Circular orbit — velocity, period, energy: Worked examples

The tools we reuse (all built in the parent note): $v = \frac{GM}{r}, T = 2 π \frac{r ^{3}}{GM}, K = \frac{GM m}{2 r}, U = - \frac{GM m}{r}, E = - \frac{GM m}{2 r}$

Here $G = 6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ is the gravitational constant, $M$ is the central body's mass (the big one you orbit), $m$ is the orbiting body's mass (the small one), $r$ is the distance from the center of the big body (NOT the altitude above its surface — that is a trap we hit deliberately). Here $K$ is kinetic energy (energy of motion), $U$ is gravitational potential energy (energy of position, negative because the orbit is bound), and $E = K + U$ is the total. We compute all three explicitly in Example 5.

The scenario matrix

Every problem about circular orbits falls into one of these cells. The examples below are labelled by cell.

Cell	What varies / what's asked	Example
A. Forward, Earth	given $r$ → find $v, T$	Ex 1
B. Backward	given $T$ → find $r$ (invert Kepler)	Ex 2
C. Altitude trap	given altitude, must add planet radius	Ex 3
D. Different central mass	orbit the Sun / another star	Ex 4
E. Energy budget	given a maneuver → find $K, U, E, Δ E$	Ex 5
F. Limiting case $r \to \infty$	what do $v, T, E$ do at the edge?	Ex 6
G. Degenerate: surface-grazing	smallest possible orbit, $r = R_{planet}$	Ex 7
H. Mass-independence check	two very different $m$ , same $r$	Ex 8
I. Exam twist	ratio problem, no numbers plugged	Ex 9

Constants we reuse: $M_{\oplus} = 5.97 \times 1 0^{24} kg, R_{\oplus} = 6.37 \times 1 0^{6} m, M_{⊙} = 1.989 \times 1 0^{30} kg$

Example 1 — Cell A: forward, Earth

Worked example A satellite orbits Earth at radius

$r = 7.00 \times 1 0^{6}$ m. Find $v$ and $T$ . Forecast: Guess the speed before reading on. Is it closer to 1 km/s, 8 km/s, or 30 km/s? And is the period minutes or hours?

Step 1 — Speed. Plug into $v = GM / r$ : $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{7.00 \times 1 0 ^{6}} = 5.691 \times 1 0^{7} \approx 7.54 \times 1 0^{3} m/s$ Why this step? This is the ONLY speed at which gravity exactly equals the centripetal need $m v^{2} / r$ at this radius — the balance that defines the orbit.

Step 2 — Period. Use $T = 2 π r / v$ : $T = \frac{2 π ( 7.00 \times 1 0 ^{6} )}{7540} \approx 5.83 \times 1 0^{3} s \approx 97 min$ Why this step? One lap = circumference over speed. Pure definition, no new physics.

Verify: $v \approx 7.5$ km/s (near the famous ISS ~7.7 km/s — good, since this radius is close to the ISS's). Units: $(m^{3} s^{- 2}) / m = m^{2} s^{- 2} = m/s$ ✓. Period ~1.6 h is sensible for low orbit.

Example 2 — Cell B: backward, invert Kepler

Worked example A satellite around Earth has period

$T = 2.00$ hours $= 7200$ s. What is its orbital radius $r$ ? Forecast: Longer period than ISS (~1.5 h) — so should $r$ be bigger or smaller than $7 \times 1 0^{6}$ m?

Step 1 — Recall where the two-form Kepler law comes from. The parent note wrote $T = 2 π r^{3} / GM$ . Square both sides to clear the root: $T^{2} = (2 π)^{2} \cdot \frac{r ^{3}}{GM} = \frac{4 π ^{2}}{GM} r^{3}$ Why this step? Squaring turns the awkward square-root form into a clean $T^{2} \propto r^{3}$ form that is easy to invert for $r$ . The two forms are the same law.

Step 2 — Rearrange to isolate $r$ : $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3}$ Why this step? We were handed $T$ and asked for $r$ ; Kepler's third law is the one equation linking them, so we solve it for $r$ .

