Circular orbit — velocity, period, energy: Visual walkthrough

Step 1 — What "moving in a circle" secretly requires

WHAT. Picture a ball on the end of a string, whirled around your hand at a steady speed. It traces a circle. Now ask: is anything pulling on it? Yes — the string. If you cut the string, the ball flies off in a straight line, tangent to the circle.

WHY. This is the deepest fact of circular motion: an object moving in a circle is constantly being yanked toward the center. Without that inward yank, it would go straight (that is what "no force" does — Newton's first law). So circular motion is not "no force"; it is a specific inward force at every instant. We give that inward direction a name: centripetal (Latin for "center-seeking").

PICTURE. The blue ball moves along the circle. The green arrow is its velocity — always tangent, pointing the way it is heading. The orange arrow is the inward pull that bends the path. Cut the pull (dashed gray) and it escapes straight.

Step 2 — Building the inward acceleration $v^{2} / r$ from a tiny arc

WHAT. Before we can say how big the inward force is, we must find how big the inward acceleration is — because turning is nothing but the velocity vector changing direction, and "acceleration" just means "rate the velocity changes." Let us construct $v^{2} / r$ with our own hands, using two triangles.

Watch the body travel a tiny slice of the circle in a short time $Δ t$ (Greek letter delta, meaning "a small change in"). It sweeps a small angle $Δ θ$ at the center.

Position triangle (left of the figure). The two radii (each length $r$ ) to the start and end points, plus the little chord between them, form a skinny triangle with tip-angle $Δ θ$ . For a tiny slice the chord is essentially the arc length, and arc $=$ radius $\times$ angle, so the chord $\approx r Δ θ$ .
Velocity triangle (right of the figure). The velocity arrow has the same length $v$ at start and end (steady speed) but has rotated by the very same angle $Δ θ$ (because it stays tangent, and the tangent turns exactly as much as the radius does). Lay the two velocity arrows tail-to-tail: they form a skinny triangle, also with tip-angle $Δ θ$ , whose two equal sides have length $v$ . The short third side is the change in velocity $Δ v$ , and it points inward, toward the center. Same reasoning as before: $Δ v \approx v Δ θ$ .

WHY these two triangles. They are the same shape — both isosceles with the identical tip-angle $Δ θ$ — so their short-side / long-side ratios match:

$\frac{Δ v}{v} = Δ θ = \frac{chord}{r} = \frac{v Δ t}{r}$

(using chord $\approx$ distance travelled $= v Δ t$ ). Rearranging,

$acceleration a \frac{Δ v}{Δ t} = \frac{v ^{2}}{r}$

That left-hand ratio is acceleration (change in velocity per unit time), and it points inward because $Δ v$ did. In the limit of a truly tiny slice ( $Δ t \to 0$ ) the approximations become exact:

$a_{c} = \frac{v ^{2}}{r} (directed toward the center)$

Term by term: $v^{2}$ on top (faster body turns its velocity more per second, and the arrows are longer and swing faster — hence the square); $r$ on the bottom (a wider circle turns the velocity through a smaller angle per metre travelled).

PICTURE. Left: the position triangle, two radii $r$ and the chord $v Δ t$ , tip-angle $Δ θ$ . Right: the velocity triangle, two arrows of length $v$ , tip-angle $Δ θ$ , short inward side $Δ v$ . The matching angle is highlighted in green on both.

Now the force. Newton says force $=$ mass $\times$ acceleration. Multiply our freshly built $a_{c}$ by the body's mass $m$ :

$F_{c} = m a_{c} = \frac{m v ^{2}}{r}$

$m$ (kilograms) — a heavier body needs more pull to be turned equally.
$v^{2}$ — double the speed, quadruple the required pull.
$r$ on the bottom — a wider, gentler circle needs less pull.

Step 3 — Who pays the bill? Gravity, and only gravity

WHAT. In space there is no string. The only force on an orbiting satellite is gravity — the pull of the big body (mass $M$ , e.g. Earth) on the small body (mass $m$ , the satellite). Newton's law of universal gravitation gives its size at distance $r$ :

$F_{g} = \frac{G M m}{r ^{2}}$

Term by term:

$G$ = the universal gravitational constant, $6.674 \times 1 0^{- 11}$ (a fixed number of nature; it sets the strength of gravity everywhere).
$M$ = mass of the big central body (kg) — bigger planet, stronger pull.
$m$ = mass of the small orbiting body (kg) — heavier satellite, more pull on it.
$r^{2}$ = distance squared, on the bottom — the inverse-square law: go twice as far and gravity drops to a quarter. (See Newton's Law of Universal Gravitation.)

