Circular Motion and Centripetal Acceleration

Orbital mechanics is now infrastructure. GPS satellites circle at $r \approx 26{,}600\ \mathrm{km}$ with periods of half a sidereal day, and every position fix your phone computes rests on equation (7) plus relativistic corrections; Starlink and its competitors fly thousands of spacecraft in low orbits near $7.6\ \mathrm{km/s}$, where equation (6) sets both their speed and their ninety-odd-minute periods. Mission designers speak Kepler fluently: a transfer to Mars is a half-ellipse whose flight time comes from the third law applied to the transfer orbit's semimajor axis. Closer to the ground, the same $\omega^2 r$ that spins samples in a laboratory centrifuge at thousands of $g$ is the design equation for proposed rotating space habitats, where $\omega^2 r = g$ at the rim would manufacture artificial gravity. The same equations weigh the invisible. Stars orbiting the compact radio source at the Milky Way's center — one of them, S2, swings through its 16-year orbit at up to $7700\ \mathrm{km/s}$ — reveal, through $M = 4\pi^2 a^3/(GT^2)$, a central mass of about four million Suns confined to a region smaller than the solar system: the supermassive black hole Sagittarius A*, whose measurement earned the 2020 Nobel Prize in Physics. From a cornering car to a black hole scale, it is one balance: inward force equals $mv^2/r$.

Whirl a ball on a string and feel the physics in your hand. At every instant the ball "wants" to continue straight along the tangent — Newton's first law is not suspended just because the path is curved. The string's tension yanks the ball inward, continuously bending the velocity back toward the circle. Let go, and the answer to "what was the force doing?" is revealed by what stops happening: the ball flies off along the tangent line, not radially outward. Nothing was ever pushing it out; something was always pulling it in. Why $v^2/r$, with the speed appearing twice? Over a short time $\mathrm{d}t$ the velocity vector, of length $v$, rotates through the same small angle $\mathrm{d}\theta = \omega\,\mathrm{d}t$ that the position sweeps. The tip of the velocity vector traces a tiny arc of length $v\,\mathrm{d}\theta$, directed toward the center. Divide by $\mathrm{d}t$: $a = v\omega = v^2/r$. One factor of $v$ because a faster ball turns its velocity vector through the arc more quickly; a second factor because a faster ball has a longer velocity vector to turn. Faster motion on a tighter circle is punished on both counts — which is why the required force grows so viciously with speed.

Now stretch the picture to a planet. Replace the string with gravity and the hand with the Earth: the Moon is the ball, perpetually deflected from its tangent line by just the right amount. Newton checked this quantitatively in 1666: the Moon sits about $60$ Earth radii away, and its measured centripetal acceleration, $(2\pi/T)^2 r \approx 2.7\times10^{-3}\ \mathrm{m/s^2}$, is smaller than $g$ by a factor of $3600 = 60^2$ — exactly the inverse-square prediction. The force that curves a thrown apple and the force that curves the Moon are one law. Newton drew the whole idea as a single thought experiment: fire a cannonball horizontally from a high mountain, then again faster, and faster still. Slow shots curve into the ground nearby; faster ones land farther away; and at one particular speed the ball falls toward the Earth exactly as fast as the ground curves away beneath it — the fall never ends, it just keeps missing:

Angular language first, because a circle has a natural clock that straight-line coordinates hide. Every point on a spinning wheel — hub to rim — sweeps the same angle in the same time, so the honest rate to track is not metres per second but radians per second. That one rate, plus the radius, then hands back the ordinary speed of any chosen point, and one full turn ($2\pi$ radians) defines the period:

where: $\omega$ — angular velocity in $\mathrm{rad/s}$; $v$ — the (tangential) speed; $T$ — the period, the time for one revolution; $f = 1/T$ — the frequency, revolutions per unit time. Read $v = \omega r$ as the exchange rate between the two languages: same $\omega$, bigger circle, faster ride. Every result below is written twice — a $v$-form and an $\omega$-form — and this line is the dictionary that converts between them; reach for the $\omega$-forms whenever the problem hands you a period or a rotation rate rather than a speed.

The next two derivations are the same game played with different pieces: identify which real force can point at the center, and make it pay the bill (3). They are the two classic "force-aiming" problems of mechanics.

