Circular orbit — velocity, period, energy: Foundations

This page builds every letter and squiggle the parent note uses, starting from nothing. If a symbol appears in the parent topic without explanation, it gets explained here — with a picture.

The cast of characters (in build order)

We introduce symbols one at a time. Each new one is allowed to lean only on the ones before it.

1. $r$ — the radius (distance to the center)

Picture it. In the figure below the big body sits at the middle and the satellite rides on a circle. The straight spoke from the center out to the satellite is $r$ .

Figure — Circular orbit — velocity, period, energy

The orbit map: the big central body at the center, the small orbiting body, the radius-spoke $r$ , and the speed arrow pointing along the circle. (We name the masses and the speed in the next two sections.)

Why the topic needs it. Gravity weakens with distance, and a bigger circle is a longer trip. Both effects are controlled by this single number $r$ , so it appears in every formula.

2. $m$ and $M$ — the two masses

Picture it. Return to the orbit map above: the big disk at the center is now named $M$ (Earth, Sun, ...) and the little dot going around is $m$ (satellite, planet, moon). We assume $M ≫ m$ (read: "much, much bigger"), so the big body barely moves.

Why the topic needs it. Gravity's strength depends on the product of the two masses. A famous punchline of this chapter is that $m$ eventually cancels out of the speed — a feather and a bus orbit alike — but you can only see it cancel if you first know what it is.

3. $v$ — the orbital speed

Picture it. In the orbit map the orange arrow tangent to the circle is $v$ . It always points along the direction of travel — sideways relative to the spoke $r$ , never toward or away from the center.

Why the topic needs it. $v$ is the headline answer: "how fast must I go to orbit here?" Everything else (period, energy) is built from $v$ .

4. Gravity's pull — $F_{g} = \frac{GM m}{r ^{2}}$

Now we assemble our first formula from the symbols above, plus one new constant.

Read it piece by piece.

$GM m$ on top: more mass of either body → stronger pull. $G$ just scales it.
$r^{2}$ on the bottom: double the distance and the pull drops to a quarter (because $2^{2} = 4$ ). This is the inverse-square law.

The inverse-square law: as the satellite moves outward the gravity arrow shrinks like $1/ r^{2}$ — at double the distance it is only a quarter as strong.

Why $r^{2}$ and not $r$ ? Picture gravity spreading out from the center like paint sprayed onto ever-bigger spheres. The paint that covered a small sphere must now coat a sphere with area growing as $r^{2}$ , so it thins out as $1/ r^{2}$ .

Why the topic needs it. This is the supply side of the orbit: the actual force available to hold the satellite on its circle. See Newton's Law of Universal Gravitation.

5. The centripetal requirement — $F_{c} = \frac{m v ^{2}}{r}$

This is not a new force — it is a demand. It says: "if you want radius $r$ at speed $v$ , you must provide this much inward force, from somewhere."

The centripetal demand: at every point the orange velocity arrow is turning, and turning requires the plum arrow of inward force $m v^{2} / r$ pointing to the center.

Read the formula.

Faster ( $v$ bigger) → the arrow swings harder → more force needed. And $v$ is squared, so doubling speed quadruples the demand.
Tighter circle ( $r$ smaller) → sharper turn → more force. So $r$ is on the bottom.

Why the topic needs it. This is the demand side. See Centripetal Force and Uniform Circular Motion.

6. The balance — solving for $v$ step by step

The whole topic is one sentence: supply = demand. Gravity is the only force available, so it must provide exactly the centripetal amount: $gravity supplies \frac{GM m}{r ^{2}} = circle demands \frac{m v ^{2}}{r}$

Now watch the algebra unfold — this is the derivation the parent note leans on, done slowly.

Step 1 — cancel the satellite mass $m$ . It appears once on each side, so divide both sides by $m$ : $\frac{GM}{r ^{2}} = \frac{v ^{2}}{r}$ This is why the answer will not care about $m$ : it is gone before we even finish.

