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TurtleMeister
Oct27-09, 03:05 PM
According to Wikipedia, Big G is a constant of proportionality. When speaking of proportionality one thinks of two quantities which can vary in such a way as to have a constant ratio of something. If that is true, then in the case of Big G, what are the quantities, and what is the something?

A.T.
Oct27-09, 04:43 PM
According to Wikipedia, Big G is a constant of proportionality. When speaking of proportionality one thinks of two quantities which can vary in such a way as to have a constant ratio of something. If that is true, then in the case of Big G, what are the quantities, and what is the something?
So you have already read this:
http://en.wikipedia.org/wiki/Gravitational_constant
including
http://upload.wikimedia.org/math/7/c/d/7cdf733b81cd2b83d434160241d6023c.png
and
http://upload.wikimedia.org/math/9/8/1/981c4a7801639525969c8d798aca28ce.png
and you ask what is proportional to what, and what the units of G are?

TurtleMeister
Oct27-09, 08:36 PM
Thank you for the reply A.T. I think I'm understanding the proportionality better now. However, I am still confused about something. In Newton's second law of motion there is no constant of proportionality (F = ma). Here there is a direct relationship between force, mass, and acceleration. But in the universal law of gravitation, the proportionality G is required. Why is that? I understand the inverse square of the distance. But why does it appear that the mass used in the universal law of gravitation is different from the mass used in the second law of motion?

Nabeshin
Oct27-09, 09:08 PM
I believe the reason for the lack of constant of proportionality in F=ma is due to how force is defined. For example, having a definition of mass and acceleration, we define the force of 1N to be that which is produces a 1m/s^2 acceleration of a 1kg body. Now, we COULD have defined force differently, perhaps using the universal law. In this case, we would say something like 1N is the force two 1kg bodies exert on each other at a distance of 1m. I think then you would actually get the constant of proportionality, G, in f=ma somewhere.

Cheers.

mikelepore
Oct27-09, 09:16 PM
For gravity, "G" is inside "a". To line up the terms, F = (m)(a) corresponds to F = (m1) (G m2/r^2). The last thing in parentheses is the acceleration of m1.

TurtleMeister
Oct27-09, 10:44 PM
I believe the reason for the lack of constant of proportionality in F=ma is due to how force is defined. For example, having a definition of mass and acceleration, we define the force of 1N to be that which is produces a 1m/s^2 acceleration of a 1kg body. Now, we COULD have defined force differently, perhaps using the universal law. In this case, we would say something like 1N is the force two 1kg bodies exert on each other at a distance of 1m. I think then you would actually get the constant of proportionality, G, in f=ma somewhere.
Are you saying that 1N in the universal law of gravitation has (or could have) a different definition than 1N in the second law of motion? I must be misunderstanding you because that doesn't make any sense.
For gravity, "G" is inside "a". To line up the terms, F = (m)(a) corresponds to F = (m1) (G m2/r^2). The last thing in parentheses is the acceleration of m1.
Yes, I understand that G is used for a force due to gravity. I guess I should rephrase the question.

Disregarding the inverse square of the distance - I understand that, why is a constant of proportionality required for a force on an object due to gravity and not for a force on the same object due to acceleration (such as that produced by a rocket)?

Nabeshin
Oct27-09, 10:53 PM
Are you saying that 1N in the universal law of gravitation has (or could have) a different definition than 1N in the second law of motion? I must be misunderstanding you because that doesn't make any sense.


That is indeed what I'm saying, but note that the choice of the symbol N is a bit deceiving, since we think of 1 Newton = 1 kg*m/s^2. This would be a different unit, but corresponding to our concept of force. In particular, it would have dimensions mass^2/length^2.

A.T.
Oct28-09, 03:55 AM
why is a constant of proportionality required for a force on an object due to gravity and not for a force on the same object due to acceleration (such as that produced by a rocket)?
That is just an artifact of the used unit system. You can use a unit system where G = 1, so it doesn't appear in the formula:
http://en.wikipedia.org/wiki/Geometrized_unit_system

TurtleMeister
Oct28-09, 03:04 PM
That is interesting A.T. But I'm not sure it clears things up for me. For example, just like G, pi is a constant of proportionality. If we change the unit system in an equation so that pi does not appear then we have not erased what it represents, we have only hidden it. The proportionality is still there. Also, if we change the unit system for universal gravitation then wouldn't we also have to change it for the second law of motion thereby putting us right back where we started from with the discrepancy?

