Motion in Gravitational Fields¶
Syllabus mapping
PY-12-01 — Advanced Mechanics. Covers: Newton's law of universal gravitation; gravitational field strength at a distance from a mass; modelling circular orbits as an approximation to real (elliptical) orbits; Kepler's laws; the relationship between orbital radius and orbital period (law of periods); a secondary-source investigation comparing satellite types.
Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.
You need to know: how gravitational force and field strength depend on distance (not just the constant \(9.8\ \text{m s}^{-2}\) you've used so far), how to model an orbit as uniform circular motion, and how to relate orbital radius to orbital period using Kepler's law of periods.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Circular Motion — centripetal force and acceleration, applied here to gravity specifically
- Forces and Motion — the distinction between mass and weight
Core¶
Newton's law of universal gravitation¶
Every pair of masses in the universe attracts each other with a force:
Symbols: \(F\) — gravitational force between the two masses; \(G\) — the universal gravitational constant (\(6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\)); \(M\), \(m\) — the two masses; \(r\) — distance between their centres (not their surfaces).
This is the general law. \(F_w=mg\), from Forces and Motion, is just this equation evaluated specifically at Earth's surface — a special case, not a separate rule.
Gravitational field strength¶
Rearranging, the field strength (force per unit mass, equivalently the local acceleration due to gravity) at distance \(r\) from a mass \(M\) is:
Unlike the constant \(9.8\ \text{m s}^{-2}\) used throughout Year 11 (which only applies near Earth's surface), \(g\) genuinely changes with distance — it's roughly \(9.8\ \text{m s}^{-2}\) at sea level, noticeably less at the top of a tall mountain, and much less again in low Earth orbit (still enough to keep the ISS falling around the planet, just not enough to feel like "down" the way it does on the ground).
Worked example — field strength at altitude
Find Earth's gravitational field strength at an altitude of \(400\ \text{km}\) (roughly the ISS's orbit). Earth's mass is \(5.97\times10^{24}\ \text{kg}\) and its radius is \(6.37\times10^6\ \text{m}\).
Remember \(r\) is measured from Earth's centre, so \(r = 6.37\times10^6 + 4.00\times10^5 = 6.77\times10^6\ \text{m}\):
Still nearly \(90\%\) of surface gravity — the ISS isn't "beyond gravity," it's in free fall, constantly falling toward Earth but moving sideways fast enough to keep missing it. That's what an orbit actually is.
Near the surface, treat the field as uniform
Over the small height changes in a typical Year 11 problem (a building, a hill, a thrown ball), \(g\) barely changes, so \(\Delta U=mg\Delta h\) from Momentum and Energy is a perfectly good approximation. At orbital-scale distances, that approximation breaks down and you need the full \(1/r^2\) form above — this is exactly why the syllabus separates "near-surface" and "orbital" gravity into different years.
Modelling orbits as uniform circular motion¶
A stable orbit is gravity supplying exactly the centripetal force needed for circular motion at that radius and speed — set Newton's law of gravitation equal to the centripetal force requirement:
Notice \(m\) (the orbiting object's own mass) cancels completely — orbital speed at a given radius doesn't depend on the mass of the thing orbiting, only on the mass being orbited and the radius. Solving for orbital velocity:
Real orbits are actually ellipses, not perfect circles (that's Kepler's first law, below) — treating them as circular is a simplifying approximation that works well for most HSC-level problems, especially for orbits with low eccentricity like most satellites.
Kepler's laws¶
Kepler described planetary motion (later explained by Newton's gravitation) with three laws:
- Law of ellipses: orbits are ellipses, with the central body at one focus, not the centre.
- Law of equal areas: a line from the central body to the orbiting body sweeps out equal areas in equal times — meaning an object moves faster when closer to what it's orbiting, and slower when farther away.
- Law of periods: for objects orbiting the same central mass, the ratio of the cube of orbital radius to the square of orbital period is the same constant for all of them.
The law of periods is the one you'll actually calculate with. Setting \(v=2\pi r/T\) equal to \(v_{orb}\) from above and squaring:
Since the right-hand side depends only on \(G\) and \(M\) (the mass being orbited), every satellite of the same central body — regardless of its own mass or exactly which orbit it's in — gives the same value of \(r^3/T^2\). This makes comparing two satellites of the same planet a ratio problem, no need to know \(M\) at all.
Worked example — comparing two satellites
Satellite A orbits Earth at radius \(r_A = 7.0\times10^6\ \text{m}\) with period \(T_A = 96\ \text{minutes}\). Satellite B orbits at radius \(r_B = 4.2\times10^7\ \text{m}\) (geostationary altitude). Find \(T_B\).
