Circular Motion & Gravitation
High SchoolMotion in a circle
An object in uniform circular motion travels at constant speed but its velocity constantly changes direction — so it accelerates. This centripetal acceleration points toward the centre:
By Newton's second law a net inward centripetal force Fc = mv²/r is required. This is not a new force — it is provided by tension, gravity, friction or the normal force, depending on the situation.
Newton's law of universal gravitation
Every mass attracts every other mass:
For an orbit, gravity supplies the centripetal force: GMm/r² = mv²/r. This yields orbital speed v = √(GM/r) and, with v = 2πr/T, Kepler's third law T² ∝ r³.
There is no centrifugal force
The outward "centrifugal force" you feel on a merry-go-round is not a real force — it is your body's inertia trying to go straight while the seat pushes you inward. In an inertial frame, the only real horizontal force is the inward (centripetal) one. Remove it — cut the string — and the object flies off along a tangent, not radially outward.
Orbiting is "perpetual falling". The Moon is continuously accelerating toward Earth (falling), but its tangential velocity carries it sideways just fast enough that it never gets closer — it falls around the Earth. Astronauts are weightless not because gravity is absent (it's ~90% of surface value at the ISS) but because they and their craft fall together — a state of free fall.
F = GMm/r² — governs both falling fruit and planetary motion.Worked example — geostationary orbit height
At what altitude must a satellite orbit to stay above a fixed point on Earth's equator? (M = 5.97×10²⁴ kg, G = 6.674×10⁻¹¹, period T = 24 h = 86 400 s.)
- Gravity provides centripetal force:
GMm/r² = m(2π/T)²r. Mass m cancels. - Rearrange for orbital radius:
r³ = GMT² / (4π²). - Numerator:
GMT² = 6.674×10⁻¹¹ × 5.97×10²⁴ × (86 400)² = 2.97×10³⁰. - Divide by
4π² = 39.48:r³ = 7.53×10²⁸, sor = 4.22×10⁷ m. - Subtract Earth's radius
6.37×10⁶ m: altitude≈ 3.59×10⁷ m ≈ 35 900 km.
This famous ~36 000 km figure is where all TV and weather satellites sit, appearing motionless in the sky.
A 0.50 kg ball on a 1.2 m string is whirled at 3.0 m·s⁻¹ horizontally. Find the tension.
Tension provides the centripetal force: T = mv²/r = 0.50 × 3.0² / 1.2 = 0.50 × 9.0 / 1.2 = 3.75 N. (If whirled in a vertical circle, weight would add or subtract at top/bottom.)
Why doesn't the Moon fall into the Earth if gravity pulls it inward?
It is falling — continuously — but it also moves sideways at about 1 km·s⁻¹. The inward fall and sideways motion combine so the Moon curves around Earth at constant distance. Gravity changes its direction, not its speed.
If Earth's radius halved but mass stayed the same, how would your weight change?
Surface gravity g = GM/R². Halving R quarters R², so g quadruples. Your weight would become four times larger — though your mass is unchanged.
Two planets orbit a star; one is 4× farther out. Compare their periods.
Kepler's third law: T² ∝ r³. If r increases 4×, then T² increases 4³ = 64×, so T increases √64 = 8×. The outer planet's year is eight times longer.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1A 0.20 kg ball on a 0.50 m string moves at 4.0 m/s in a circle. Find the tension.
- a = v^2/r = 4.0^2 / 0.50 = 32 m/s^2
- F = m*a = 0.20 * 32
Example 2Estimate the orbital speed near Earth's surface. (g = 9.8, R = 6.4e6 m)
- Gravity provides centripetal force: g = v^2/R
- v = sqrt(g*R) = sqrt(9.8 * 6.4e6)
- = sqrt(6.27e7)
Now you try
Work each one out first, then tap to reveal the worked answer.