A-Level H2 Physics: Circular Motion & Gravitation — One Equation Behind Every Orbit and Loop-the-Loop Question

Why This Topic Quietly Costs So Many Marks

Circular motion and gravitation sit in different chapters of the H2 Physics syllabus, and most students prepare for them as if they were unrelated. That is the first mistake. Every circular motion question — a ball on a string, a car on a banked road, a satellite orbiting the Earth — is asking exactly one question: what force, or combination of forces, is providing the centre-seeking force? Once you can answer that question cleanly for any setup, the rest is algebra.

The reason students lose marks here is not that the physics is hard. It is that they treat “centripetal force” as a new, mysterious force to add into their diagrams, rather than recognising it as the net result of the real forces already present (gravity, tension, normal force, friction). This single misunderstanding is behind the majority of wrong answers we see in tuition — from forgetting the weight component at the top of a vertical circle, to double-counting forces on a banked track.

The One Idea That Unifies Everything

For any object moving in a circle at constant speed, the net force acting on it always points toward the centre of the circle, and its magnitude is:

F_{net, toward centre} = mv²/r = mrω²

That is it. There is no separate “centripetal force” pushing or pulling from nowhere — centripetal force is a label for whatever combination of real forces (gravity, tension, normal reaction, electrostatic attraction, etc.) happens to be doing that job at that instant. The entire syllabus's worth of circular motion questions collapses into one procedure:

Identify every real force acting on the object.
Draw a clean free-body diagram at the specific point in the circle the question asks about.
Resolve forces along the radius (toward the centre is positive) and find the net inward force.
Set that net inward force equal to mv²/r (or mrω²).
Solve for whatever the question wants — speed, tension, minimum angle, and so on.

Gravitation is simply the special case where the force providing that centripetal pull is gravity itself, governed by Newton's Law of Gravitation:

F = GMm / r²

Combine these two equations — F = mv²/r and F = GMm/r² — and every satellite, orbit, and “gravitational field strength at height h” question in the H2 syllabus follows from setting them equal to each other. This is the “aha moment” we aim for in lessons: students stop seeing gravitation as a separate topic and start seeing it as circular motion with one specific force plugged in.

The Core Toolkit (Know These Cold, Derive Everything Else)

Angular speed: ω = v/r = 2π/T
Centripetal acceleration: a = v²/r = rω²
Centripetal force: F = mv²/r = mrω²
Newton's Law of Gravitation: F = GMm/r²
Gravitational field strength: g = GM/r² (this is force per unit mass, and it is NOT always 9.81 m s^-2 — that value only holds at Earth's surface)
Gravitational potential: φ = −GM/r (always negative, zero defined at infinity)
Total energy of an orbiting satellite: E = −GMm/2r (negative, because the satellite is bound)

Notice these are not eight separate things to memorise — they are two ideas (F = mv²/r and F = GMm/r²) combined and rearranged in different ways depending on what the question gives you.

Worked Examples: Applying the Framework

Example 1 — Vertical Circular Motion (the classic “bucket at the top” question)

A bucket of water is swung in a vertical circle of radius 0.80 m. Find the minimum speed at the top of the circle for the water to stay in the bucket.

Follow the framework. At the top of the circle, the forces acting on the water are its weight (mg, pointing down, i.e. toward the centre) and the normal reaction from the bucket base (N, also toward the centre, since the bucket is above the water). Both forces point toward the centre here, so:

mg + N = mv²/r

The water is on the verge of leaving the bucket precisely when N = 0 — the bucket base can only push, not pull, so the minimum speed occurs when the reaction force drops to zero and gravity alone provides the centripetal force:

mg = mv²/r ⇒ v_min = √(gr) = √(9.81 × 0.80) ≈ 2.80 m s^-1

Students who memorise “v = √(gr) at the top” without understanding why N = 0 at minimum speed cannot adapt when the question instead asks for tension in a string at the bottom of the circle, where weight now points away from the centre:

T − mg = mv²/r ⇒ T = mg + mv²/r

Same framework, different point in the circle, different sign for gravity's contribution. This is exactly why step 2 — drawing the diagram at the specific point asked — is non-negotiable.

Example 2 — Banked Track (no friction)

A car travels around a circular banked track of radius 100 m at the correct speed such that no friction is required. The bank is inclined at angle θ to the horizontal. Find v in terms of θ, r and g.

Two forces act on the car: weight (mg, straight down) and the normal reaction (N, perpendicular to the track surface). Resolve these into vertical and horizontal (radial) components.

