If the apple falls, why doesn’t the Moon?

Before Newton the answer seemed simple: heaven and Earth followed different rules. On Earth, things fall. In the heavens, the stars and planets turn in perfect circles forever. Newton’s great idea was to distrust that separation: perhaps the Moon is falling all the time.

Newton’s cannon

The apple tree in the orchard of Woolsthorpe Manor that tradition identifies as “Newton's tree”.
The apple tree in the orchard of Woolsthorpe Manor that tradition identifies as “Newton’s tree”.David Smith · Wikimedia Commons · CC BY-SA 2.0

In a draft for the Principia, Newton proposed a thought experiment. Imagine a cannon on top of a very high mountain, firing horizontally. With little powder, the ball lands nearby. With more, it lands further away. But the Earth is round: if the ball is fast enough, the ground curves away beneath it exactly as fast as it falls. The ball falls forever and never reaches the ground. It is in orbit.

Near the Earth’s surface that speed is about 7.9 km/s, or 28,000 km/h. It is the speed of satellites in low orbit. One complete turn takes about 84 minutes, ignoring air resistance, which in practice forces satellites to fly well above the atmosphere.

The Moon is that cannonball. It falls towards the Earth at every instant, but moves sideways so fast that it never arrives.

The inverse-square law

For the idea to work, gravity must weaken with distance. Newton assumed it decreases with the square of the distance. In today’s notation:

$$F=G\,\frac{Mm}{r^2}.$$

Here $M$ and $m$ are the masses, $r$ is the distance between their centres, and $G$ is a universal constant. Doubling the distance divides the force by $2^2=4$. Tripling it divides by $9$.

The Moon test

Newton did the calculation that turned this assumption into science. The Moon is about 60 Earth radii from the centre of the Earth. If gravity falls off with the square of the distance, the Moon’s acceleration should be $60^2=3600$ times smaller than that of an apple at the surface:

$$\frac{9.81}{3600}\approx0.00272\ \text{m/s}^2.$$

Now let us measure the Moon’s acceleration another way, using only astronomy. It travels a circle of radius $r\approx384{,}400$ km in $T\approx27.3$ days. The acceleration of a body moving in a circle is

$$a=\frac{4\pi^2r}{T^2}=\frac{4\pi^2\cdot3.844\times10^8}{(27.32\cdot86{,}400)^2}\approx0.00272\ \text{m/s}^2.$$

The two numbers agree. The force that brings down the apple and the force that holds the Moon are the same. Newton wrote that he found the results to “answer pretty nearly”. In 1666, however, he was using a poor value for the Earth’s radius, and the agreement was not perfect. Only with better measurements, years later, did the calculation close for good.

Kepler, derived in four lines

Decades earlier, Kepler had found in astronomical data that the square of a planet’s period is proportional to the cube of the size of its orbit. Nobody knew why. With gravitation and the second law, the explanation fits in a few lines, at least for circular orbits.

Gravity is the force that keeps the body on its curve:

$$\frac{GMm}{r^2}=\frac{mv^2}{r}\quad\Longrightarrow\quad v^2=\frac{GM}{r}.$$

Since one turn has length $2\pi r$, the period is $T=\dfrac{2\pi r}{v}$. Squaring and substituting $v^2$:

$$T^2=\frac{4\pi^2}{GM}\,r^3.$$

That is Kepler’s third law. For elliptical orbits the calculation is longer, and Newton did it in the Principia, showing too that the orbit under an inverse-square force is an ellipse, the answer he had given Halley in 1684.

Compare radii of circular orbits around the same central mass. Larger orbits are slower: the period grows as r to the power 3/2.

A universal law

The boldest word in the theory is universal: the same law holds for apples, moons, planets and comets. In 1846, irregularities in the orbit of Uranus led Urbain Le Verrier to calculate where an unknown planet ought to be. Astronomers pointed their telescope there and found Neptune. It was discovered with pencil and paper before it was seen.

Exercise: if the radius of an orbit triples, how does the period change?

Since $T^2\propto r^3$, we have $T\propto r^{3/2}$. The period is multiplied by $3^{3/2}=3\sqrt3\approx5.2$.

Exercise: at what distance from the Earth's centre does gravity drop to 1/4 of its surface value?

The force falls off as $1/r^2$. For it to be 4 times smaller, the distance must double: 2 Earth radii, or about 12,700 km from the centre.

Sources: Newton Project timeline · Newton and the Principia, MacTutor.

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