The Moon Is Falling Towards Earth All the Time and Always Missing

A mango dropping from a tree and the Moon circling Earth look like completely different motions. They are the same force at work — gravity — acting over very different distances.

Planets follow precise rules in their orbits, and one law of force explains all of them.

This part covers Kepler's laws, the universal law of gravitation, adding gravitational forces, and measuring the gravitational constant $G$.

**Planets move in ellipses with the Sun at one focus, the line from the Sun to a planet sweeps out equal areas in equal times, and the square of the orbital period is proportional to the cube of the semi-major axis, $T^2\propto a^3$.

Why equal areas.** In a short time $dt$, the line to the planet sweeps area $dA=\frac{1}{2}{r}^{2}d\theta$, so

$$\frac{dA}{dt}=\frac{1}{2}{r}^2\omega =\frac{L}{2m}$$

Gravity points along the line to the Sun, so it exerts no torque about the Sun; $L$ is constant, and so is the areal speed.

Worked example 1 — the third law. A planet's semi-major axis is $4$ times Earth's:

$$\frac{T^2}{T_E^2}=4^3=64\Rightarrow T=8\text{ Earth years}$$

Worked example 2 — speed in orbit. If a planet's farthest distance from the Sun is $3$ times its nearest distance, then $v_{near}{r}_{near}=v_{far}{r}_{far}$ gives $v_{far}=\frac{v_{near}}{3}$.

An everyday example. A stone whirled on a string that winds around your finger speeds up as the string shortens — the same angular momentum rule that makes planets faster near the Sun.

The substance. The law of areas holds for any central force, not only gravity.

**Every pair of point masses attracts along the line joining them with a force $F=\frac{G{m}_1{m}_2}{r^2}$, where $G=6.67\times 10^{-11}$ N m${}^2$ kg${}^{-2}$ and $r$ is the distance between their centres.

For uniform spheres, the whole mass can be treated as concentrated at the centre.

Worked example 1 — two students.** Two $50$ kg students stand $1$ m apart:

$$F=\frac{6.67\times 10^{-11}\times 50\times 50}{1^2}\approx 1.67\times 10^{-7}\text{ N}$$

Worked example 2 — Earth on a bag. With $M_E=6\times 10^{24}$ kg and $R=6.4\times 10^6$ m, a $1$ kg bag feels

$$F=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 1}{(6.4\times 10^6{)}^2}\approx 9.77\text{ N}$$

Worked example 3 — Earth and Moon. With $M_m=7.4\times 10^{22}$ kg and $r=3.84\times 10^8$ m:

$$F\approx \frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 7.4\times 10^{22}}{(3.84\times 10^8{)}^2}\approx 2.0\times 10^{20}\text{ N}$$

Halving the distance makes the force $4$ times larger.

An everyday example. You and the friend sitting beside you attract each other right now, but the force is far too small to notice against the Earth's pull.

The substance. **The bag pulls Earth up with the same $9.77$ N** — Earth just has too much mass to show any acceleration.

**The net gravitational force on a mass is the vector sum of the forces from each other mass acting on it separately, ${\vec{F}}_{net}={\vec{F}}_1+{\vec{F}}_2+\dots$.

Worked example 1 — along a line.** A $1$ kg mass is at the origin, with $2$ kg at $x=1$ m and $3$ kg at $x=-2$ m.

$$F_{+x}=\frac{G(1)(2)}{1^2}=2G{F}_{-x}=\frac{G(1)(3)}{2^2}=0.75G$$

$$F_{net}=1.25G\approx 8.34\times 10^{-11}\text{ N towards }+x$$

Worked example 2 — at right angles. $1$ kg masses at $(1,0)$ and $(0,1)$ m each pull a $1$ kg mass at the origin with force $G$:

$$F_{net}=\sqrt{2}G\approx 9.43\times 10^{-11}\text{ N at }45°$$

Worked example 3 — symmetry. Equal masses at the corners of an equilateral triangle exert zero net force on a mass at its centre.

Worked example 4 — the neutral point. Earth has about $81$ times the Moon's mass. At distance $x$ from Earth along the line to the Moon (separation $d$), the pulls cancel when

$$\frac{81}{x^2}=\frac{1}{(d-x{)}^2}\Rightarrow \frac{x}{d-x}=9\Rightarrow x=0.9d$$

An everyday example. A spacecraft travelling to the Moon passes a point where Earth's and the Moon's pulls exactly balance.

The substance. Forces add as vectors — two equal pulls at $90°$ give $\sqrt{2}$ times one pull, not twice.

**G is measured with a torsion balance: large spheres attract small spheres on a light rod hung from a fine wire, the tiny gravitational couple twists the wire through an angle $\theta$, and balancing it against the wire's restoring torque $k\theta$ gives $G$.**

For small spheres of mass $m$ on a rod of length $L$, each at distance $d$ from a large sphere of mass $M$:

$$\frac{GMm}{d^2}L=k\theta \Rightarrow G=\frac{k\theta d^2}{MmL}$$

Worked example 1 — how tiny the effect is. Suppose $M=20$ kg, $m=0.05$ kg, $d=0.1$ m and $L=0.5$ m:

$$F=\frac{6.67\times 10^{-11}\times 20\times 0.05}{0.01}=6.67\times 10^{-9}\text{ N}\tau =FL\approx 3.3\times 10^{-9}\text{ N m}$$

With $k=1\times 10^{-7}$ N m/rad, the twist is only $\theta \approx 0.033$ rad, about $1.9°$.

Worked example 2 — weighing the Earth. Using $g=\frac{GM}{R^2}$ with $g=9.8$ m/s${}^2$:

$$M=\frac{g{R}^2}{G}=\frac{9.8\times (6.4\times 10^6{)}^2}{6.67\times 10^{-11}}\approx 6.0\times 10^{24}\text{ kg}$$

An everyday example. A hanging mobile toy that twists at the slightest touch shows how a fine suspension can reveal very small forces.

The substance. **$G$ is a universal constant, while $g$ changes from place to place** — do not mix them up.