# What is the nature of gravity constant?

1. Nov 11, 2007

### freesom

what is the nature of gravity constant?

i just ask that?

and if we can know the nature of it

can we understand the meaning of gravity?

2. Nov 11, 2007

### Staff: Mentor

Huh? The gravitational constant? It is number. A physical constant.

Physicists look for order, they don't look for "meaning". "Meaning" doesn't really mean anything anyway.

3. Nov 12, 2007

### rbj

since it is unit-dependent, there is no meaning in the numerical value that you normally see (except, i s'pose, knowing that using the anthropometric units we usually use, gravity can be said to be a "weak force" because G appears to be very small in these human-oriented units). but that meaning really has to do with scale that seems meaningful to humans, which is sorta arbitrary.

it's not just a number, it's a dimensionful number. one that normally has units attached to it. so the value of the number attached to those units change if the units change. we can define a set of units to make the numerical value of G anything we want. Planck units (and some other definitions of "natural units") are chosen so that G = 1.

oh, Russ, certainly sometimes they look for meaning. no? isn't that what physicists do when they try to simplify models (like Maxwell's eqs.)? or come up with a completely different paradigm (like "gravity = curvature of space-time"). that stuff is about "meaning", isn't it?

4. Nov 12, 2007

### yogi

The units of Big G, the so called "Gravitational Constant" are most interesting: cubic meters per second squared per kilogram. Translated, this means Volumetric Acceleration per unit mass. The first thing that might occur to one looking at this is why would the volumetric accelertion be the same for a much smaller universe, say one only the size of a basketball as compared to one having a Hubble radius of 10^26 meters. The other thing that should be suggested is how do you get the number 6.67 x 10^-11 out of what we think we know about the size and rate of expansion of the cosmos. Specifically, how does the volumetric acceleration come into being if the Hubble expansion rate is uniform?

5. Nov 15, 2007

### gamburch

The gravitational constant is equal to (3Pi/4)H(squared)/density of matter in the universe. I hope this helps.

6. Nov 15, 2007

### yogi

Hi gamburch

How did you arrive at this - I assume you used Friedmann's relationship that relates G and density. If the universe has critical density mass - then the factor would be 3pi/8 rather than 3pi/4.

Yogi

7. Nov 15, 2007

### rbj

in addition, gamburch, can you use the LaTeX markup so we what you are talking about? and can you identify all of your symbols? is it this:

$$G = \frac{3 \pi}{4} \frac{H^2}{\rho}$$ ??

where $\rho$ is the "density of matter in the universe". if so, can you identify H?

8. Nov 15, 2007

### pervect

Staff Emeritus
This appears to be an incorrect derivation from The Friedmann equations.

The correct expression would be

$$G = \frac{3 H^2}{8 \pi \rho}$$

In any event, this equation is not always true - it requires that the cosmological constant $\Lambda$ be zero, that the spatial curvature of the universe K=0, and of course the assumptions that GR is correct and the cosmological principle holds so that the universe is homogenoeus and isotropic.

H here would be Hubble's constant.

Last edited: Nov 15, 2007
9. Nov 15, 2007

### pervect

Staff Emeritus
10. Nov 16, 2007

### gamburch

Yogi: Thanks for your interest. Gravitational potential has the units of velocity squared. If one uses the value of the density of matter of the universe published by Peeblers some time ago, one can compute the gravitational potential locally arising from all matter in the universe by merely integrating over a sphere of size equal to the speed of light divided by the Hubble constant. This computation yields a number close to the speed of light squared. I like this way of doing things because it rather nicely explains the relationship of mass to energy; the potential times the matter being the cost of its assembly.

If in the computation one substitutes c/H for the radius one has c squared on both sides yielding a relationship between the gravitational constant, the density of the universe and Hubble's constant. The constant in the equation comes from the volume of the sphere and has no magic. A more careful worker than I would have accounted for the Doppler effect on the local field so the criticisms of my simple computation are probably very apt. I did not get the equation from elsewhere, however. What you see is what you get.