Everything so far assumed a frozen speedometer. Release that assumption — a car accelerating out of a bend, a pendulum bob swinging through its arc — and a second acceleration component appears along the path while the centripetal one keeps bending it. The tangential component is driven by the angular acceleration $\alpha$:

where: $a_t$ — the tangential acceleration, the piece that changes the speed; $\alpha$ — the angular acceleration in $\mathrm{rad/s^2}$. The two components are perpendicular by construction — $a_c$ points at the center, $a_t$ along the tangent — so they never share work: $a_c$ turns the velocity, $a_t$ stretches it, and the total acceleration has magnitude $a = \sqrt{a_c^2 + a_t^2}$, tilting away from the pure inward direction whenever the speed is changing. Both components fall out of one master formula. In plane polar coordinates — the natural language for anything circling, developed fully in the vectors-and-coordinates article — twice differentiating the position $\vec r = r\,\hat r$ with the rotating unit vectors $\dot{\hat r} = \dot\theta\,\hat\theta$, $\dot{\hat\theta} = -\dot\theta\,\hat r$ gives

where: $\hat r, \hat\theta$ — the outward radial and forward tangential unit vectors riding with the particle. On a circle the radius is constant, so $\dot r = \ddot r = 0$ and the Coriolis term $2\dot r\dot\theta$ dies: what survives is exactly $-r\dot\theta^2\,\hat r + r\ddot\theta\,\hat\theta$ — the centripetal term (2) and the tangential term (8), derived in one stroke. Equation (9) is the bridge to everything beyond circles: keep $\dot r \neq 0$ and it governs spirals, orbits with changing radius, and the fictitious forces of rotating frames.

Which formula answers which question

1 · The unbanked curve (confidence builder)

A car rounds a flat curve of radius $45\ \mathrm{m}$. The static friction coefficient between tires and dry asphalt is $\mu_s = 0.80$. Find the maximum safe speed, and repeat for an icy day with $\mu_s = 0.25$. (Take $g = 9.8\ \mathrm{m/s^2}$.)

Step 1. Identify the inward force. On a flat road the only horizontal force available is static friction (static, because the contact patch doesn't slide sideways while the car grips): $f_s \le \mu_s N$.

Step 2. Vertical balance: $N = mg$, so the friction ceiling is $\mu_s mg$.

Step 3. Demand the centripetal force from (3): $\dfrac{mv^2}{r} \le \mu_s m g$. The mass cancels — a $1500\ \mathrm{kg}$ SUV and a $900\ \mathrm{kg}$ hatchback share the same limit.

Step 4. Solve: $v_{\max} = \sqrt{\mu_s g r} = \sqrt{0.80 \times 9.8 \times 45} = \sqrt{352.8} = \boxed{18.8\ \mathrm{m/s} \approx 68\ \mathrm{km/h}}$.

Step 5. Icy day: $v_{\max} = \sqrt{0.25 \times 9.8 \times 45} = \sqrt{110.3} = \boxed{10.5\ \mathrm{m/s} \approx 38\ \mathrm{km/h}}$ — the limit falls only as $\sqrt{\mu_s}$, but the required friction grows as $v^2$: taking the icy curve at the dry-day speed demands $3.2$ times more grip than the ice can give.

2 · Parking a satellite over one spot (exam level)

A geostationary satellite must hover over a fixed point on the equator, orbiting once per sidereal day, $T = 86{,}164\ \mathrm{s}$. Find its orbital radius, altitude, and speed. (Use $GM_E = 3.986\times10^{14}\ \mathrm{m^3/s^2}$, $R_E = 6371\ \mathrm{km}$.)

Step 1. The requirement: match Earth's rotation, so $\omega = 2\pi/T = 2\pi/86{,}164 = 7.292\times10^{-5}\ \mathrm{rad/s}$.

Step 2. Gravity supplies the centripetal force. Using the angular form of (3): $\dfrac{GM_E m}{r^2} = m\omega^2 r$, so $r^3 = \dfrac{GM_E}{\omega^2}$.

Step 3. Compute: $r^3 = \dfrac{3.986\times10^{14}}{(7.292\times10^{-5})^2} = \dfrac{3.986\times10^{14}}{5.317\times10^{-9}} = 7.497\times10^{22}\ \mathrm{m^3}$, giving $r = \boxed{4.217\times10^{7}\ \mathrm{m} \approx 42{,}200\ \mathrm{km}}$.

Step 4. Altitude above the surface: $h = r - R_E = 42{,}170 - 6371 \approx \boxed{35{,}800\ \mathrm{km}}$ — the crowded "Clarke belt" where broadcast and weather satellites park.

