13.2 Isaac Newton

A popular image has Newton thinking of the idea of gravity after an apple fell on his head. This amusing story is at least close to the truth. Newton himself said that the “notion of gravitation” came to him as he “sat in a contemplative mood” and “was occasioned by the fall of an apple.” It occurred to him that, perhaps, the apple was attracted to the center of the earth but was prevented from getting there by the earth’s surface. And if the apple was so attracted, why not the moon? In other words, perhaps gravitation is a universal force between all objects in the universe! This is not shocking today, but no one before Newton had ever thought that the mundane motion of objects on earth had any connection at all with the stately motion of the planets through the heavens.

A photo of Sir Isaac Newton. — Isaac Newton, 1642–1727.

Suppose the moon’s circular motion around the earth is due to the pull of the earth’s gravity. Then, as you learned in Chapter 8 and is shown in FIGURE 13.3, the moon must be in free fall with the free-fall acceleration $g_{at moon}$ .

The figure illustrates the circular orbit of the Moon around the Earth in the counterclockwise direction.

FIGURE 13.3 Full Alternative Text

NOTE The symbol $g_{moon}$ is the free-fall acceleration caused by the moon’s gravity—that is, the acceleration of a falling object on the moon. Here we’re interested in the acceleration of the moon by the earth’s gravity, which we’ll call $g_{at moon}$ .

The centripetal acceleration of an object in uniform circular motion is

a_{r} = g_{at moon} = \frac{v_{m}^{2}}{r_{m}}

(13.1)

The moon’s speed is related to the radius $r_{m}$ and period $T_{m}$ of its orbit by $v_{m} =$ $circumference/period = 2 π r_{m} / T_{m}$ . Combining these, Newton found

g_{at moon} = \frac{4 π^{2} r_{m}}{T_{m}^{2}} = \frac{4 π^{2} (3.84 \times 10^{8} m)}{{(2.36 \times 10^{6} s)}^{2}} = 0.00272 {m/s}^{2}

Astronomical measurements had established a reasonably good value for $r_{m}$ by the time of Newton, and the period $T_{m} = 27.3$ days $(2.36 \times 10^{6} s)$ was quite well known.

The moon’s centripetal acceleration is significantly less than the free-fall acceleration on the earth’s surface. In fact,

\frac{g_{at moon}}{g_{on earth}} = \frac{0.00272 {m/s}^{2}}{9.80 {m/s}^{2}} = \frac{1}{3600}

This is an interesting result, but it was Newton’s next step that was critical. He compared the radius of the moon’s orbit to the radius of the earth:

\frac{r_{m}}{R_{e}} = \frac{3.84 \times 10^{8} m}{6.37 \times 10^{6} m} = 60.2

NOTE We’ll use a lowercase r, as in $r_{m}$ , to indicate the radius of an orbit. We’ll use an uppercase R, as in $R_{e}$ , to indicate the radius of a star or planet.

Newton recognized that ${(60.2)}^{2}$ is almost exactly 3600. Thus, he reasoned:

If g has the value 9.80 at the earth’s surface, and
If the force of gravity and g decrease in size depending inversely on the square of the distance from the center of the earth,
Then g will have exactly the value it needs at the distance of the moon to cause the moon to orbit the earth with a period of 27.3 days.

His two ratios were not identical, but he found them to “answer pretty nearly” and knew that he had to be on the right track.

I deduced that the forces which keep the planets in their orbs must be reciprocally as the squares of their distances from the centers about which they revolve; and thereby compared the force requisite to keep the Moon in her orb with the force of gravity at the surface of the Earth; and found them answer pretty nearly.

Isaac Newton