Step 2 — clear the $r$ on the bottom. Multiply both sides by $r$ . On the left $\frac{r}{r ^{2}} = \frac{1}{r}$ ; on the right the $r$ cancels entirely: $\frac{GM}{r} = v^{2}$

Step 3 — undo the square. $v^{2}$ means " $v$ times itself"; to get $v$ back we take the square root of both sides (the square root is the question "what number, times itself, gives this?"). Squaring loses a sign, so algebraically there are two answers, $+ GM / r$ and $- GM / r$ . But $v$ here is a speed — the length of the velocity arrow — and a length is never negative. So we keep only the positive root and discard the negative one: $v = + \frac{GM}{r}$

7. $T$ — the period, and $2 π r$ — the circumference

Picture it. Unroll the circle from the orbit map into a straight line of length $2 π r$ . The satellite walks that whole line at speed $v$ , so the time is distance ÷ speed: $T = \frac{2 π r}{v} .$

Why the topic needs it. This one definition, plus $v = GM / r$ , gives Kepler's Third Law $T^{2} \propto r^{3}$ — see Kepler's Three Laws.

8. $K$ , $U$ , $E$ — kinetic, potential, and total energy

Why is $U$ negative? We choose $U = 0$ infinitely far away (no pull left). Falling inward from there releases energy, so any bound position sits below zero — hence the minus sign and the $1/ r$ : closer in means more negative.

Build $E$ explicitly. For an orbiting satellite we already found $v^{2} = \frac{GM}{r}$ , so the kinetic energy becomes $K = \frac{1}{2} m v^{2} = \frac{1}{2} m \cdot \frac{GM}{r} = \frac{GM m}{2 r} .$ Now add the potential energy $U = - \frac{GM m}{r}$ : $E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = \frac{GM m}{2 r} - \frac{2 GM m}{2 r} = - \frac{GM m}{2 r} .$

The energy ladder: the deep teal level is $U = - GM m / r$ , the orange arrow is the kinetic energy $K = GM m /2 r$ lifting you off the bottom, and their sum $E = - GM m /2 r$ (plum) sits halfway up — still below the $E = 0$ escape line.

Equipment checklist

Test yourself — cover the right side. If any answer is fuzzy, reread that section above.

What does

r

measure, and from where?

The distance from the center of the big body out to the satellite — not the altitude above its surface.

Why must

r

exceed the central body's physical radius?

Otherwise the "orbit" is underground; a real orbit only exists outside the central body, where these formulas hold.

What are

m

and

M

m

= small orbiting mass,

M

= large central mass (in kg), with

M ≫ m

Difference between speed and velocity in a circle?

Speed (length of the arrow) is constant; velocity (arrow + direction) changes because the direction keeps turning.

What is

G

, its units, and do you ever solve for it?

The gravitational constant

6.674 \times 1 0^{- 11} m^{3} kg^{- 1} s^{- 2}

; a fixed number of nature you look up, never solve for.

Why is

r

squared in

F_{g} = GM m / r^{2}

Gravity spreads over spheres of area growing as

r^{2}

, so it thins as the inverse square of distance.

Is centripetal force a new kind of force?

No — it's a requirement (

m v^{2} / r

) that some real force (here, gravity) must supply.

When solving

v^{2} = GM / r

, why keep only the positive root?

v

is a speed (a length), which can never be negative, so the negative root is discarded.

Which mass cancels when you solve for

v

, and at which step?

The satellite mass

m

, in Step 1, by dividing both sides of

GM m / r^{2} = m v^{2} / r

m

What is the circumference of a circle of radius

r

2 π r

, with

π \approx 3.14159

How do you get period from speed?

T = \frac{2 π r}{v}

= (distance per lap) ÷ (speed).

How does

E = - GM m / (2 r)

come from

K

and

U

Add

K = GM m /2 r

and

U = - GM m / r

\frac{1}{2 r} - \frac{1}{r}

(times

GM m

)

= - \frac{1}{2 r}

Why is

U

negative?

We set

U = 0

at infinity; falling inward releases energy, so bound positions sit below zero.

What does the sign of total energy

E

tell you?

E < 0

bound (trapped),

E = 0

just escapes,

E > 0

flies away free.

Connections

Newton's Law of Universal Gravitation — where $F_{g} = GM m / r^{2}$ comes from.
Centripetal Force and Uniform Circular Motion — where $m v^{2} / r$ comes from.
Kepler's Three Laws — $T^{2} \propto r^{3}$ built from these symbols.
Gravitational Potential Energy — the origin of $U = - GM m / r$ .
Escape Velocity — the $E = 0$ boundary.
Elliptical Orbits and the Vis-viva Equation — generalizes speed and energy.
Virial Theorem — explains $U = 2 E = - 2 K$ .