Circular Orbits — Velocity, Period, and Altitude
Circular Orbits
A circular orbit is the simplest type of orbit, where a satellite moves at constant speed in a perfect circle around a central body. For this to happen, the gravitational force must provide exactly the centripetal force needed for circular motion. Setting G M m over r squared equal to m v squared over r gives us the orbital velocity: v equals the square root of G M over r. This beautifully simple equation tells us that orbital speed depends only on the central body's mass and the orbital radius — not on the satellite's mass.
The implications are profound. At low Earth orbit, about 400 kilometers altitude, the orbital speed is roughly 7.7 kilometers per second — about 27,500 kilometers per hour. This means the ISS completes an orbit in just 92 minutes, experiencing 16 sunrises and sunsets every day. Higher orbits are slower: at geostationary altitude of 35,786 kilometers, the orbital period is exactly 24 hours, so the satellite appears to hover over one spot on the equator.
Kepler's third law, which Newton derived from his gravitational law, states that the square of the orbital period is proportional to the cube of the orbital radius. Specifically, T squared equals 4 pi squared over G M times r cubed. This powerful relationship allows us to determine the mass of any celestial body by observing the orbit of a satellite around it.
The elegance of orbital mechanics lies in how a few simple equations — rooted in nothing more than Newton's second law and his law of gravitation — can predict the motion of every satellite, planet, and moon in the universe.
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{\mu}{r}}$$
where $r$ is the orbital radius (center of body to satellite).
$$T = 2\pi \sqrt{\frac{r^3}{GM}} = 2\pi \sqrt{\frac{r^3}{\mu}}$$
$T^2 \propto r^3$: farther orbits take disproportionately longer.
An orbit with $T = 24$ hours (86,400 s), placed on the equatorial plane. The satellite appears stationary from Earth. Altitude: $\approx 35{,}786$ km.
Find the orbital velocity and period of the ISS at altitude 408 km.
Step 1: Find $r$.
$$r = 6371 + 408 = 6779 \text{ km} = 6.779 \times 10^6 \text{ m}$$
Step 2: Compute velocity.
$$v = \sqrt{\frac{3.986 \times 10^{14}}{6.779 \times 10^6}} = \sqrt{5.879 \times 10^7} = 7{,}668 \text{ m/s} \approx 7.67 \text{ km/s}$$
Step 3: Compute period.
$$T = \frac{2\pi r}{v} = \frac{2\pi (6.779 \times 10^6)}{7668} = 5{,}553 \text{ s} \approx 92.6 \text{ min}$$
Derive the altitude of geostationary orbit.
Step 1: Set $T = 86{,}400$ s and solve for $r$.
$$T = 2\pi\sqrt{\frac{r^3}{\mu}} \implies r^3 = \frac{\mu T^2}{4\pi^2}$$
Step 2: Substitute.
$$r^3 = \frac{(3.986 \times 10^{14})(86400)^2}{4\pi^2} = \frac{2.975 \times 10^{24}}{39.48} = 7.534 \times 10^{22}$$
Step 3: Take cube root.
$$r = (7.534 \times 10^{22})^{1/3} = 4.216 \times 10^7 \text{ m} = 42{,}164 \text{ km}$$
Step 4: Altitude $= 42{,}164 - 6{,}371 = 35{,}793$ km $\approx 35{,}786$ km.
The Moon orbits Earth with $T = 27.3$ days and $r = 384{,}400$ km. Use this to verify Earth's mass.
Step 1: From $T^2 = 4\pi^2 r^3/(GM)$, solve for $M$.
$$M = \frac{4\pi^2 r^3}{G T^2}$$
Step 2: Convert: $T = 27.3 \times 86400 = 2.359 \times 10^6$ s, $r = 3.844 \times 10^8$ m.
$$M = \frac{4\pi^2 (3.844 \times 10^8)^3}{(6.674 \times 10^{-11})(2.359 \times 10^6)^2} = \frac{2.243 \times 10^{27}}{3.714 \times 10^2} \approx 6.04 \times 10^{24} \text{ kg}$$
Close to the accepted $5.972 \times 10^{24}$ kg — the small discrepancy comes from rounding.
Practice Problems
Show Answer Key
1. $r = 6571$ km. $v = \sqrt{3.986 \times 10^{14}/6.571 \times 10^6} = 7788$ m/s. $T = 2\pi \times 6.571 \times 10^6/7788 = 5301$ s $= 88.4$ min.
2. $r = 26571$ km. $v = \sqrt{\mu/r} = 3874$ m/s. $T = 2\pi r/v = 43{,}078$ s $= 717.9$ min $= 11.97$ hr.
3. $T = 7200$ s. $r^3 = \mu T^2/(4\pi^2) = 1.312 \times 10^{22}$. $r = 2.360 \times 10^7$ m. Alt $= 23600 - 6371 = 17{,}229$ km.
4. $T \propto r^{3/2}$. $(2r)^{3/2} = 2^{3/2} r^{3/2} = 2\sqrt{2} \cdot r^{3/2}$. ✓
5. $T = 5400$ s. $r = (\mu T^2/4\pi^2)^{1/3} = 6{,}654$ km. Alt $= 6654 - 6371 = 283$ km.
6. $r = 42164$ km. $v = \sqrt{\mu/r} = 3075$ m/s $= 3.07$ km/s.
7. $T = 2\pi\sqrt{(9.376 \times 10^6)^3/4.283 \times 10^{13}} = 2\pi \times 4394 = 27{,}609$ s $= 7.67$ hr.
8. $v = \sqrt{G \cdot 4M/(2R)} = \sqrt{2GM/R} = \sqrt{2} \cdot v_E$. About $41\%$ faster.
9. $v = \sqrt{\mu/r} \propto r^{-1/2}$. Velocity decreases as the square root of radius.
10. $r = 7371$ km. Circumference $= 2\pi \times 7371 = 46{,}311$ km.