Step 5. Speed: $v = \omega r = 7.292\times10^{-5} \times 4.217\times10^{7} = \boxed{3.07\ \mathrm{km/s}}$.

Step 6. Cross-check with the inverse-square law: the centripetal demand is $a_c = v^2/r = (3075)^2/4.217\times10^{7} = 0.224\ \mathrm{m/s^2}$, while gravity supplies $g\,(R_E/r)^2 = 9.8 \times (6371/42{,}170)^2 = 0.224\ \mathrm{m/s^2}$ ✓ — the books balance. Notice there is exactly one geostationary radius: equation (7) welds period to radius, so "orbit in 24 hours" is not a choice about speed but a statement about altitude. A satellite any lower must orbit faster than the Earth turns; any higher, slower.

3 · The conical pendulum, with numbers (skills check)

A $0.30\ \mathrm{kg}$ bob whirls on a $1.2\ \mathrm{m}$ string that holds a steady $25°$ from the vertical. Find the period, the bob's speed, and the string tension — and verify the force balance. (Take $g = 9.8\ \mathrm{m/s^2}$.)

Step 1. Period from (5): $T = 2\pi\sqrt{\dfrac{L\cos\phi}{g}} = 2\pi\sqrt{\dfrac{1.2 \times 0.906}{9.8}} = 2\pi\sqrt{0.111} = 2\pi \times 0.333 = \boxed{2.09\ \mathrm{s}}$.

Step 2. Geometry and speed: the circle's radius is $r = L\sin\phi = 1.2 \times 0.423 = 0.507\ \mathrm{m}$; the angular rate is $\omega = 2\pi/T = 3.00\ \mathrm{rad/s}$; so $v = \omega r = 3.00 \times 0.507 = \boxed{1.52\ \mathrm{m/s}}$.

Step 3. Tension from the vertical balance: $T_{\mathrm{str}} = \dfrac{mg}{\cos\phi} = \dfrac{0.30 \times 9.8}{0.906} = \boxed{3.24\ \mathrm{N}}$ — larger than the bob's weight ($2.94\ \mathrm{N}$), because the string must both hold the bob up and pull it around.

Step 4. Verify the centripetal books: the horizontal tension component is $T_{\mathrm{str}}\sin\phi = 3.24 \times 0.423 = 1.37\ \mathrm{N}$, while the demand (3) is $\dfrac{mv^2}{r} = \dfrac{0.30 \times (1.52)^2}{0.507} = 1.37\ \mathrm{N}$ ✓ — the horizontal tension supplies exactly the required centripetal force.

Texts. Morin, Introduction to Classical Mechanics, Ch. 3 (using F = ma) and Ch. 10 · Goldstein, Classical Mechanics, 3rd ed., Ch. 1 · Landau & Lifshitz, Mechanics, Vol. 1, §5 · Taylor, Classical Mechanics, Ch. 1 & 9 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 1 & 9. All equations follow from first principles — differentiation of the rotating position vector and Newton's second law — with no empirical constants beyond the standard symbolic $g$ and $G$. The key intermediate results are the polar unit-vector derivatives $\dot{\hat r}=\dot\theta\hat\theta$, $\dot{\hat\theta}=-\dot\theta\hat r$ and the general polar acceleration $(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta+2\dot r\dot\theta)\hat\theta$, from which the banked-curve result $v=\sqrt{gr\tan\beta}$ and the conical-pendulum period $T=2\pi\sqrt{L\cos\phi/g}$ follow as special cases. Worked-example numbers ($v_{\max} = 18.8$ and $10.5\ \mathrm{m/s}$; $r = 4.217\times10^7\ \mathrm{m}$, $h \approx 35{,}800\ \mathrm{km}$, $v = 3.07\ \mathrm{km/s}$; conical pendulum $T = 2.09\ \mathrm{s}$, $v = 1.52\ \mathrm{m/s}$, $T_{\mathrm{str}} = 3.24\ \mathrm{N}$) are arithmetic recomputed from the boxed formulas; the Moon-test ratio $(2\pi/2.36\times10^6)^2 \times 3.84\times10^8 \approx 2.7\times10^{-3}\ \mathrm{m/s^2} \approx g/3600$ confirms the inverse-square scaling.

Builds on: Velocity and Acceleration · Vectors in Kinematics and Coordinate Systems · Position, Displacement, and Reference Frames · Newton's Second Law. Leads to: Friction Forces · Normal Force · Gravitational Force Near Earth's Surface · Relative Motion and Galilean Transformations.