How Newton derived his law of gravitation

1. Nov 13, 2004

QuantumDefect

I was wondering if any of you could explain how Newton derived his Law of Gravitation: F=G*m_1*m_2/r^2 ? If this question is to vague tell me, thanks guys.

2. Nov 13, 2004

arivero

It is not too vague, but too long. A first indication is that he relied strongly in the concept of force (and inertial mass) from Hooke's law as well as his (F=ma), and a second one come from Galileo' experiment about free fall.

3. Nov 14, 2004

Newton relied heavily on Keplar's three laws of Planetary motion. These were deduced from the data that Tycho Brahe obtained over many years. These laws seemed to be correct, but didn't explain WHY objects orbited as they did.

Newton devised laws of circular motion and realised that the same force that made things fall down on Earth, was responsiple for the Centripetal force that keep the planets in orbit. This was his 'Eureka' moment - realising that the two forces were one and the same.

Newton then showed mathematically that the gravitational force between two objects was proportional to their masses and inversely proportional to the square of the distances between them.

To get a value for the constant 'G' he estimated the density of planet Earth and from that estimated its mass.

Clever chap.........!

4. Nov 14, 2004

Manchot

I didn't know that Newton ever tried to estimate G. I thought that Cavendish was the first to do so?

5. Nov 15, 2004

dextercioby

It was definitely Cavendish...

6. Nov 15, 2004

Integral

Staff Emeritus
Cavendish designed the experiment to directly measure G. Why do you doubt that Newton made an estimate? Do you really think that he would come up with a constant, then leave it without a value?

7. Nov 15, 2004

dextercioby

It could have been ignorance,but i don't remember seeing in the "Principia" the constant of proportionality between the gravitational attraction force and the product of the 2 masses devided by the square of the distance between them...Not to mention some numerical values...
Incidentally though,Cavendish's famous experiment came 100 years after Halley's first publication of Newton's "Principia"...In 1787,to be exact.

Last edited: Nov 15, 2004
8. Nov 15, 2004

dextercioby

In high-school i learned the other way around.It was Cavendish who first calculated the Earth's mass (i have no recallection of the density) using Newton's law,because he was the first man who knew both the Earth's radius (mean radius),both his constant (of course,he didn't name it that way) G.And that was done in 1787,shortly before Lagrange's first definition of the meter (do u rememeber it?? :tongue2: ),which is an evidence to support the fact that Cavendish kney Earth's mean radius.

9. Nov 15, 2004

Yes, Cavendish was the first to calculate the Earths mass (and big 'G'), but as Integral points out above - do you really thing Newton wouldn't have tried to estimate its value?

It isn't that hard to make an estimate of the Earth's mass is it? The Earth is mosly made of rock and it must be Iron rich due to it having a magnetic field. Estimate the proportions of each, measure the density of an average rock, add a bit for the iron rich core and then use the Earths dimensions to work out its mass. From this you get big 'G'.

10. Nov 16, 2004

dextercioby

As i said above,i didn't see any numerical values for G,or even G (or any other latter) itself in the "Philosophiae Naturalis Principia Mathematica".Maybe it's in other Newton's writing,or maybe it isn't al all.

So far,i have no evidence Isaac Newton knew the mass of planet Earth,by this simple method,or any other one...

11. Nov 16, 2004

Orion1

Newtonian Nexus...

According to my records, Newton derived the law of gravitation as follows:

$$F = M \frac{dv}{dt}$$
$$a = s \frac{d^2}{dt^2}$$
$$\frac{dv}{dt} = s \frac{d^2}{dt^2}$$
$$F = M \left( s \frac{d^2}{dt^2} \right)$$
$$F = Ma$$

$$s \frac{d^2}{dt^2} = \frac{F_m}{M_m}$$

$$s_m \frac{d^2}{dt^2} = G \frac{M_e}{r_m^2}$$

$$s_m \frac{d^2}{dt^2} = \frac{F_m}{M_m} = G \frac{M_e}{r_m^2}$$

$$F_m = M_m s \frac{d^2}{dt^2} = G \frac{M_e M_m}{r_m^2}$$

$$g = G \frac{M_e}{r_e^2}$$ - Terra
$$a_m = G \frac{M_e}{r_m^2}$$ - Lunar
$$GM_e = gr_e^2 = a_m r_m^2$$
$$\frac{g}{a_m} = \left( \frac{r_m}{r_e} \right)^2$$
$$F_e = G \frac{M_e M_m}{r_m^2}$$
$$G = \frac{g r_e^2}{M_e}$$

Last edited: Nov 16, 2004