Orbital velocity for circular orbit — derivation: Worked examples

This is the practice arena for the parent derivation. We already earned the one formula that runs the whole show:

Below we hit every kind of question this single formula can produce.

The scenario matrix

Think of a formula's "scenarios" as the different shapes of question a teacher or reality can pose. Here every input can be given, hidden, doubled, sent to zero, or sent to infinity — and each such move is a cell:

Cell	What is varied / tested	Example
A. Surface orbit	$h = 0$ , so $r = R$ — the $g R$ shortcut	Ex 1
B. High orbit	$h \neq = 0$ , must use $r = R + h$	Ex 2
C. Ratio/scaling	speeds compared, no numbers plugged	Ex 3
D. Solve backwards	given $v_{o}$ , find $r$ (or $h$ )	Ex 4
E. Different planet	$M, R$ not Earth's — general $GM$	Ex 5
F. Limiting cases	$r \to \infty$ , $r \to R$ , " $r$ inside planet"	Ex 6
G. Real-world word problem	altitude quoted, find speed + period + ground track	Ex 7
H. Exam twist	mixes escape velocity / mass cancellation trap	Ex 8
I. Kepler link	combine with $v = 2 π r / T$	Ex 9

The nine examples below cover all nine cells. Figures carry the geometry where geometry matters.

Ex 1 — Cell A: surface orbit (the shortcut)

Ex 2 — Cell B: high orbit, $r = R + h$

Worked example Statement

A satellite orbits at altitude $h = 3.6 \times 1 0^{7} m$ (roughly geostationary height). Find $v_{o}$ . Use $GM = g R^{2}$ with $g = 9.8$ , $R = 6.4 \times 1 0^{6}$ .

Forecast: Higher than Ex 1 or lower? By roughly what factor?

Step 1 — Build $r$ from the centre. Why this step? The formula demands centre-distance. $r = R + h = 6.4 \times 1 0^{6} + 3.6 \times 1 0^{7} = 4.24 \times 1 0^{7} m$ .

Step 2 — Compute $GM$ . Why this step? We aren't given $G$ and $M$ separately, but $GM = g R^{2}$ . $GM = 9.8 \times (6.4 \times 1 0^{6})^{2} = 9.8 \times 4.096 \times 1 0^{13} = 4.014 \times 1 0^{14} m^{3} / s^{2}$

Step 3 — Apply the master formula. Why this step? We now have both ingredients the formula needs — $GM$ from Step 2 and $r$ from Step 1 — so this is where they combine to deliver the actual speed. $v_{o} = \frac{GM}{r} = \frac{4.014 \times 1 0 ^{14}}{4.24 \times 1 0 ^{7}} = 9.47 \times 1 0^{6} \approx 3.08 \times 1 0^{3} m/s$

Verify: ~3.1 km/s, well below LEO's 7.9 km/s, as expected since $r$ is ~6.6× larger and $v_{o} \propto 1/ r$ : $7.9/ 6.6 \approx 3.1$ ✓. Forecast: lower.

Ex 3 — Cell C: pure scaling, no numbers

The figure below plots that $1/ r$ curve: find P (yellow dot) at radius $r$ with speed 1, then trace along the blue curve to Q (red dot) at $9 r$ — its height on the curve has dropped to exactly $1/3$ . The dashed guide lines make the "9× out, 3× down" trade visible at a glance.

Figure — Orbital velocity for circular orbit — derivation

Ex 4 — Cell D: solve backwards (find $r$ from $v_{o}$ )

Worked example Statement

A probe is measured moving at $v_{o} = 5.0 \times 1 0^{3} m/s$ in circular orbit around Earth. At what altitude $h$ is it? ( $GM = 3.986 \times 1 0^{14}$ , $R = 6.4 \times 1 0^{6}$ .)

Forecast: Higher or lower than LEO (which was 7.9 km/s)? Since it's slower, guess high or low altitude?

Step 1 — Rearrange the master formula for $r$ . Why this step? Now the unknown is inside the root; square both sides to free it. $v_{o}^{2} = \frac{GM}{r} \Rightarrow r = \frac{GM}{v _{o}^{2}}$

Step 2 — Plug in. Why this step? With $r$ now isolated on the left, substituting the measured speed and Earth's $GM$ turns the rearranged formula into an actual centre-distance number. $r = \frac{3.986 \times 1 0 ^{14}}{( 5.0 \times 1 0 ^{3} ) ^{2}} = \frac{3.986 \times 1 0 ^{14}}{2.5 \times 1 0 ^{7}} = 1.594 \times 1 0^{7} m$

Step 3 — Convert to altitude. Why this step? $r$ is from the centre; altitude subtracts $R$ . $h = r - R = 1.594 \times 1 0^{7} - 6.4 \times 1 0^{6} = 9.54 \times 1 0^{6} m \approx 9, 540 km$

Verify: Slower than LEO ⇒ higher orbit ⇒ positive altitude ✓. Back-check: $GM / r = 3.986 \times 1 0^{14} /1.594 \times 1 0^{7} = 2.5 \times 1 0^{7} = 5000$ m/s ✓.