TurtleMeister
Oct29-09, 01:13 AM
Thank you A.T., Nabeshin, and mikelepore for the responses.

I'm still unclear about what BigG is a proportionality of. All the text I have read does not go into detail. It appears to have the units of force, but force of what? The two masses?

I know that there are three types of mass, or properties of mass. Active, passive, and inertial. I know that the m in F = ma is inertial. But I'm not sure about F = (GMm) / r2. I'm thinking active and passive. That would explain the discrepancy between universal gravitation and the second law of motion. But what about G? Does it represent the proportionality between the active and passive mass? Wouldn't a proportionality between the masses violate the equivalence principle?

ideasrule
Oct29-09, 01:25 AM
Forget about the equivalence principle, active/passive masses, and the like. GMm/r^2 is part of Newtonian equation, so general relativity is irrelevant.

There were no units of force at the time of Newton. The SI unit of force, the newton, was chosen so that F=ma. It could have been chosen so that F=Mm/r^2; then F=kma because the force needed to accelerate 1 kg at 1 m/s^2 isn't the same as the force between 2 1-kg objects separated by 1 m.

jtbell
Oct29-09, 01:40 AM
I'm still unclear about what BigG is a proportionality of. All the text I have read does not go into detail. It appears to have the units of force, but force of what? The two masses?

G is the constant of proportionality between force (which has units of newtons as defined by F = ma) and Mm/r^2 (which has units of kg^2/m^2). Accordingly, G has units of N*m^2/kg^2 in order to make the units balance on both sides of F = GMm/r^2.

In principle, we could define the gravitational force between two 1-kg masses separated by 1 m to be a new unit of force, and call it, say, the "Cavendish" (Cav). In Cavendish units, the law of gravitation would read simply F = Mm/r^2. The second law of motion would read F = Cma, where C is a proportionality constant with units of Cav/(kg*m/sec^2) = Cav*sec^2/(kg*m).

TurtleMeister
Oct29-09, 04:43 PM
Thanks very much for the replies. I think you guys are saying the same thing as A.T. Sorry, I guess I'm not being very clear on what I'm asking. I understand that the equations can be manipulated by changing the unit system. However, doing so does not change the fact that there is a proportionality difference between the second law of motion and the universal law of gravitation. And that is what my question is about. For example:

a = 3
b = 4
C = 0.25 'constant of proportionality

ab = 12
Cab = 3

C tells me that something in the second equation (either a or b) is 1/4 what it is in the first equation. So in the universal law of gravitation, what is G telling me?

A.T.
Oct29-09, 06:42 PM
For example:

a = 3
b = 4
C = 0.25 'constant of proportionality

ab = 12
Cab = 3

It seems you don't understand what a constant of proportionality (http://en.wikipedia.org/wiki/Constant_of_proportionality) is. In this example C is not a constant of proportionality between two variables.


C tells me that something in the second equation (either a or b) is 1/4 what it is in the first equation.

They could also be both 1/2 of the first, or any other combination that gives 1/4 as product. So C tells you little here.

TurtleMeister
Oct29-09, 06:57 PM
It seems you don't understand what a constant of proportionality is. In this example C is not a constant of proportionality between two variables.
I thought it was probably something I was not understanding. Could you possibly give me an example (similar to mine) where C is a constant of proportionality?

A.T.
Oct29-09, 07:05 PM
I thought it was probably something I was not understanding. Could you possibly give me an example (similar to mine) where C is a constant of proportionality?
I gave you the link: http://en.wikipedia.org/wiki/Constant_of_proportionality#Direct_proportionality

With your letters: a = C * b makes C the constant of proportionality between the variables a and b.

DaleSpam
Oct29-09, 07:14 PM
F = G mM/rē
ma = G mM/rē
a = G M/rē
a rē/M = G

Is that what you want?

TurtleMeister
Oct29-09, 08:09 PM
Ok, I think I understand now. G is the constant of proportionality between the force and the mass? Can I assume that G is so small because of the very weak gravitational force of the mass? And the reason it's not needed for the second law of motion is that the second law of motion does not rely on a gravitational field?

