Apply Newton's law of universal gravitation and the field model to orbital motion.

What this dot point is asking

This dot point links gravity to circular motion, showing how the same inverse-square force that makes an apple fall also keeps the Moon and artificial satellites in orbit.

Newton's law of universal gravitation

Newton proposed that any two point masses attract each other along the line joining them with a force proportional to the product of their masses and inversely proportional to the square of their separation:

Here $G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ is the universal gravitational constant and $r$ is the distance between the centres of the masses. The inverse-square dependence is crucial: doubling the separation reduces the force to a quarter.

The gravitational field

It is often more useful to describe gravity as a field. The gravitational field strength $g$ at a point is the force per unit mass placed there:

where $M$ is the mass creating the field. Near Earth's surface this evaluates to about $9.8\ \text{N kg}^{-1}$, the same number as the free-fall acceleration in $\text{m s}^{-2}$. Field strength is a vector pointing toward the mass that creates it, and it falls off as the inverse square of distance from the centre.

The field model is powerful because it separates the source of the field from the object that responds to it. Once you know $g$ at a point, the force on any mass placed there is simply $F = mg$, with no need to revisit the inverse-square calculation. This is the same logic used for electric fields, and it lets you treat field strength as a property of the space itself, mapped out by field lines that point toward the source mass.

Comparing gravity with Coulomb's law

Newton's law of gravitation and Coulomb's law share the same inverse-square form, $F \propto \dfrac{1}{r^2}$, with a product of source quantities on top. The differences are instructive: gravity acts on mass, which is always positive, so it is always attractive, while the electric force acts on charge of either sign and can attract or repel. Gravity is also vastly weaker: between two protons the Coulomb force exceeds the gravitational force by about $10^{36}$ times. Recognising this shared structure lets you carry over the same ratio techniques between the two topics.

Orbital motion

For a satellite in a stable circular orbit, gravity is the only force acting and it points toward the central body, so gravity provides exactly the centripetal force needed:

The satellite mass $m$ cancels, leaving the orbital speed:

A higher orbit means a slower speed. Substituting $v = \dfrac{2\pi r}{T}$ and rearranging gives Kepler's third law:

so $T^2$ is proportional to $r^3$ for all satellites of the same central body.

A geostationary satellite is a special case: it has a period of exactly one day so it stays above the same point on the equator. Setting $T = 86400\ \text{s}$ in Kepler's third law gives an orbital radius of about $4.2 \times 10^7\ \text{m}$.

When tackling orbital questions, decide first whether you need the field equation or the orbit equation. Use $g = \dfrac{GM}{r^2}$ for field strength and weight, and use $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ whenever the body is actually orbiting, since that is what links gravity to circular motion.