Ex 5 — Cell E: a different planet (Mars)

Worked example Statement

Mars has $M = 6.42 \times 1 0^{23} kg$ , radius $R = 3.39 \times 1 0^{6} m$ . Find the orbital speed of a low Mars orbit ( $h \approx 0$ ). $G = 6.67 \times 1 0^{- 11}$ .

Forecast: Mars is smaller and lighter than Earth — faster or slower than Earth's 7.9 km/s?

Step 1 — Use $GM$ directly. Why this step? We're given $M$ and $G$ separately here (no $g$ handed to us), so compute $GM$ . $GM = 6.67 \times 1 0^{- 11} \times 6.42 \times 1 0^{23} = 4.282 \times 1 0^{13} m^{3} / s^{2}$

Step 2 — Low orbit ⇒ $r = R$ , then apply the formula. Why this step? "Low orbit" means altitude $h \approx 0$ , so $r = R + 0 = R$ — exactly the surface-orbit condition of Ex 1, but for Mars. Feeding Mars's own $R$ into the master formula gives its low-orbit speed. $v_{o} = \frac{GM}{R} = \frac{4.282 \times 1 0 ^{13}}{3.39 \times 1 0 ^{6}} = 1.263 \times 1 0^{7} \approx 3.55 \times 1 0^{3} m/s$

Verify: ~3.6 km/s, less than half of Earth's — consistent with Mars's weaker gravity ✓. Units as in Ex 1. Forecast: slower.

Ex 6 — Cell F: limiting & degenerate cases

Worked example Statement

Discuss what $v_{o} = GM / r$ predicts as (a) $r \to \infty$ , (b) $r \to R$ from above, (c) if someone naively sets $r < R$ (inside the planet).

Forecast: Does speed go to zero, infinity, or a finite value as the orbit gets huge?

Step 1 — Case $r \to \infty$ . Why this step? The denominator inside the root grows without bound. $lim_{r \to \infty} \frac{GM}{r} = 0$ An infinitely distant orbit needs essentially zero speed — gravity there is vanishingly weak, so the required centripetal force is tiny.

Step 2 — Case $r \to R^{+}$ . Why this step? This is the fastest possible real orbit — hugging the surface. Recall the surface relation $GM = g R^{2}$ (Earth's surface gravity times its radius squared); substituting it lets us swap the two equivalent forms: $v_{o} \to \frac{GM}{R} = \frac{g R ^{2}}{R} = g R \approx 7.9 km/s (Earth)$ This is the maximum circular orbital speed for Earth; you can't circle faster than skimming the ground (and even that ignores air).

Step 3 — Case $r < R$ (inside the planet). Why this step? Test the domain limit. The bare formula still spits out a number, but it's physically meaningless: below the surface, only the mass inside radius $r$ pulls you (shell theorem), so $M$ is no longer the whole planet. The formula as written assumes all of $M$ sits inside your orbit. So $r \geq R$ is the valid domain.

Verify: Monotonic decrease confirmed: $\frac{d v _{o}}{d r} = - \frac{1}{2} GM r^{- 3/2} < 0$ for all $r > 0$ , so speed strictly falls as $r$ grows, hitting its ceiling at $r = R$ and $\to 0$ at infinity ✓.

The figure below draws exactly these three regions on the $v_{o} (r)$ curve: the green dot marks the surface ceiling at $r = R$ (fastest orbit), the yellow arrow follows the curve sliding toward zero as $r \to \infty$ , and the red shaded band $r < R$ flags the forbidden interior where the whole-mass assumption breaks.

Ex 7 — Cell G: real-world word problem (speed, period, ground track)

Worked example Statement

The International Space Station orbits at altitude $h = 4.0 \times 1 0^{5} m$ (400 km). Find its speed, how long one orbit takes, and how far its ground track advances per orbit along the equator ( $R = 6.4 \times 1 0^{6}$ m). Use $GM = 3.986 \times 1 0^{14}$ .

Forecast: People say "the ISS circles Earth every 90 minutes." True? Guess yes/no.

Step 1 — Centre-distance. Why this step? The classic trap: 400 km is altitude $h$ , not $r$ . $r = R + h = 6.4 \times 1 0^{6} + 4.0 \times 1 0^{5} = 6.8 \times 1 0^{6} m$ .