WHY. Crucially, gravity points toward the center — straight at the big body. That is exactly the centripetal direction from Steps 1–2. So gravity is perfectly positioned to be the inward pull the circle demands. It is the only candidate, so it must be the one paying the bill.

PICTURE. The Earth at center; the satellite on its circle. The orange arrow (gravity) points from satellite to Earth's center — the same direction as the "center-seeking" arrow of Step 1.

Step 4 — Set the bill equal to the payment

WHAT. The circle demands $\frac{m v ^{2}}{r}$ (Step 2). Gravity offers $\frac{GM m}{r ^{2}}$ (Step 3). For a steady circular orbit at constant radius, these must be exactly equal — no leftover, no shortfall:

$gravity supplies \frac{GM m}{r ^{2}} = circle demands \frac{m v ^{2}}{r}$

WHY equal, not just "greater than"? If gravity were more than the demand, the surplus pull would drag the satellite inward — the radius would shrink (it spirals down). If gravity were less, the body would swing outward — the radius grows (it flies into a wider ellipse). Only exact balance keeps $r$ constant. That single equation is the entire physics of a circular orbit.

PICTURE. A balance scale: gravity on one pan, the centripetal requirement on the other, level. Beside it, the two failure modes — too much pull spirals in (red), too little flies out (blue).

Step 5 — Cancel and solve for the speed

WHAT. Now just do algebra on the balance equation. Both sides contain the satellite mass $m$ — divide it out. Both sides contain $r$ on the bottom (one power); multiply through to leave one:

\;\;\xrightarrow{\;\div\, m\;}\;\; \frac{GM}{r^2}=\frac{v^2}{r} \;\;\xrightarrow{\;\times\, r\;}\;\; \frac{GM}{r}=v^2$$ Take the (positive) square root: $$\boxed{\,v=\sqrt{\dfrac{GM}{r}}\,}$$ Term by term in the answer: - $G,M$ on top — a bigger/stronger central body means faster orbits. - $r$ on the bottom, under a square root — bigger orbit → **smaller** speed, falling off like $1/\sqrt{r}$. - **$m$ is gone.** It cancelled. A feather and a bus at the same altitude orbit at the *same* speed. **WHY the cancellation matters.** A heavier satellite feels more gravity (good, more payment) but also demands more centripetal force (bigger bill) — and both scale with $m$ identically, so $m$ disappears. This is the same reason all objects fall at the same rate. **PICTURE.** The curve $v=\sqrt{GM/r}$: high and steep for small $r$, gently sinking for large $r$. Two dots — feather and bus — sit on the *same* point at the same $r$. > [!mistake] "Higher orbit → faster" > *Feels right:* more altitude, more energy. *Truth:* the $r$ is on the **bottom** under the root, so higher = **slower**. The extra energy is potential, not kinetic. (Full story in the parent note.) --- ## Step 6 — From speed to period (one free lap) **WHAT.** A period $T$ is the time for one full loop. One loop covers the circle's circumference $2\pi r$ (distance) at speed $v$. Time = distance ÷ speed: $$T=\frac{2\pi r}{v}$$ Now substitute the $v$ we just earned: $$T=\frac{2\pi r}{\sqrt{GM/r}} =2\pi r\sqrt{\frac{r}{GM}} =2\pi\sqrt{\frac{r^3}{GM}}$$ - $2\pi r$ = the lap distance (circumference). - dividing by $\sqrt{GM/r}$ and tidying pulls one extra $r$ under the root, giving $r^3$. Square both sides to see the pattern: $$T^2=\frac{4\pi^2}{GM}\,r^3\quad\Longrightarrow\quad T^2\propto r^3$$ **WHY this is famous.** That last line is ==Kepler's Third Law== — and here it drops straight out of Newton's gravity, with the proportionality constant $4\pi^2/GM$ handed to us for free. See [[Kepler's Three Laws]]. **PICTURE.** Three orbits of growing radius; the bar chart beside them shows $T$ shooting up faster than $r$ — the $r^{3/2}$ steepness. --- ## Step 7 — Degenerate & limiting cases (where the formula is tested) **WHAT.** A formula is only trustworthy if it behaves sanely at the extremes. Check the edges: - **$r\to\infty$ (infinitely far):** $v=\sqrt{GM/r}\to 0$ and $T=2\pi\sqrt{r^3/GM}\to\infty$. Far away, gravity is feeble, so you creep around forever. ✓ Sensible. - **$r\to$ surface radius (lowest possible orbit):** $v$ is *largest* — this is the minimum orbital speed a planet allows, grazing the surface. For Earth, about $7.9$ km/s. - **$r\to 0$:** $v\to\infty$ — unphysical, but honestly so: you cannot orbit *at* the center; the math screams to warn you. - **$M\to 0$ (no central body):** $v\to0$, $T\to\infty$ — no pull, no bending, the "orbit" degenerates into a straight line at infinite radius. ✓ - **Compare with [[Escape Velocity]]:** set total energy to zero and you get $v_{\text{esc}}=\sqrt{2GM/r}=\sqrt{2}\,v_{\text{circ}}$. So circular speed is exactly $1/\sqrt2\approx0.707$ of escape speed at the same radius — the orbit lives *below* the escape line. **WHY show these.** Every reader will eventually wonder "what if $r$ is huge / tiny / the planet vanishes?" — here are all of them, none left to chance. **PICTURE.** The $v(r)$ curve again, now with two lines: the orange circular-speed curve and, above it by a factor $\sqrt2$, the red escape-speed curve. The gap between them is the family of bound elliptical orbits (see [[Elliptical Orbits and the Vis-viva Equation]]). --- ## The one-picture summary Everything above, compressed into two clean panels. **First**, the single idea that runs the whole show — gravity *is* the string, so its inward pull equals the circle's demand, and cancelling $m$ hands you the speed: **Second**, the three results that then fall out for free, each a simple function of the radius $r$: > [!recall]- Feynman retelling — the whole walkthrough in plain words > Whirl a ball on a string: it only stays on the circle because the string keeps yanking it inward. Why must the yank be $mv^2/r$? Draw two tiny triangles — one made of the radii to two nearby points, one made of the velocity arrows at those two points. They're the same skinny shape (same little corner angle), so the velocity's inward "kink" per second works out to exactly $v^2/r$; times the mass gives the force the circle needs. In space there's no string — the *only* pull is gravity, $GMm/r^2$, and lucky for us it already points straight at the center, exactly where the circle needs a pull. So gravity plays the role of the string. Set the "yank you need" equal to the "yank gravity gives," cross out the satellite's own weight $m$ (it cancels because a heavier satellite needs a bigger yank *and* gets a bigger yank equally), and you're left with $v=\sqrt{GM/r}$: bigger orbit, slower speed. Ask how long one loop takes — circumference over speed — and you get $T=2\pi\sqrt{r^3/GM}$, which is exactly Kepler's law $T^2\propto r^3$. Check the edges: infinitely far → crawling and eternal; too fast (by a factor $\sqrt2$) → you break free. All of it is one idea: *an orbit is falling in a circle, at the one speed where gravity's pull is precisely the pull the circle demands.* > [!mnemonic] Carry it in your pocket > **"Bill = Payment."** The circle bills $mv^2/r$; gravity pays $GMm/r^2$. Equal them, cancel $m$, and read off $v=\sqrt{GM/r}$. --- ## Connections - [[Centripetal Force and Uniform Circular Motion]] — the source of the $mv^2/r$ demand (Steps 1–2). - [[Newton's Law of Universal Gravitation]] — the $GMm/r^2$ payment (Step 3). - [[Kepler's Three Laws]] — $T^2\propto r^3$ emerges in Step 6. - [[Escape Velocity]] — the $\sqrt2$ line in Step 7. - [[Elliptical Orbits and the Vis-viva Equation]] — what lives between circular and escape speeds. - [[Gravitational Potential Energy]] · [[Virial Theorem]] — carry the story into energies (parent note).