--- title: "Orbits: Period, Speed and Kepler's Third Law" description: "For a small satellite in a circular orbit around a much larger body, gravity supplies the centripetal force. Derive orbital speed and period, compare orbits around the same central mass, and calculate" canonical: https://lightmysky.com/learn/science/orbits-period-speed-and-keplers-third-law-mt_PLKLykGF8Z source: https://lightmysky.com/learn/science/orbits-period-speed-and-keplers-third-law-mt_PLKLykGF8Z.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Orbits: Period, Speed and Kepler's Third Law For a small satellite in a circular orbit around a much larger body, gravity supplies the centripetal force. Derive orbital speed and period, compare orbits around the same central mass, and calculate a geostationary satellite's radius and height above Earth's surface. Subject: Science · Area: Space Systems & Earth's History · Ages 17 to 18 Page: https://lightmysky.com/learn/science/orbits-period-speed-and-keplers-third-law-mt_PLKLykGF8Z ## Ready when they can - Derives v = √(GM/r) by equating GMm/r² with mv²/r for a small satellite in a circular orbit - Derives T² = 4π²r³/(GM), explains its straight-line plot and compares periods around the same central mass - States the conditions for a geostationary orbit and calculates its radius and height above Earth's surface ## Lesson: Circular orbits and geostationary satellites Let a small satellite of mass m circle a much larger spherical body of mass M. Treat M as fixed and ignore drag and other bodies. G is the gravitational constant; r is distance from the body's centre. Gravity supplies the inward force: GMm/r² = mv²/r. Cancel m and multiply by r: v² = GM/r, so v = √(GM/r). For this circular-orbit model, satellite mass does not affect speed. Around the same body, a larger circular radius gives a lower speed. *(drawing: These are successive forms of the same equation for a circular orbit.)* Period T is the time for one orbit, so v = 2πr/T. Square every factor in the fraction: (2πr/T)² = 4π²r²/T². Substitute for v²: GM/r = 4π²r²/T². Multiply both sides by rT² to clear the denominators: GMT² = 4π²r³. Divide by GM: T² = 4π²r³/(GM). For fixed M, T² is proportional to r³. Plot T² vertically against r³ horizontally: a straight line through the origin with gradient 4π²/(GM). **Example.** For two small satellites in circular orbits around the same body, (T_B/T_A)² = (r_B/r_A)³. Cube the radius ratio, take its positive square root, then multiply by T_A to find T_B. For a radius ratio of 2, the period ratio is √8, about 2.83. These comparisons describe circular orbits, not speeds during an engine burn. *(drawing: Cube the radius factor, then take the square root to find the period factor.)* A geostationary satellite moves eastward in a circular orbit in Earth's equatorial plane. Its period matches Earth's sidereal day, one rotation relative to distant stars: about 23 h 56 min 4 s, or 86,164 s. It stays above one point on the equator. Matching the period alone gives a geosynchronous orbit. An inclined or elliptical orbit with that period is not geostationary. From T² = 4π²r³/(GM), multiply by GM, divide by 4π², then take the cube root: r = ∛(GMT²/(4π²)). In a × 10^n, multiply a by 10 to the power n; use a calculator's exponent-entry key. Enter GM × T² / (4 × π²), then take the cube root. Use GM in m³/s² and T in seconds; r is metres from the centre. Divide by 1,000 for km, then subtract R in km: h = r - R. Earth's geostationary radius is about 42,200 km and height about 35,800 km. **Recap.** For small satellites in circular orbits, v = √(GM/r) and T² = 4π²r³/(GM). Geostationary motion needs a circular equatorial orbit moving eastward with Earth's sidereal rotation period. *Lesson, questions and flashcards written for this dataset. OpenStax College Physics, Astronomy and University Physics Volume 1 were consulted for physics facts and teaching design.* ## Practice 8 questions on this page, each with its working shown. ## Needs first - [Circular Motion: Angular Speed and Centripetal Force](https://lightmysky.com/learn/science/circular-motion-angular-speed-and-centripetal-force-mt_PW1G6nYStq) - [Newton's Law of Gravitation and Gravitational Field Strength](https://lightmysky.com/learn/science/newtons-law-of-gravitation-and-gravitational-field-strength-mt_uUYXl3L0oW) ## Optional connections - [Orbital Mechanics](https://lightmysky.com/learn/science/orbital-mechanics-mt_1nl396BaVa) - [Changing Ideas About Space](https://lightmysky.com/learn/science/changing-ideas-about-space-mt_KiRn5lnRgj) ## Opens up - [The Two-Body Problem, Orbital Energy and Kepler's First Two Laws](https://lightmysky.com/learn/science/the-two-body-problem-orbital-energy-and-keplers-first-two-laws-mt_CjHFftkB6M) - [Coulomb's Law and the Electric Field of a Point Charge](https://lightmysky.com/learn/science/coulombs-law-and-the-electric-field-of-a-point-charge-mt_p7FuzIsQch)