Step 3 — Plug in: $r = (\frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 7200 ) ^{2}}{4 π ^{2}})^{1/3} = (5.230 \times 1 0^{20})^{1/3} \approx 8.06 \times 1 0^{6} m$ Why this step? The cube root undoes the $r^{3}$ ; we take it last so the arithmetic inside stays clean.

Verify: $8.06 \times 1 0^{6} > 7 \times 1 0^{6}$ ✓ — a longer period means a bigger, slower orbit, matching " $T$ grows with $r$ ." Sanity: cube it back, $(8.06 \times 1 0^{6})^{3} = 5.24 \times 1 0^{20}$ ✓.

Example 3 — Cell C: the altitude trap

Worked example A weather satellite orbits at

altitude 800 km above Earth's surface. Find $v$ . Forecast: The naive mistake is to use $r = 800$ km $= 8 \times 1 0^{5}$ m. Guess how wrong that answer would be.

Step 1 — Convert altitude to orbital radius. The formulas use distance from Earth's center: $r = R_{\oplus} + h = 6.37 \times 1 0^{6} + 8.00 \times 1 0^{5} = 7.17 \times 1 0^{6} m$ Why this step? Gravity's $1/ r^{2}$ is measured from the center of mass, not from the ground. Skipping $R_{\oplus}$ is the single most common orbit error.

Step 2 — Speed with the correct $r$ : $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{7.17 \times 1 0 ^{6}} = 5.557 \times 1 0^{7} \approx 7.45 \times 1 0^{3} m/s$ Why this step? Same master formula as Ex 1; the only subtlety was getting $r$ right.

Verify: If you'd wrongly used $r = 8 \times 1 0^{5}$ m you'd get $v = 4.98 \times 1 0^{8} \approx 2.23 \times 1 0^{4}$ m/s — three times too big and physically impossible for LEO. The correct 7.45 km/s sits right in the low-orbit band ✓.

Example 4 — Cell D: different central mass (orbit the Sun)

Worked example Earth orbits the Sun at

$r = 1.496 \times 1 0^{11}$ m. Predict Earth's orbital speed and check the period is one year. Forecast: Earth's orbital speed around the Sun — 8 km/s like a satellite, or much faster?

Step 1 — Use $M = M_{⊙}$ , not $M_{\oplus}$ . The central body is now the Sun: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1.989 \times 1 0 ^{30} )}{1.496 \times 1 0 ^{11}} = 8.872 \times 1 0^{8} \approx 2.98 \times 1 0^{4} m/s$ Why this step? $M$ is always the mass you orbit. Swap the Sun in and the same formula gives heliocentric speeds.

Step 2 — Period: $T = \frac{2 π ( 1.496 \times 1 0 ^{11} )}{2.98 \times 1 0 ^{4}} \approx 3.15 \times 1 0^{7} s$ Why this step? Same distance-over-speed. Convert: $3.15 \times 1 0^{7} s \div (86400 s/day) \approx 365$ days.

Verify: $v \approx 29.8$ km/s is the textbook Earth orbital speed ✓, and $T \approx 365$ days = one year ✓. The formula that gave ISS's 92 minutes also gives Earth's year — same physics, different $M$ and $r$ .

Example 5 — Cell E: energy budget for a maneuver (K, U, E all shown)

Worked example A

$m = 1200$ kg satellite orbits Earth at $r_{1} = 6.80 \times 1 0^{6}$ m. (a) Find its $K$ , $U$ , and $E$ . (b) Raise it to $r_{2} = 1.02 \times 1 0^{7}$ m — how much energy $Δ E$ must you supply? Forecast: Higher orbit = more total energy, so the raise costs energy. But does the satellite end up faster or slower?

Step 1 — Kinetic energy at $r_{1}$ . Use $K = \frac{GM m}{2 r _{1}}$ : $K_{1} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 6.80 \times 1 0 ^{6} )} \approx 3.514 \times 1 0^{10} J$ Why this step? $K$ is the energy of motion; plugging $v^{2} = GM / r$ into $\frac{1}{2} m v^{2}$ gives this compact form — no need to compute $v$ first.