DaleSpam
Oct29-09, 08:59 PM
Can I assume that G is so small because of the very weak gravitational force of the mass?G is small because of our choice of units. The only fundamental constants whose values have any physical significance are the dimensionless ones like the fine structure constant. All others are purely an artifact of the chosen system of units.

TurtleMeister
Oct29-09, 09:16 PM
So we could change our system of units so that G was large in relation to M? If so, what would happen to F = ma?

DaleSpam
Oct29-09, 09:20 PM
So we could change our system of units so that G was large in relation to M?Yes, although IMO it doesn't make sense to "compare apples and oranges" like that.If so, what would happen to F = ma?Nothing.

TurtleMeister
Oct29-09, 09:46 PM
I'll give that some thought. It's starting to look like G is not so magical after all. :)

I would to thank everyone who has responded in this thread. You've been very helpful to me. I may return later if I have more questions.

Turtle

Rasalhague
Oct29-09, 09:48 PM
A couple of simpler examples:

1 inch = 0.0254 m
0.0254 = inch * m^-1

0.0254 is the constant of proportionality which relates inches to meters. It's dimensionless because inches and meters have the same dimension, length. The energy of a photon is Planck's constant times the frequency:

E = h*f
h = E*f^-1

Energy is measured in joules, frequency in cycles per second. Planck's constant is 6.626 * 10^-34 J*s. It consists of a number and the appropriate units to relate energy to frequency. Energy and frequency aren't measured in the same units as each other, which is why Planck's constant needs the right units to balance the equation.

Likewise with the gravitational constant G:

m*a = G*M*m*r^-2
G = a * M^-1 * r^2

And G = 6.674 * 10^-11 m^3 * s^-2 * kg^-1. It's the constant of proportionality between force and M*m*r^-2. If we used some system of units other than SI units, these constants might have different values. Coulomb's constant plays the corresponding role for electrical force: m*a = k*Q*q*r^-2, where k stands for Coulombs constant, Q is the charge of what is exerting the force, and q the charge of what is affected by the force.

What do you mean by "the discrepancy between universal gravitation and the second law of motion"?

TurtleMeister
Oct29-09, 10:03 PM
What do you mean by "the discrepancy between universal gravitation and the second law of motion"?
Just a poor choice of words. It looked like a discrepancy because of my misunderstanding. Thanks for the examples.

TurtleMeister
Oct31-09, 03:12 AM
Ok, I've become confused again and I need your help. In the manual for the Pasco torsion balance there is a procedure for measuring the gravitational constant. The equation used is:

G = (tr2) / (2dm1m2)

which is derived from the law of universal gravitation, with t being the torque on the torsion ribbon, and d being the length of the pendulum bob arm.

G = (Fr2) / (m1m2)

It would appear from these equations that G actually represents a proportionality between the objects mass and the force of it's gravitational field. For example, increasing r2 will cause F to decrease by the same amount (inverse square law). Increasing the mass of m1 and/or m2 will cause the same increase in F. So G always stays the same (constant ratio).

I've never seen G described as a constant of proportionality between an objects inertial mass and it's gravitational mass. If this is correct then why is it not described that way. Or, what am I doing wrong?

DaleSpam
Oct31-09, 06:08 AM
I've never seen G described as a constant of proportionality between an objects inertial mass and it's gravitational mass. If this is correct then why is it not described that way. Or, what am I doing wrong?It is not correct. An object's inertial mass is equal to its gravitational mass. Also, G is not dimensionless whereas the ratio of the inertial mass to the gravitational mass is dimensionless (and equal to 1).

TurtleMeister
Oct31-09, 10:48 AM
Also, r is not a constant. So any change in r will change the ratio of mass to force. I guess I was up too late last night. Thanks for the reply.

gutti
Nov4-09, 07:06 PM
Yous are all getting confused.
There is a difference between G and g.
g changes and is the force per kilogram due to gravity.
G is a constant (meaning it never changes)

And i could be wrong on this point, but i think it was kepler that first calculated it t^2/r^3 in his third law of planetary motion, where r is the distance between an object and the sun and t is its period.

TurtleMeister
Nov4-09, 10:21 PM
Actually there was no confusion between Big G and Little g. The confusion was caused by me being an idiot when it comes to math. :) But I appreciate the input anyway.