CIRCULAR MOTION
An object traveling in circular motion, making a circle of radius r, must be pulled inward toward the center of the circle to keep it from rushing away. The inward acceleration a caused by whatever force keeps it in circular motion is related to the circle’s radius, and to the speed of the object v , by a simple formula recognized in the 17th century by Christoph Huygens. The formula can be written in various equivalent ways:
The first form is particularly instructive: it says that in circular motion, acceleration is to velocity as velocity is to radius. [The reasons behind this relation are discussed at the end of this post from 2022.]Furthermore, the speed of an object in steady circular motion is the circle’ s circumference 2πr divided by the time T (the period) that it takes to complete one cycle. This allows us to obtain a relation between the period, the radius and the acceleration:
The radius of the Earth is about 4000 miles (6400 km); the circular orbit of a near-Earth satellite, just above the atmosphere, would have an orbital radius only a few percent larger, perhaps 4200 miles (6700 km). As for the acceleration at or near Earth’s surface, that was measured by Galileo: in the absence of air resistance, the speed of a falling object increases every second by almost 10 meters per second (about 32 feet per second). Equations (1) and (2) allow us to calculate the speed v of the orbiting satellite: about 5 miles per second (8 km/second). The period T of the orbit can then be obtained from equation (2): it takes about 90 minutes = 1.5 hours = 1/16 days for the satellite to orbit the Earth. Even though there were no near-Earth satellites in Newton’s day, he could imagine them and calculate what their speed and period would be.
IF WEIGHT WERE CONSTANT
Suppose that the acceleration a due to Earth’s gravity were a constant, independent of an object’s distance r from the Earth’s center. (This would be true if weight were the same as mass, since an object’s mass is intrinsic to the object, and thus independent of its distance from Earth or anything else.) Then both v and T , according to equations (1) and (2), would grow as the square root of r. We can write this using the “is proportional to” symbol, “∝”:
But this would lead to a wrong conclusion. The Moon is roughly 60 times further from Earth’s center compared to us (or to a near-Earth satellite.) The square root of 60 is about 7.5. If equation (3) were correct, then the Moon’s speed in its orbit would be 7.5 x (5 miles per second), which is about 40 miles per second (64 km/second) — and it would then orbit the Earth in an amount of time equal 7.5 x 1.5 hours, which is just under half a day. This prediction is a total failure! The Moon orbits the Earth about once a month.There are two possible responses to this revelation.
- the gravity that holds us to the ground is not the force that keeps the Moon in its orbit.the acceleration a of an object due to Earth’s gravity changes with the object’s distance r.
Newton guessed the second option. It was a guess because no experiments available at Newton’s time could actually move far enough away from the Earth’s surface, or measure gravity precisely enough, to say whether this was true or not. Let’s see that it was correct.
IF WEIGHT DECREASED WITH DISTANCE FROM EARTH
If instead a decreases as 1/r2, then the relation between v, T and r would become
This last relation between period and radius was familiar to Newton: it was Kepler’s law for the orbits of the planets [a law which you can confirm through do-it-yourself astronomy, as I have explained in this sequence of posts: 1, 2, 3 ; see also here] . So if gravity’s force indeed causes an acceleration that falls as the inverse of distance-squared, and if the Sun’s gravity holds the planets in orbit, and if Earth’s gravity holds the Moon, then equation (4) should apply to all objects orbiting the Earth. That means that the Moon’s speed can be inferred from that of a near-Earth satellite (real or imagined.) The speed should be (1/7.5) x (5 miles per second), or about 2/3 of a mile per second (1 km/second). Meanwhile its period would be given by multiplying 1.5 hours = 1/16 days by 603/2 = (7.5)3 = 465. This works!! 465 x (1/16 days) is roughly one month!That is how Newton obtained evidence that
- the Earth’s gravitational pull holds the Moon in orbit,the Sun’s gravitational pull holds the planets in orbit,gravity’s pull decreases as the square of the inverse of the distance from the pulling object (specifically, if the object is ball-shaped, from its center),and so an object’s weight (which depends on its location) is conceptually different from its mass (which is independent of its location)