Step 2 — Potential energy at $r_{1}$ . Use $U = - \frac{GM m}{r _{1}}$ : $U_{1} = - \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{6.80 \times 1 0 ^{6}} \approx - 7.029 \times 1 0^{10} J$ Why this step? $U$ is the energy of position, negative because the orbit is bound (you'd need to add energy to reach infinity).

Step 3 — Total energy at $r_{1}$ . Add them: $E_{1} = K_{1} + U_{1} \approx 3.514 \times 1 0^{10} - 7.029 \times 1 0^{10} = - 3.514 \times 1 0^{10} J$ Why this step? Note $E_{1} = - K_{1}$ and $U_{1} = 2 E_{1}$ — the virial relations from the parent note, a built-in check.

Step 4 — Energy to raise the orbit = difference of totals. $Δ E = E_{2} - E_{1} = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ $= 2.390 \times 1 0^{17} \times (1.4706 \times 1 0^{- 7} - 9.8039 \times 1 0^{- 8}) = 2.390 \times 1 0^{17} \times 4.902 \times 1 0^{- 8} \approx 1.17 \times 1 0^{10} J$ Why this step? You supply exactly the gap between the two total energies; the $K$ and $U$ pieces rearrange automatically inside the orbit.

Verify: $E_{1} = - K_{1}$ ✓ and $U_{1} = 2 E_{1}$ ✓ (virial theorem holds). $Δ E > 0$ ✓ (raising costs energy). Twist: $v_{1} = GM / r_{1} = 7.65$ km/s, $v_{2} = GM / r_{2} = 6.25$ km/s — the satellite ends up slower despite gaining total energy ✓ ( $K$ drops from $3.51 \times 1 0^{10}$ to $2.34 \times 1 0^{10}$ J; the surplus went into $U$ ).

Example 6 — Cell F: the limit $r \to \infty$

Worked example What happens to

$v$ , $T$ , and $E$ as the orbital radius grows without bound? Forecast: Which of the three heads to zero, which blows up, and which one creeps up toward a ceiling?

Step 1 — Speed. $v = GM / r$ . As $r \to \infty$ , the fraction $\to 0$ , so $lim_{r \to \infty} v = 0.$ Why this step? Reading a limit off a formula: whatever sits under $r$ in a denominator vanishes as $r$ grows.

Step 2 — Period. $T = 2 π r^{3} / GM \to \infty$ (numerator grows without bound). Why this step? $r^{3}$ dominates; the lap takes forever because the orbit is enormous and the speed crawls.

Step 3 — Energy. $E = - GM m /2 r \to 0^{-}$ (approaches zero from below). Why this step? $E$ is negative and its magnitude $\propto 1/ r$ shrinks. In the figure below, the violet curve $E$ climbs toward the dashed orange line $E = 0$ but never crosses it, while the magenta curve $v$ sags toward zero.

Figure: horizontal axis is orbital radius $r$ (in millions of metres). The magenta curve is orbital speed $v$ in km/s, dropping toward zero as $r$ grows. The violet curve is total energy per kilogram $E$ in MJ, rising toward — but never reaching — the dashed orange line at $E = 0$ , the escape boundary.

Verify: The limit $E \to 0^{-}$ is exactly the escape boundary. $E = 0$ means the body is barely unbound (at rest at infinity) — the doorway to Escape Velocity. A very distant orbit is almost free ✓.

Example 7 — Cell G: degenerate surface-grazing orbit

Worked example What is the speed and period of a hypothetical satellite skimming Earth's surface,

$r = R_{\oplus} = 6.37 \times 1 0^{6}$ m (ignore air)? This is the smallest possible circular orbit. Forecast: Faster or slower than the ISS? This is the fastest a circular orbit can be — why?

