QCE Physics - Unit 3 - Gravity and motion

Orbital Mechanics | QCE Physics

Learn QCE Physics orbital mechanics, including universal gravitation, gravitational fields, Kepler's laws and circular orbit calculations.

Updated 2026-06-15 - 4 min read

QCAA official coverage - Physics 2025 v1.3

Exact syllabus points covered

  1. Describe the Law of Universal Gravitation.
  2. Solve problems involving the magnitude of the gravitational force between two masses using $F = \frac{GMm}{r^2}$.
  3. Describe the concept of gravitational fields.
  4. Solve problems involving the gravitational field strength at a distance from an object using $g = \frac{F}{m} = \frac{GM}{r^2}$.
  5. State the three laws of planetary motion.
  6. Describe the relationship between the Law of Universal Gravitation and uniform circular motion and recognise this as the third law of planetary motion.
  7. Solve problems involving the third law of planetary motion using $\frac{T_a^2}{r_a^3} = \frac{T_b^2}{r_b^3} = \frac{4\pi^2}{GM}$.

Orbital mechanics explains why planets, moons and satellites move the way they do. In QCE Physics, the core idea is that gravity supplies the inward force for orbital motion.

Gravity as centripetal force in an orbit

Original Sylligence diagram for physics orbital mechanics map.

Gravity as centripetal force in an orbit

Newton's law of universal gravitation

Newton's law of universal gravitation states that any two masses attract each other with a force:

$ F = \frac{GMm}{r^2} $

where $G$ is the gravitational constant, $M$ and $m$ are the masses, and $r$ is the distance between their centres. The force is attractive, acts along the line joining the centres of mass and follows an inverse-square relationship.

If distance doubles, the gravitational force becomes one quarter as large. If one mass doubles, the force doubles. This is the same kind of proportional reasoning that appears in data interpretation questions.

Gravitational fields

A gravitational field describes the force per unit mass at a point:

$ g = \frac{F}{m} $

Combining this with universal gravitation gives:

$ g = \frac{GM}{r^2} $

This field strength depends on the source mass $M$ and the distance from its centre. It does not depend on the test mass placed there. A $1\ \mathrm{kg}$ object and a $10\ \mathrm{kg}$ object at the same point in the same gravitational field have the same field strength, though the heavier object experiences a larger weight force.

Kepler's three laws

Kepler's laws describe planetary motion:

  1. Planets move in elliptical orbits with the central body at one focus.
  2. A line from the planet to the central body sweeps out equal areas in equal times.
  3. The square of orbital period is proportional to the cube of orbital radius or semi-major axis.

For circular-orbit QCE problems, the third law is usually written as:

$ \frac{T_a^2}{r_a^3} = \frac{T_b^2}{r_b^3} $

For an object orbiting a central mass $M$:

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

This equation shows that the period depends on orbital radius and the central mass, not the mass of the orbiting object.

Why gravity gives Kepler's third law

For a circular orbit, gravity provides the centripetal force:

$ \frac{GMm}{r^2} = \frac{mv^2}{r} $

The orbiting mass $m$ cancels:

$ \frac{GM}{r^2} = \frac{v^2}{r} $

Using $v=2\pi r/T$ leads to:

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

That is the physics connection between universal gravitation and Kepler's third law. Kepler's laws describe the pattern; Newton's gravitation explains why the pattern exists.

Worked example

Interpreting orbital data

Graphs of $T^2$ against $r^3$ should be linear for objects orbiting the same central mass. The gradient is:

$ \frac{4\pi^2}{GM} $

That means the central mass can be inferred from orbital data. If the graph is not linear, check whether the distances are measured from the centre, whether the orbit is circular enough for the model, and whether all objects orbit the same central mass.

Quick check

Sources