Circular Orbits & Orbital Speed

A satellite is any object that orbits another under the influence of gravity. The Moon orbits Earth, Earth orbits the Sun, and thousands of artificial satellites orbit Earth every day — enabling GPS, weather forecasting, and global communications. For a satellite to maintain a stable circular orbit, it must travel at precisely the right speed: too slow and it spirals inward; too fast and it escapes into space. This delicate balance is not a coincidence — it arises directly from Newton's law of gravitation combined with the requirements of circular motion. The key insight is that gravity provides exactly the centripetal force needed to keep the satellite moving in a circle. Because gravitational force depends on distance from the planet's centre, the required orbital speed also depends on altitude. Satellites in low orbits must move faster than those in high orbits — a counter-intuitive result that becomes clear once you work through the mathematics. The International Space Station, orbiting at roughly 400 km altitude, completes one lap of Earth in about 92 minutes at nearly 7.7 km/s. The orbital period — the time taken to complete one full orbit — is closely related to orbital radius through Kepler's third law. Understanding the relationship between orbital speed, radius, and period is fundamental to designing satellite missions, predicting spacecraft trajectories, and appreciating the mechanics of the Solar System itself.

Key takeaways

Setting Up the Force Balance
Deriving Orbital Speed
Finding the Orbital Period
Geostationary vs. Low Earth Orbits

Setting Up the Force Balance

For a satellite of mass m orbiting a planet of mass M at orbital radius r (measured from the planet's centre), gravity must supply the centripetal force. Newton's law of gravitation gives the gravitational force as F = GMm/r². The centripetal force required for circular motion is F = mv²/r. Setting these equal: GMm/r² = mv²/r. Notice that the satellite's mass m cancels from both sides — orbital speed is independent of the satellite's own mass.

Deriving Orbital Speed

After cancelling m and rearranging GMm/r² = mv²/r, we get v² = GM/r, so v = √(GM/r). This is the orbital speed formula. It tells us that orbital speed decreases as orbital radius increases — satellites higher up move more slowly. For Earth (M ≈ 5.97 × 10²⁴ kg, G = 6.67 × 10⁻¹¹ N m² kg⁻²), a satellite at r = 6.8 × 10⁶ m (about 400 km above the surface) has orbital speed v ≈ 7.7 km/s.

Finding the Orbital Period

The orbital period T is the circumference of the orbit divided by the orbital speed: T = 2πr/v. Substituting v = √(GM/r) gives T = 2πr/√(GM/r) = 2π√(r³/GM). This is Kepler's Third Law in disguise: T² ∝ r³. Squaring both sides, T² = 4π²r³/(GM). A larger orbit means a longer period, both because the path is longer and because the satellite travels more slowly. Geostationary satellites (T = 24 hours) must orbit at r ≈ 42,200 km from Earth's centre.

Geostationary vs. Low Earth Orbits

A geostationary orbit has T = 24 hours, so the satellite stays fixed above one point on the equator — ideal for TV broadcasts and weather monitoring. Low Earth Orbit (LEO) satellites like the ISS orbit at 300–600 km altitude with periods of about 90 minutes. They cover the whole globe but cannot stay fixed over one location. The required altitude for a geostationary orbit can be found by setting T = 86,400 s in T² = 4π²r³/(GM) and solving for r, yielding r ≈ 4.22 × 10⁷ m.

The Key Equation

v = \sqrt{\frac{GM}{r}}, \qquad T = 2\pi\sqrt{\frac{r^3}{GM}}

v — Orbital speed (m s⁻¹): The linear speed of the satellite along its circular orbit
G — Gravitational constant (N m² kg⁻²): Universal constant, G = 6.674 × 10⁻¹¹ N m² kg⁻²
M — Mass of central body (kg): Mass of the planet (or star) being orbited — NOT the satellite's mass
r — Orbital radius (m): Distance from the centre of the planet to the satellite
T — Orbital period (s): Time taken for the satellite to complete one full orbit

Worked Example

Problem. A communications satellite orbits Earth in a circular geostationary orbit. Earth's mass is M = 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N m² kg⁻². (a) Calculate the orbital radius required for a 24-hour period. (b) Calculate the orbital speed of the satellite.

r \approx 4.22 \times 10^{7}\text{ m}, \quad v \approx 3070\text{ m s}^{-1}

Answer. Geostationary orbital radius ≈ 4.22 × 10⁷ m; orbital speed ≈ 3,070 m s⁻¹ (≈ 3.07 km/s).

Figure

Diagram

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · English

Newton formulated the Universal Law of Gravitation (F = GMm/r²) in his 1687 masterwork Principia Mathematica, providing the mathematical foundation for understanding orbital motion. By combining this with his laws of motion, Newton showed that the Moon's orbit — and indeed all planetary orbits — could be explained by the same gravitational force that causes apples to fall. His work directly underpins every formula used to calculate orbital speed and period. Newton famously used the concept of a cannon fired from a very tall mountain to explain orbits — if fired fast enough, the cannonball would fall around Earth continuously, never hitting the ground. This 'Newton's cannon' thought experiment is essentially how we understand satellite orbits today, nearly 340 years later.

Newton's Law of Gravitation — Orbital speed is derived directly by setting gravitational force equal to centripetal force — you must understand F = GMm/r² first.
Gravitational Field Strength — The gravitational field at the orbital radius determines the acceleration that keeps the satellite in orbit, linking g to orbital mechanics.
Centripetal Force and Acceleration — Circular orbital motion requires centripetal force; understanding a = v²/r is essential to deriving the orbital speed formula.
Kepler's Third Law — The period–radius relationship T² ∝ r³ is Kepler's Third Law, which emerges directly from the orbital period equation derived here.