Vertically, there is no acceleration, so: N cosθ = mg

Horizontally (toward the centre), the horizontal component of N provides the entire centripetal force since there's no friction: N sinθ = mv²/r

Dividing the two equations eliminates both N and m — a pattern worth recognising, since it appears repeatedly in banking and conical pendulum problems:

tanθ = v²/(rg) ⇒ v = √(rg tanθ)

Example 3 — Geostationary Satellite

Find the orbital radius of a geostationary satellite. (M_Earth = 5.97 × 10²⁴ kg, G = 6.67 × 10^-11 N m² kg^-2)

Here, gravity itself provides the centripetal force, so set the two force equations equal:

GMm/r² = mrω²

The mass of the satellite, m, cancels — a useful check that you have set up the equation correctly, since orbital radius should never depend on the satellite's own mass. Rearranging:

r³ = GM/ω²

A geostationary satellite has a period T equal to one sidereal day, 86,164 s (not exactly 24 hours — a common but usually acceptable simplification at H2 level unless the question specifies otherwise). Using ω = 2π/T:

r³ = GMT²/4π² = (6.67×10^-11 × 5.97×10²⁴ × 86164²) / 4π²

Solving gives r ≈ 4.22 × 10⁷ m, roughly 42,200 km from Earth's centre — around 35,800 km above the surface, which matches the value used by real communications satellites. Being able to derive this from first principles, rather than recalling “36,000 km” as a fact, is exactly what distinguishes a Band 1 answer from a Band 3 one when the examiner changes the planet or the numbers.

Example 4 — Gravitational Field Strength Above the Surface

Given that g at Earth's surface (radius R) is 9.81 m s^-2, find g at a height of 2R above the surface (i.e. at distance 3R from the centre).

Since g = GM/r², g is inversely proportional to r². At the surface, r = R; at the new point, r = 3R (height 2R above a surface at radius R gives a distance of R + 2R = 3R from the centre — a distinction students frequently get wrong by using r = 2R instead).

g_new/g_surface = (R/3R)² = 1/9 ⇒ g_new = 9.81/9 ≈ 1.09 m s^-2

Exam Traps That Cost the Most Marks

Treating “centripetal force” as an extra force in the diagram. It never appears as its own arrow — it is the resultant of the real forces you've already drawn. Adding an extra arrow labelled F_c is one of the most common errors we correct at Intuitional.
Forgetting the direction of weight relative to the centre. At the top of a vertical circle, weight points toward the centre; at the bottom, it points away. Getting this sign wrong flips the entire answer.
Using g = 9.81 m s^-2 at altitude or on another planet. This value is specific to Earth's surface. Any question involving height above the surface, another planet, or a satellite requires g = GM/r² with the correct r measured from the centre of the mass, not the surface.
Confusing radius from centre with height above surface. As in Example 4, r = R + h, not r = h. This single substitution error is extremely common under exam pressure.
Assuming uniform circular motion when the question describes non-uniform motion. If speed is changing (e.g. a pendulum swinging, not a satellite in stable orbit), there is also a tangential component of force — the mv²/r equation only accounts for the radial (centripetal) component.
Sign errors with gravitational potential and energy. Both φ and total orbital energy E are negative by convention (zero defined at infinite separation). Students often drop the negative sign when calculating energy required to move a satellite to a higher orbit, leading to answers with the wrong sign or wrong physical interpretation (do work on the system vs. energy released).
Not distinguishing weight from gravitational force in orbit. A common conceptual trap: students think an orbiting astronaut is “weightless” because gravity has vanished. In reality, gravity is very much present and is exactly what provides the centripetal force keeping the astronaut in orbit — the sensation of weightlessness comes from being in free fall together with the spacecraft, not from an absence of gravitational force.

How to Practise This Properly

Do not practise by sorting questions into “vertical circle questions” and “satellite questions” and memorising a formula for each type. That approach falls apart the moment a question combines the two ideas — for instance, asking for the minimum speed of a satellite skimming just above a planet's surface, which is really a vertical circular motion setup with gravity as the sole radial force.

Instead, practise the five-step framework on every question, regardless of the surface story:

List every real force acting on the object.
Draw the free-body diagram at the exact point specified.
Resolve forces along the radius, taking toward-the-centre as positive.
Set the net radial force equal to mv²/r.
Substitute in gravitation (F = GMm/r²) only if gravity is the force providing that pull.

Once this becomes automatic, past-year questions on circular motion and gravitation — no matter how the numbers or context are dressed up — start to look like variations on the same five steps rather than a fresh problem each time. That shift, from “which formula do I use” to “what is the physics actually doing here,” is the difference we aim to build in every H2 Physics student who walks through our doors in Teck Whye.