Step 2 — Speed. Why this step? Speed is the direct output of the master formula, and everything downstream (period, ground track) is built from it, so we compute it first. $v_{o} = \frac{3.986 \times 1 0 ^{14}}{6.8 \times 1 0 ^{6}} = 5.862 \times 1 0^{7} \approx 7.66 \times 1 0^{3} m/s$

Step 3 — Period from geometry. Why this step? One lap covers circumference $2 π r$ at constant speed, so $T = \frac{2 π r}{v _{o}}$ . $T = \frac{2 π \times 6.8 \times 1 0 ^{6}}{7.66 \times 1 0 ^{3}} \approx \frac{4.273 \times 1 0 ^{7}}{7.66 \times 1 0 ^{3}} \approx 5.58 \times 1 0^{3} s$

Step 4 — Ground track shift. Why this step? While the ISS completes one loop, Earth itself spins eastward underneath. The equator's surface moves at $v_{eq} = \frac{2 π R}{T _{day}}$ with $T_{day} = 86400$ s, so in time $T$ the ground beneath slides by $Δ x = v_{eq} T$ — that's why the ISS passes over different longitudes each orbit. $v_{eq} = \frac{2 π \times 6.4 \times 1 0 ^{6}}{86400} \approx 465 m/s, Δ x = 465 \times 5580 \approx 2.60 \times 1 0^{6} m \approx 2600 km$

Verify: $5580 s \div 60 \approx 93$ minutes ✓ — matches the famous "~90 minute" ISS orbit. The ground track shifts ~2600 km westward relative to the ground each pass, which is why the ISS traces those slanted sine-like paths on tracking maps ✓. Forecast: yes.

Ex 8 — Cell H: exam twist (escape-velocity + mass trap)

Worked example Statement

Two satellites orbit Earth at the same radius $r$ : satellite X has mass $1000 kg$ , satellite Y has mass $2000 kg$ . (a) Which is faster? (b) If X's orbital speed is $v_{o}$ , what speed would it need to escape from that radius?

Forecast: Does the heavier one orbit slower, faster, or the same?

Step 1 — Mass cancellation. Why this step? In the derivation $\frac{GM m}{r ^{2}} = \frac{m v _{o}^{2}}{r}$ , the satellite mass $m$ cancels. So $v_{o}$ depends only on $M$ (Earth) and $r$ — not on X or Y's mass. $v_{X} = v_{Y} = \frac{GM}{r}$ They travel at exactly the same speed. (a) answered: neither is faster.

Step 2 — Escape from the same radius. Why this step? Escape velocity is a separate result: $v_{esc} = 2 GM / r = 2 v_{o}$ . $v_{esc} = 2 v_{o} \approx 1.414 v_{o}$

Verify: If, say, $v_{o} = 7.66 km/s$ (Ex 7 radius), escape needs $1.414 \times 7.66 \approx 10.8 km/s$ ✓. The trap "heavier ⇒ faster" is false; the trap "orbital = escape" is off by exactly $2$ . See Escape Velocity.

Ex 9 — Cell I: Kepler link (period vs radius)

Worked example Statement

Show that combining $v_{o} = GM / r$ with $v_{o} = 2 π r / T$ gives Kepler's Third Law $T^{2} \propto r^{3}$ , then use it: if orbit A has period $T_{A}$ at radius $r$ , what is the period at radius $8 r$ ?

Forecast: Longer or shorter period at bigger radius? By what factor for $8 r$ ?

Step 1 — Equate the two speed expressions. Why this step? Both equal $v_{o}$ , so they equal each other, letting us eliminate $v$ and expose the $T$ – $r$ relationship. $\frac{2 π r}{T} = \frac{GM}{r}$

Step 2 — Solve for $T$ . Square both sides: $\frac{4 π ^{2} r ^{2}}{T ^{2}} = \frac{GM}{r} \Rightarrow T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ Why this step? This is exactly Kepler's Third Law: $T^{2} \propto r^{3}$ . See Kepler's Third Law.

Step 3 — Apply the ratio for $8 r$ . Why this step? Since $T \propto r^{3/2}$ , multiplying $r$ by 8 multiplies $T$ by $8^{3/2}$ . $\frac{T _{B}}{T _{A}} = 8^{3/2} = (8)^{3} = (2.828)^{3} \approx 22.6$

Verify: $8^{3/2} = 8^{3} = 512 \approx 22.6$ ✓. Bigger orbit ⇒ much longer period — the far Moon takes ~27 days while LEO takes ~90 min. Forecast: longer, factor ~22.6.

Active recall

Recall Test the scenarios

$h = 0$ shortcut speed for Earth? ⇒ $g R \approx 7.9$ km/s.
Given $v_{o}$ , how do you get $r$ ? ⇒ $r = GM / v_{o}^{2}$ .
Fastest possible circular orbit occurs at which $r$ ? ⇒ $r = R$ (surface-skimming).
As $r \to \infty$ , $v_{o} \to$ ? ⇒ $0$ .
Does doubling satellite mass $m$ change $v_{o}$ ? ⇒ No — $m$ cancels.
$T$ vs $r$ relation from the formula? ⇒ $T^{2} \propto r^{3}$ .

Connections

Orbital velocity for circular orbit — derivation — the parent this drills.
Newton's Law of Universal Gravitation
Centripetal Force and Uniform Circular Motion
Escape Velocity — the $2$ cousin (Ex 8).
Kepler's Third Law — derived in Ex 9.
Acceleration due to gravity g and GM = gR² — the $GM = g R^{2}$ swap used throughout.
Geostationary Orbit — the $h \approx 3.6 \times 1 0^{7}$ m case of Ex 2.