Step 1 — Speed at the minimum radius: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.37 \times 1 0 ^{6}} = 6.254 \times 1 0^{7} \approx 7.91 \times 1 0^{3} m/s$ Why this step? Smallest $r$ gives the largest $v$ (since $v \propto r^{- 1/2}$ ). No orbit can be faster than a surface-grazer.

Step 2 — Period: $T = \frac{2 π ( 6.37 \times 1 0 ^{6} )}{7910} \approx 5.06 \times 1 0^{3} s \approx 84 min$ Why this step? Smallest orbit → shortest period. This ~84 min is the famous "minimum orbital period" for Earth.

Verify: $v \approx 7.9$ km/s ≈ "first cosmic velocity" ✓. It is larger than Ex 1's 7.54 km/s (which was at bigger $r$ ) ✓, confirming smaller $r$ ⇒ faster. Any real orbit ( $r > R_{\oplus}$ ) is slower and longer-period than this bound.

Example 8 — Cell H: mass-independence

Worked example At the same radius

$r = 8.0 \times 1 0^{6}$ m, compare a $1$ kg cubesat and a $4.2 \times 1 0^{5}$ kg space station. Do they orbit at the same speed? Forecast: The station is 420,000× heavier. Surely it moves differently?

Step 1 — Write the balance and cancel $m$ : $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow \frac{GM}{r} = v^{2}$ Why this step? Gravity scales with $m$ , and the centripetal need also scales with $m$ ; the two $m$ 's cancel exactly.

Step 2 — Compute the one speed both share: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{8.0 \times 1 0 ^{6}} = 4.981 \times 1 0^{7} \approx 7.06 \times 1 0^{3} m/s$ Why this step? Since $m$ dropped out, there is a single orbital speed at this $r$ — the same for feather and freighter.

Verify: Both give 7.06 km/s ✓. Their energies do differ: $E = - GM m /2 r$ scales with $m$ , so the station's $∣ E ∣$ is $4.2 \times 1 0^{5}$ times the cubesat's — but the trajectory (speed, period) is identical.

Example 9 — Cell I: the exam twist (ratio, no plugging)

Recall Which cell is which — self-test

Given only the radius of a satellite around Earth, which formulas do you reach for? ::: $v = GM / r$ , then $T = 2 π r / v$ (Cell A). Given a period and asked for radius, what do you invert? ::: $r = (GM T^{2} /4 π^{2})^{1/3}$ (Cell B). A problem gives "altitude 500 km" — what must you do first? ::: Add the planet radius: $r = R + h$ (Cell C). How do you get $K$ , $U$ , $E$ once you know $r$ ? ::: $K = GM m /2 r$ , $U = - GM m / r$ , $E = - GM m /2 r$ (Cell E). As $r \to \infty$ , what do $v$ , $T$ , $E$ approach? ::: $v \to 0$ , $T \to \infty$ , $E \to 0^{-}$ (Cell F). The smallest possible circular orbit has $r = ?$ and is the fastest or slowest? ::: $r = R_{planet}$ ; it's the fastest circular orbit (Cell G). Does doubling the satellite's mass change $v$ or $T$ ? ::: No — $m$ cancels; only $E, K, U$ scale with $m$ (Cell H). If $r$ quadruples, $T$ multiplies by? ::: $4^{3/2} = 8$ (Cell I).

Connections

Parent: Circular orbit — velocity, period, energy — the four master formulas used here.
Newton's Law of Universal Gravitation — the $GM m / r^{2}$ we start every balance from.
Centripetal Force and Uniform Circular Motion — the $m v^{2} / r$ side of the balance.
Kepler's Three Laws — inverting $T^{2} \propto r^{3}$ powers Cells B and I.
Gravitational Potential Energy — the $U = - GM m / r$ behind Cell E.
Escape Velocity — the $E = 0$ ceiling that Cell F creeps toward.
Elliptical Orbits and the Vis-viva Equation — where non-circular maneuvers live.
Virial Theorem — the $E = - K$ , $U = 2 E$ bookkeeping behind the energies.