What is projectile motion?

WAPhysicsSyllabus dot point

How does a gravitational field govern projectile, circular and orbital motion?

Model gravitation as a field and apply it to projectile motion, uniform circular motion and satellite orbits

A focused answer to the WACE Year 12 Physics Unit 3 dot point on gravitation and orbital motion. Newton's law of gravitation, gravitational field strength, projectile and circular motion, and deriving orbital speed and Kepler's third law for satellites.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

WACE wants you to treat gravity as a field, not just a force, and then use that single idea to explain three kinds of motion: projectiles, objects moving in circles, and satellites in orbit. The mathematics is the same Newtonian framework throughout, so the skill being tested is choosing the right relationship and combining them cleanly.

Newton's law of universal gravitation

Every mass attracts every other mass with a force

F=G\frac{m_1m_2}{r^2}

where $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ and $r$ is the centre-to-centre separation. The force is always attractive and obeys an inverse-square law: double the separation and the force falls to a quarter.

The gravitational field strength is the force per unit mass at a point,

g=\frac{F}{m}=G\frac{M}{r^2}

where $M$ is the mass producing the field. Near Earth's surface this evaluates to about $9.8\ \text{N kg}^{-1}$ . Notice $g$ depends only on the source mass and distance, not on the test mass, which is why all objects fall with the same acceleration.

Projectile motion

A projectile moves in a uniform downward field, so it has constant horizontal velocity and constant vertical acceleration $g$ . The two directions are independent. Horizontally,

x=u_x t

and vertically,

v_y=u_y-gt,\qquad y=u_y t-\tfrac{1}{2}gt^2.

Split the launch velocity into $u_x=u\cos\theta$ and $u_y=u\sin\theta$ , solve the vertical equation for the time of flight, then feed that time into the horizontal equation for range.

Uniform circular motion

An object moving in a circle of radius $r$ at constant speed $v$ has a centripetal acceleration directed at the centre,

a=\frac{v^2}{r},\qquad F_{net}=\frac{mv^2}{r}.

This is not a new force; some real force (tension, friction, normal force or gravity) provides it. The speed relates to the period $T$ by $v=\frac{2\pi r}{T}$ .

Satellites and orbits

For a satellite, gravity is the only force and it supplies the centripetal force. Setting them equal,

G\frac{Mm}{r^2}=\frac{mv^2}{r}.

The satellite mass cancels, giving the orbital speed

v=\sqrt{\frac{GM}{r}}.

Substituting $v=\frac{2\pi r}{T}$ and rearranging gives Kepler's third law,

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

so $T^2\propto r^3$ for any object orbiting the same central mass. A geostationary satellite is the special case where $T$ equals one sidereal day, fixing its orbital radius.

Orbital speed and period of a low satellite

Satellite at 400 km altitude

Take Earth's mass $M=5.97\times10^{24}\ \text{kg}$ and radius $6.37\times10^{6}\ \text{m}$ . The orbital radius is $r=6.37\times10^{6}+4.00\times10^{5}=6.77\times10^{6}\ \text{m}$ .

Orbital speed:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6.77\times10^{6}}}.

The numerator is $3.98\times10^{14}$ , so $v=\sqrt{5.88\times10^{7}}=7.67\times10^{3}\ \text{m s}^{-1}$ , about $7.7\ \text{km s}^{-1}$ .

Period:

T=\frac{2\pi r}{v}=\frac{2\pi(6.77\times10^{6})}{7.67\times10^{3}}=5.55\times10^{3}\ \text{s}\approx 92\ \text{min}.

That 90-minute figure matches a real low-Earth orbit, a good sanity check.

Energy in orbit and gravitational field graphs

You should also be able to read field and force graphs. A graph of $g$ against $r$ falls off as $1/r^2$ ; the area under a force-distance graph relates to work done, which connects to gravitational potential energy. Be ready to describe how kinetic and gravitational potential energy trade off as an orbit changes radius.

Discuss this in the community ->