Since both orbit Earth, \(r^3/T^2\) is the same constant for each:
Close to \(24\) hours, as expected — this is essentially the geostationary-orbit calculation, which is why geostationary satellites sit at this specific altitude and no other.
Practical¶
Secondary-source investigation: compare low Earth orbit and geostationary orbit satellite uses — e.g. altitude, period, coverage area, and typical applications (LEO: imaging, ISS, some communications; geostationary: weather, broadcast television, some communications). Use the law of periods above to calculate and compare their orbital speeds and periods directly, rather than just quoting figures.
Advanced
Why circular orbits need a minimum speed for a given radius, and vice versa. From \(v_{orb}=\sqrt{GM/r}\): for any chosen radius there's exactly one speed that gives a stable circular orbit. Too slow at that radius and gravity overpowers the required centripetal force, pulling the orbit inward (toward an elliptical or decaying path); too fast and it would need more inward force than gravity alone supplies, so the orbit widens instead. This is the same idea as the conical pendulum in Circular Motion, just with gravity playing the role tension played there.
Geostationary orbit as a special case. A geostationary satellite has a period matching Earth's rotation (\(\approx24\ \text{hours}\)) and orbits directly above the equator, so it appears stationary from the ground — useful for satellite dishes that never need to re-aim. Its radius is fixed by the law of periods (worked example above) — there's only one altitude where this works, unlike low Earth orbit, which covers a whole range of altitudes and periods.
Extension
Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Kepler's laws were discovered empirically (by fitting Tycho Brahe's naked-eye observational data), decades before Newton derived them theoretically from the inverse-square gravitation law. Deriving the law of periods from Newton's law of gravitation plus circular motion, as done above, is one of the clean, satisfying moments in the whole course where two completely different starting points converge on the same result — worth a depth study exploring that history, or extending the derivation to genuinely elliptical orbits (which requires the semi-major axis in place of \(r\), and calculus beyond this course to derive properly, but is well documented and a good stretch for a student who's found this section's algebra comfortable).
Video/visual resources¶
- 🎥 Khan Academy — TODO: source a Newton's law of gravitation / orbital mechanics explainer. Must clearly connect \(F=GMm/r^2\) to circular motion and derive \(v_{orb}=\sqrt{GM/r}\), not just state formulas. Essential.
- 🎥 Physics High — TODO: check for a NSW-syllabus-aligned gravitation and orbits video, ideally covering the law of periods with a worked satellite-comparison example matching the one above. Nice-to-have alongside the Khan Academy one above.
- 🖥️ NASA Eyes on the Solar System or similar orbital visualisation tool — TODO: source a good interactive that shows real (elliptical) planetary orbits and lets students see Kepler's law of equal areas in action (faster near the central body, slower far away).
Check yourself¶
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Find Earth's gravitational field strength at the Moon's orbital radius, \(r=3.84\times10^8\ \text{m}\), using \(M_{Earth}=5.97\times10^{24}\ \text{kg}\).
Answer
\(g = \dfrac{GM}{r^2} = \dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{(3.84\times10^8)^2} \approx 2.70\times10^{-3}\ \text{m s}^{-2}\) — a tiny fraction of surface gravity, consistent with the Moon being held in orbit rather than falling straight in.
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Two satellites orbit the same planet. Satellite X has orbital radius \(r\) and satellite Y has orbital radius \(4r\). How many times longer is Y's period than X's?
Answer
\(\dfrac{r_X^3}{T_X^2} = \dfrac{r_Y^3}{T_Y^2} \implies T_Y = T_X\sqrt{\left(\dfrac{r_Y}{r_X}\right)^3} = T_X\sqrt{4^3} = T_X\sqrt{64} = 8T_X\) — eight times longer.
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Explain why an astronaut aboard the ISS experiences apparent weightlessness, even though Earth's gravitational field strength at that altitude is still around 90% of its surface value.
Answer
Apparent weightlessness isn't the absence of gravity — it's because the astronaut and the station are both in free fall, accelerating toward Earth at the same rate (\(g\) at that altitude) while also moving sideways fast enough to continually "miss" the planet, which is what an orbit is. With nothing pushing back against the astronaut (no floor supplying a normal force, since the whole station is falling too), there's no sensation of weight — the same effect as the brief weightless feeling at the top of a fairground drop tower, just sustained indefinitely because the sideways speed keeps the fall going forever instead of ending in a landing.