizontal axis is distance from the center of the earth, the vertical axis is gravitational force. You can see the force is proportional to the distance for a bit, then changes to be inversely proportional to the square of the distance
 
  • #14
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Thank you very much, OS.

I do have some questions to ask but would like to wait for some time.

Where can I get codes to write things like this:
LaTeX Code: M=\\frac{4}{3} \\pi p r^3 .
LaTeX Code: \\frac{M}{r^2}=\\frac{4/3 \\pi p r^3}{r^2} = Kr

Do we also have to use such codes on CAS softwares?

What does this mean: Euclid alone has looked on beauty bare?

Thanks a lot for all the help.
 
  • #15
Office_Shredder
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It's called Latex. If you type [.tex] and [./tex] tags (without the periods) in your post, whatever you type in between is interpreted as Latex code. There are extensive examples online of how you can type latex, but for some specific ones you can just click on the code in our posts and a box will pop up telling you what we typed to generate it
 
  • #16
tiny-tim
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hi jackson6612! :smile:
You okayed it. But suppose we know that at other decimal places beyond one decimal places relation doesn't hold. Then, how can we be sure that the relation will always hold for one decimal place to infinity?
we can't :smile:
What does this mean: Euclid alone has looked on beauty bare?
it's a http://poetry.about.com/od/poems/l/blmillayeuclid.htm"

(http://en.wikipedia.org/wiki/Edna_St._Vincent_Millay#cite_note-20" recommends Sinclair, N. et al. (2006). Mathematics and the Aesthetic. New York: Springer)

Euclid alone has looked on Beauty bare

Sonnet from The Harp-Weaver and Other Poems (1923)

Euclid alone has looked on Beauty bare.
Let all who prate of Beauty hold their peace,
And lay them prone upon the earth and cease
To ponder on themselves, the while they stare
At nothing, intricately drawn nowhere
In shapes of shifting lineage; let geese
Gabble and hiss, but heroes seek release
From dusty bondage into luminous air.
O blinding hour, O holy, terrible day,
When first the shaft into his vision shone
Of light anatomized! Euclid alone
Has looked on Beauty bare. Fortunate they
Who, though once only and then but far away,
Have heard her massive sandal set on stone.

sooo … what do you think it means? :wink:
 
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  • #17
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I'm not a science or maths student. So, please try to be explanatory. Thanks.

OS: Thanks about the Latex.

The only part of the mass of the earth that counts is the part between you and the center of the earth
I have checked [tex]4/3 \pi r^3[/tex] is sphere's volume. Suppose, I'm standing inside a deep underground well, then where my feet touched the earth's surface, that entire part of earth will be cut along that level. Then, we aren't left with spherical earth. Where am I going wrong? Let me further clarify what I'm saying. I think equator divides earth into two hemi-spheres. Suppose my feet are just parallel to that equator line (which makes the well's depth equal to earth's radius), then all the upper hemi-sphere would not be included in the calculation. Please guide me.

Tim: Thanks.

I said:
You okayed it. But suppose we know that at other decimal places beyond one decimal places relation doesn't hold. Then, how can we be sure that the relation will always hold for one decimal place to infinity?
Your reply:
we can't
I'm sure there would be a lot of physics and chemistry laws whee things are derived using such relationships. If you can't even be convinced that if it holds to infinity for 1st decimal place, then how it be rigorously correct? I have just checked that Coulomb's law uses this relationship where the force between charges is proportional to the distance between them. If you aren't sure about first decimal place, then strictly speaking it's false. Please guide me.

And that "Euclid alone has looked on Beauty bare". I believe "bare" is functioning as an adjective here. What definition fits well in your opinion?

bare:
http://www.merriam-webster.com/dictionary/bare
 
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  • #18
tiny-tim
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If you can't even be convinced that if it holds to infinity for 1st decimal place, then how it be rigorously correct? I have just checked that Coulomb's law uses this relationship where the force between charges is proportional to the distance between them. If you aren't sure about first decimal place, then strictly speaking it's false.
(no, in Coulomb's law the force between charges is inversely proportional to the square of the distance)

it's not false, it's accurate enough :smile:
And that "Euclid alone has looked on Beauty bare". I believe "bare" is functioning as an adjective here. What definition fits well in your opinion?
i didn't join this forum to imdulge in textual analysis of poetry :rolleyes:

but you carry on if you want to :smile:
 

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