11. Nov 17, 2007

### yogi

Interesting Gamburch. I have seen the same formula derived using Hubble's law v = Hr, the procudure the poster used was to differentiate and plug in the value v = c at the Hubble limit getting dv/dt = Hc and then re substituting c = Hr giving an effective acceleration = (H^2)R If I remember, the fellow actually got the paper published in an electronics magazine - probably because he extrapolated the result in terms of electric fields, which I could never figure out. Maybe I can dig up the old thread, you might be interested.

regards

yogi

12. Nov 17, 2007

### yogi

This is a follow-up to post 11. I don't think the old thread is still available - at least I can't find it - but here is the rest of what the poster did - he sets the force of gravity GMm'/R^2 equal to the F = m'a force. For "a" he uses the (H^2)/R that he derived from from Hubbles law - so he sets (GM/R^2) = (H^2)R where M is the mass of the universe
and since M = V(rho) and V = 4/3(pi)R^3 .....vola, G = [3/4(pi)](H^2)

I am sure the math types will have nothing but condemnation for this sort of procedure but I think these simple relationships are telling us something about the universe that is being overlooked

Last edited: Nov 17, 2007
13. Nov 17, 2007

### gamburch

Thank you Yogi. I truly appreciate this recollection and will be mulling it over; however, today is the day for starting my Christmas home brew. I'll be back soon. Thanks again.

14. Nov 17, 2007

### yogi

Home Brew - sounds like something with profound cosmological intrigue. When we talk later I will show you my own derivation of G - and some consequences thereof. Seems there are many ways to skin the same cat.

Happy Holidays

Yogi

15. Dec 24, 2008

### exponent137

I have almost the same question as this one of Freesom. It is known what means G in Newtonian gravity. (for instance: acceleration 1 m from 1 kg heavy (point) ball) Maybe there is something still more clear. But, did G get any clearer meaning in Einsteinian curved space?

"Physicists look for order, they don't look for "meaning". "Meaning" doesn't really mean anything anyway."
I never agreed with this style of thinking. This style make unclear, what can be easily clear.

16. Dec 24, 2008

Staff Emeritus
I don't see how it can be made much clearer than "if you make this measurement, you'll get this result".

17. Dec 25, 2008

### Staff: Mentor

In both theories G has the same role. GR made two principle improvements over Newtonian gravity. First, it is more general, applying in situations where Newtonian gravity was wrong (Mercury) or unclear (bending light). Second, it makes the relationship between inertia and gravitation more clear (equivalence of gravitational and inertial mass).

G itself, as a dimensionful constant, can be seen as simply relating different units. You cannot really attach any physical significance to any dimensionful constant as anything other than a unit conversion. This independent of which theory you use.

Last edited: Dec 25, 2008
18. Dec 25, 2008

### exponent137

Still more elementary than acceleration, is impact on light. So maybe G which causes horizon of a black hole?

Every new interpretation of G can maybe "open eyes" for new physics.

And, how to connect G with equivalence principle. In the most simple example (elevator) G can be ignored.

Last edited: Dec 25, 2008
19. Dec 25, 2008

### ZachN

The true nature of the constant is that it is there to make up for a lack of understanding for the true mechanism for gravity. Constants are used to express values we can't explain through other means. That is the dirty secret of science.

e.g. - the resistance(drag) of an object falling through the atmosphere is proportional to the velocity. we get an equation to model the system which looks like this:

mdx/dt = mg - kv

we could use this generic equation to model the falling of most objects with decent results. However, much work has been done to study the drag on specifically shaped objects and so k can be replaced with more accurate models.

the kv term can be replaced in the 2-d case with d=qinf*CD*Afrontal. But now look at this we have a new constant CD (drag coefficient) which is essentially another dimensionless quantity to make up for more complex behavior which would be too difficult to discern at this time.

So that is the story of coefficients and constants. Whenever you see one remember that it represents something we don't yet understand.

20. Dec 25, 2008

### A.T.

Drag is proportional to velocity squared.
We do understand how drag works. But the computations for specific shapes are very complex, even for computers. The drag coefficient is a practical engineering approach.

Last edited: Dec 25, 2008