What are geostationary orbits?

A geostationary satellite stays above the same point on the equator, which requires its period to equal one sidereal day, T=8.6410^4\ s. Rearranging Kepler's third law T^2=4π^2GMr^3 for r gives a single fixed radius of about 4.210^7\ m (roughly 3.610^4\ km above the surface). The orbit must also lie in the equatorial plane and travel west to east; otherwise the satellite would drift relative to the ground. This is why communications and weather satellites that must point at a fixed dish all sit at the same altitude.

§-Syllabus dot point

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 4.89 min answerUpdated 2026-06-02

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

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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 $6.37\times10^{6}\ \text{m}$ .

Step 1: Find the orbital radius.

The orbital radius is measured from Earth's centre, so the altitude of $400\ \text{km}$ must be added to Earth's radius. Using $r$ from the centre avoids the most common error in this type of question.

r=6.37\times10^{6}+4.00\times10^{5}=6.77\times10^{6}\ \text{m}.

Step 2: Calculate the orbital speed.

Setting gravity equal to the centripetal force and cancelling the satellite mass gives $v=\sqrt{GM/r}$ . The numerator $GM=3.98\times10^{14}\ \text{m}^3\text{s}^{-2}$ .

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 $7.7\ \text{km s}^{-1}$ .

Step 3: Calculate the orbital period.

The satellite travels one full circumference $2\pi r$ in one period $T$ , so $T = 2\pi r / v$ . This is a useful sanity check: a real low-Earth orbit is close to 90 minutes.

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.

Geostationary orbits

A geostationary satellite stays above the same point on the equator, which requires its period to equal one sidereal day, $T=8.64\times10^4\ \text{s}$ . Rearranging Kepler's third law $T^2=\dfrac{4\pi^2}{GM}r^3$ for $r$ gives a single fixed radius of about $4.2\times10^7\ \text{m}$ (roughly $3.6\times10^4\ \text{km}$ above the surface). The orbit must also lie in the equatorial plane and travel west to east; otherwise the satellite would drift relative to the ground. This is why communications and weather satellites that must point at a fixed dish all sit at the same altitude.

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20237 marksA satellite orbits Earth (

M=5.97\times10^{24}\ \text{kg}

) at an altitude of

2.00\times10^{6}\ \text{m}

above the surface (Earth radius

6.37\times10^{6}\ \text{m}

). (a) Calculate the orbital speed of the satellite. (b) Calculate the orbital period. (c) Another satellite is to be placed in an orbit with twice this orbital radius. Without a full calculation, state how its period compares.

Show worked answer →

A 7 mark calculation rewards the correct radius, orbital speed, period and a Kepler scaling.

(a) Orbital speed: Orbital radius $r=6.37\times10^{6}+2.00\times10^{6}=8.37\times10^{6}\ \text{m}$ .
$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{8.37\times10^{6}}}=\sqrt{4.76\times10^{7}}=6.9\times10^{3}\ \text{m s}^{-1}.$
(b) Period: $T=\dfrac{2\pi r}{v}=\dfrac{2\pi(8.37\times10^{6})}{6.9\times10^{3}}=7.6\times10^{3}\ \text{s}$ (about $127$ minutes).
(c) Doubling the radius: From $T^2\propto r^3$ , doubling $r$ multiplies $T^2$ by $2^3=8$ , so $T$ increases by a factor of $\sqrt{8}=2.83$ .

Markers reward adding the radius, $v=\sqrt{GM/r}$ , $T=2\pi r/v$ and the $\sqrt{8}$ Kepler factor.

WACE 20205 marksExplain, using Newton's law of universal gravitation and the requirement for circular motion, why the mass of a satellite does not affect its orbital speed at a given radius.

Show worked answer →

A 5 mark explanation needs both equations and the cancellation argument.

Gravity provides the centripetal force: For a satellite the only force is gravity, and it must supply exactly the centripetal force needed for circular motion, so
$G\frac{Mm}{r^2}=\frac{mv^2}{r}.$
Cancel the satellite mass: The satellite mass $m$ appears on both sides and cancels, leaving $v=\sqrt{\dfrac{GM}{r}}$ , which depends only on the central mass $M$ and the orbital radius $r$ .
Conclusion: Because $m$ cancels, two satellites of different mass at the same radius have the same orbital speed. This is the same reason all objects fall with the same acceleration in a given gravitational field.

Markers reward equating the two forces, the cancellation of $m$ and the conclusion that speed depends only on $M$ and $r$ .