# Gravitational constant visualized

#### pugilist777

Ive recently learned about the gravitational constant. Although i am having a hard time visualizing what this actually means. I have always been stubborn in the fact that i cant just accept and idea or theory without really digging deep and understanding it on a fundamental level. I understand that this constant, and most others is a value of proportion. But im having a hard time understanding if the G constant is a value of a given region of space, or if it is a value of matter itself. From what i have seen is that the constant is a value of force between two objects, is universal, and is used in most other equations involving relativity and gravity. But i also know that Einstein's theory of gravity states that it is not a force at all, but the warping of space time. So how can the G constant be a value of force of gravity when gravity is not a force? Is there a way that i can visualize the constant better? Im decent at math, but apparently not enough. What does it mean when the kg and s have a negative square value. 6.67408 × 10-11 m3 kg-1 s-2

#### topsquark

Forum Staff
Ive recently learned about the gravitational constant. Although i am having a hard time visualizing what this actually means. I have always been stubborn in the fact that i cant just accept and idea or theory without really digging deep and understanding it on a fundamental level. I understand that this constant, and most others is a value of proportion. But im having a hard time understanding if the G constant is a value of a given region of space, or if it is a value of matter itself. From what i have seen is that the constant is a value of force between two objects, is universal, and is used in most other equations involving relativity and gravity. But i also know that Einstein's theory of gravity states that it is not a force at all, but the warping of space time. So how can the G constant be a value of force of gravity when gravity is not a force? Is there a way that i can visualize the constant better? Im decent at math, but apparently not enough. What does it mean when the kg and s have a negative square value. 6.67408 × 10-11 m3 kg-1 s-2
G is a constant. There are a number of theories kicking around that say that this constant has changed over the course of the lifetime of the Universe, but none that I know of has been completely successful. However G is a constant we use in calculating the force between two objects. Simply because we can visualize GR representing a curvature of space-time does not mean that G is going to be different if we think of it that way.

Why it has the specific value that it has today is not known.

-Dan

#### benit13

But im having a hard time understanding if the G constant is a value of a given region of space, or if it is a value of matter itself.
Neither. It's just a constant which sets the magnitude of the gravitational force between two masses held a certain distance apart. All masses in the Universe seem to interact in the same way, so there is no need for a varying value.

From what i have seen is that the constant is a value of force between two objects, is universal, and is used in most other equations involving relativity and gravity. But i also know that Einstein's theory of gravity states that it is not a force at all, but the warping of space time.
It doesn't say it isn't a force... one can conclude that the gravitational force is a natural consequence of Einstein's theory of gravity and that the theory is superior because it can be applied to objects that don't have mass (e.g. light), replicating observations. The most famous example is the observations of field stars close to the limb of the sun during the solar eclipse.

So how can the G constant be a value of force of gravity when gravity is not a force?
Sometimes gravity seems to behave like a force, so using G is useful.

If you're still puzzled by it, think of this analogy...

Let's say someone comes along and comes up with the "theory of colour" and then says that the colour of something is either "red", "green" or "blue". They apply this to the objects around them and then get satisfactory results, since all of the other colours can be obtained from mixtures of the colours. The theory becomes popular and everyone uses it.

Then someone comes along 100 years later and says "what about black?".
The old theory doesn't seem to have a good explanation of it. The new guy then realizes that a new theory of colour is required, where the red, green and blue components are not just qualitative descriptions, but actual values between 0 to 100. The colour must then be specified as RGB values, with black equal to (0,0,0), white equal to (100,100,100) and everything else being something in between.

Technically, in the new theory, "blue" is no longer correct... it should be specified as "(0,0,100)" or "(0,0,50)" or "(0,0,20)" depending on the shade of blue... but does this mean that using the description "blue" is pointless? It's convenient to be able to still use the description everyone is familiar with because it's simple to understand and describes most things well.

In the same way, Newton's theory of gravity seems to work very well for most masses. It doesn't stop being useful just because it doesn't work for light, which is ultimately why Einstein's theory is superior.

What does it mean when the kg and s have a negative square value. 6.67408 × 10-11 m3 kg-1 s-2
In simple terms, it means "per". For example:

Speed: "metres per second" is the same as "m/s" or "m s$$\displaystyle ^{-1}$$"

Acceleration: "metres per second squared" is the same as "m/s$$\displaystyle ^{2}$$" or "m s$$\displaystyle ^{-2}$$".

So the unit you have is "m$$\displaystyle ^{3}$$ kg$$\displaystyle ^{-1}$$ s$$\displaystyle ^{-2}$$" which is the same as "metres cubed per kilogram per second squared".

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#### Woody

You could almost look upon G as being a scaling factor
that converts {Mass (in kg) x Mass (in kg)} / distance (in metres) into Force (in Newtons)

The force between two masses is proportional to Mass_1 multiplied by Mass_2 divided by the distance between them squared.

G is the "constant of proportionality" which allows you to replace "proportional" by "equals".

Note that M1*M2/d^2 has units of kg^2 / m^2
While Force (Newton) has units of kg.m / s^2

This requires G to have units of m^3 kg^-1 s^-2 to make the units match across the equation.

Note that m^3 kg^-1 s^-2 is the same as m^3 / (kg s^2) {or m³/(kg.s²)}

#### pugilist777

So in other words, For 6.67408 x10-11 cubic meters of volume there is a force of 6.67408 x10-11 kg per seconds squared?

#### Woody

Making it up as I go...

No,
Your suggested relationship between volume and force is not valid.

Looking at the units of G, the closest I can get to a description is perhaps:

The rate of change of (the rate of change of (the reciprocal of density)).

kg/m³ is density so m³/kg is the reciprocal of density.
x/s² =(x/s)/s
which is the rate of change of the rate of change of x

I guess the reciprocal of density would be rarefaction,
so is it the rate of change of the rate of change of rarefaction?

In other-words the rate of change of the rate at which density is decreasing.

I guess this is part of where your alternative theory in the parallel thread comes from.

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#### benit13

So in other words, For 6.67408 x10-11 cubic meters of volume there is a force of 6.67408 x10-11 kg per seconds squared?
Don't take the units too literally. In the SI unit system, all units can be reduced down to base units (e.g. kilogram, metre, second, Coulomb, etc.), but this doesn't really yield too much additional insight. Sometimes looking at units can be quite helpful and indicative, but I don't think that's the case here.

A nice way of interpreting the gravitational constant is to just perform experiments and see how things change. For example, how does the force vary with masses? Well, it increases proportionally to the product of the two gravitating masses. i.e.

$$\displaystyle F_g \propto m_1 m_2$$

However, the force decreases by the square of the distance between the two masses, so

$$\displaystyle F_g \propto \frac{1}{r^2}$$

Combining these two relationships gives:

$$\displaystyle F_g \propto \frac{m_1 m_2}{r^2}$$

This can be made into an equality with a constant:

$$\displaystyle F_g = G \frac{m_1 m_2}{r^2}$$

Now... what value for G do we have? Well, you can determine it experimentally. If you've captured all of the possible sources of error, you will not have any other things impacting the value of G.

You can do this technique with lots of phenomena. Sometimes the number changes based on your apparatus (e.g. Hooke's spring constant) or the location of your experiment (gravitational acceleration constant). For the gravitational constant, G, it doesn't seem to vary at all.

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