Gravitational constant

1. Jan 20, 2009

chis

I am looking for some clarification on the gravitational constant.
What is the equation that produces it and what does it add to the value of a calculated gravity i.e what would the calculated gravity represent if it was removed from the equation? Also I have read it is a fudge by Newton and it's just accepted to work but has not been bottomed out.

2. Jan 20, 2009

mgb_phys

It comes from F = GMm/r^2
It's the force between two massive objects, M and m, a distance r apart.

It's interesting because since gravity is such a weak force it is very difficult to measure experimentally - it took 100years after Newton to get an answer. Even now it's only known to an accuracy of 0.01% - billions of times worse than some constants.

Newton didn't use the symbol - in his day equations weren't written like that, he would have just said that the force is proportional to the product of the masses.

The only thing large enough to give a measurable effect in his day was the Earth so he calculated a value of G*M where M is the mass of the earth, since he didn't know the mass of the Earth he couldn't work out a value of G on it's own.
Interestingly the attempts at measuring G were more aimed at getting to the M - so measuring G was really about weighing the Earth,

Last edited: Jan 20, 2009
3. Jan 20, 2009

chis

Thanks, but what is the formula to aquire the gravitaional constant not the gravitational forces between two objects?

Cheers
Chris

4. Jan 20, 2009

Staff: Mentor

It's an empirical constant. Measured, not derived.

5. Jan 20, 2009

chis

I have seen it represented as a number, is this number used in the equation?

6. Jan 20, 2009

Staff: Mentor

It's a physical constant: a number with units. G = approximately 6.67 x 10-11 N m2/kg2.

7. Jan 20, 2009

mgb_phys

Yes the number 6.673 × 10-11 m3 kg-1 s-2 is used in the equation.
It might be possible to calculate it from fundemental constants (c / mu / mass of quarks etc) but we don't have a theory yet for how.

The actual number of course depends on your units, it would be different in lbs/ft/s

8. Jan 20, 2009

atyy

Equation as given by mgb_phys, and used in the Cavendish experiment (I personally find this even more amazing than Michelson-Morley): http://www.leydenscience.org/physics/gravitation/cavend.htm

9. Jan 20, 2009

chis

Without G how inacurate would the equation be? What is G preposed to represent?

10. Jan 20, 2009

mgb_phys

The equation is perfectly accurate (ignoring general relativity effects).
What it says is that the force between two objects is proportional to the product of their masses divided by the distance between them squared

G is just the constant to put the answer in Newtons for an equation in kilograms and metres.

11. Jan 20, 2009

atyy

Yes, G is a "fudge factor" or "free parameter". The point is to (i) make your theory with as few fudge factors as possible, (ii) make sure the fudge factors can be measured. Newton's theory has only one fudge factor, and it is universal, once you measure it for one system, you know it for all other systems.

12. Jan 20, 2009

chis

Right the penny may have finally dropped. It's just used to consolidate different measurement units? Did Newton figure thi number out or did he just theorise its exsistance and it was filled in later?

Last edited: Jan 20, 2009
13. Jan 20, 2009

mgb_phys

It's not just the units - it does contain the fundemantal strength of gravity.

It's like if I told you a metal pipe weighed so much per foot, you would know that a peice twice as long weight twice as much.
So in terms of an equation: weight = k * length
the value of k depends on the units (it would be different for kg/metres and lbs/feet) but it also depends on the metal the pipe is made from (lead would be a different lbs/foot than copper.)

As I said Newton (and everybody for 100years after) couldn't measure the value - it's very small - they could measure the value of "G*Mass_of_earth" because you can get that from the movement of bodies around the Earth, notably the moon.

14. Jan 20, 2009

Janus

Staff Emeritus
It is a constant of proportionality.

It's what makes the answer come out right for any given set of units.

It's arrived at like this:

We know that the froce acting between two masses is given by

$$F = \frac{kMm}{d^2}$$

where K is some constant. At first we didn't know what value this constant had. To find out we re-arrange the formula to

$$k = \frac{Fd^2}{Mm}$$

Now we take tow known masses (m & M) put them a distance of d apart and then measure the force pulling them together. We then plug these values into the equation, which gives us our constant k. We call this constant "G".

What G works out to be depends on what units we used to measure mass, time and distance.

15. Jan 20, 2009

chis

The pipe analogy was great I think you have done it. Sorry for being so slow. So G is our best generic assumtion of a unit of gravity?

16. Jan 20, 2009

mgb_phys

Yes - except we have sort-of been lying to you!
Newton's law of gravity isn't really correct.

It's very good for typical calculations of falling balls and orbiting satellites. But close to very heavy objects (like stars) it doesn't quite give you the correct answer.
That's what Einstein did - he wrote a new more complicated theory the gives a more accurate answer.

It doesn't change the value of G (or allow us to calculate it directly) but it does say that the equation above isn't exactly correct for all cases.
It's like having to take into account the bend in the pipe when working out lbs/feet - if the pipe is short the bend is small and you can ignore it, but in some case (if the pipe is long or the object is a black hole) you need to use the more accurate theory.

17. Jan 20, 2009

chis

Ah did Einstiens refinement include density as a contributing factor?
Is G a graviton?

Last edited: Jan 20, 2009
18. Jan 20, 2009

atyy

No, density is already in Newtonian theory. See the section "Differential Form" at http://en.wikipedia.org/wiki/Gauss's_law_for_gravitational_fields. In Newton's theory, gravity travels infinitely fast. Einstein's refinement was to make gravity travel at the speed of light.

No.

19. Jan 21, 2009

Je m'appelle

Yeah, by Henry Cavendish in 1798 if I'm not wrong.

Oh, by the way, I watched a video of a MIT Classical Mechanics lecture by professor Walter Lewin, in which he affirmed that "the time it takes an apple to fall is independent of the mass of the Earth".

Is that true? Well, the apple(m) falls due to its gravitational interaction with Earth(M) by F = GMm/r^2, therefore we could apply t(g) = v/g (air resistance is not taken into consideration) to find the time it takes the apple to fall, right? However, as it is a function of "g", the bigger "g" is, the faster the apple will fall, right? But, "g" depends on the mass of the Earth (the mass of the apple is not taken into consideration as it is too small compared to Earth's), as we can see it in g = GM/r^2

So I really did not understand Mr. Lewin's point.