Deriving orbital mechanics from Newton’s laws — why satellites stay up
2026-02-26
On 4 October 1957, the Soviet Union launched Sputnik 1 into an orbit of 227 km × 941 km. The spacecraft completed one revolution in 96.2 minutes. Every number in that sentence follows from two equations: Newton’s law of gravity and the requirement that centripetal acceleration equal gravitational acceleration. Sputnik’s engineers did not need to measure orbital periods experimentally — they calculated them. The mathematics had been published in Newton’s Principia in 1687.
The essential insight is that an orbit is not a special condition requiring the suspension of physics. It is the natural outcome of falling sideways fast enough. A satellite at 400 km altitude moves at 7.7 km/s. Move slower and gravity wins — the satellite descends. Move faster and it rises. At exactly the right speed, it falls continuously and continuously misses the ground: a perpetual balance between falling and forward motion. From this balance, three things follow directly: the required orbital velocity at any altitude, the orbital period (and thus Kepler’s Third Law, that period squared is proportional to orbital radius cubed), and the altitude at which the orbital period equals Earth’s rotation period — the geostationary orbit used by communications and weather satellites. The derivation is undergraduate algebra; the result governs the placement of every spacecraft ever launched.
Why doesn’t a satellite fall down?
The International Space Station orbits at 400 km altitude, traveling at 7.7 km/s (27,600 km/h). It’s moving so fast horizontally that as gravity pulls it downward, it falls around the curve of the Earth rather than into it.
The balance: Gravity pulls inward. The satellite’s velocity carries it forward. The combination creates a circular path.
The mathematical question: What velocity is required for a circular orbit at a given altitude? How long does one orbit take?
Imagine firing a cannonball horizontally from a mountaintop: - Too slow: it arcs down and hits the ground nearby - Faster: it travels farther before landing - Orbital velocity: Fast enough that the ground curves away beneath it at the same rate it falls
Key insight: Orbit is not “escaping gravity.” It’s falling continuously but moving sideways fast enough to miss the Earth.
An object moving in a circle at constant speed is accelerating — its velocity vector constantly changes direction.
Centripetal acceleration:
a_c = \frac{v^2}{r}
Where: - v is the orbital velocity (m/s) - r is the orbital radius (distance from Earth’s center, not altitude)
This acceleration points toward the center of the circle (inward, toward Earth).
Newton’s Second Law:
F = ma_c = m\frac{v^2}{r}
For a circular orbit, gravity provides exactly this centripetal force.
The gravitational force between two masses is:
F_g = G\frac{M m}{r^2}
Where: - G = 6.674 \times 10^{-11} N·m²/kg² (gravitational constant) - M = 5.972 \times 10^{24} kg (Earth’s mass) - m is the satellite’s mass (kg) - r is the distance between centers (m)
For an orbit at altitude h above Earth’s surface:
r = R_{\oplus} + h
Where R_{\oplus} = 6.371 \times 10^6 m (Earth’s radius).
For a circular orbit:
F_g = F_c
G\frac{Mm}{r^2} = m\frac{v^2}{r}
Satellite mass cancels:
G\frac{M}{r^2} = \frac{v^2}{r}
Multiply both sides by r:
G\frac{M}{r} = v^2
Orbital velocity:
v = \sqrt{\frac{GM}{r}}
Key result: Orbital velocity depends only on Earth’s mass and the orbital radius — not on the satellite’s mass.
The satellite travels a circular path of circumference $2r$ at velocity v.
Period (time for one orbit):
T = \frac{2\pi r}{v}
Substitute v = \sqrt{GM/r}:
T = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi r \sqrt{\frac{1}{GM/r}} = 2\pi r \sqrt{\frac{r}{GM}}
T = 2\pi \sqrt{\frac{r^3}{GM}}
This is Kepler’s Third Law (for circular orbits):
T^2 = \frac{4\pi^2}{GM} r^3
In words: The square of the orbital period is proportional to the cube of the orbital radius.
Define:
\mu = GM = 3.986 \times 10^{14} \text{ m}^3/\text{s}^2
Then:
v = \sqrt{\frac{\mu}{r}}, \quad T = 2\pi\sqrt{\frac{r^3}{\mu}}
Problem: A satellite orbits at altitude h = 500 km above Earth’s surface.
(a) Orbital radius
r = R_{\oplus} + h = 6.371 \times 10^6 + 0.5 \times 10^6 = 6.871 \times 10^6 \text{ m}
(b) Orbital velocity
v = \sqrt{\frac{\mu}{r}} = \sqrt{\frac{3.986 \times 10^{14}}{6.871 \times 10^6}}
= \sqrt{5.800 \times 10^7} = 7615 \text{ m/s} \approx 7.6 \text{ km/s}
(c) Orbital period
T = 2\pi\sqrt{\frac{r^3}{\mu}} = 2\pi\sqrt{\frac{(6.871 \times 10^6)^3}{3.986 \times 10^{14}}}
= 2\pi\sqrt{\frac{3.244 \times 10^{20}}{3.986 \times 10^{14}}} = 2\pi\sqrt{8.139 \times 10^5}
= 2\pi \times 902.2 = 5668 \text{ s} \approx 94.5 \text{ minutes}
(d) Orbits per day
N = \frac{86400 \text{ s/day}}{5668 \text{ s/orbit}} \approx 15.2 \text{ orbits/day}
The satellite completes about 15 orbits per day.
Altitude: 200–2000 km
Period: 90–130 minutes
Examples: ISS (400 km), Hubble (540 km), most Earth observation satellites
Applications: High-resolution imaging, atmospheric science, human spaceflight
Goal: Match Earth’s rotation period (24 hours) so the satellite appears stationary above a fixed point on the equator.
Required period: T = 86400 s (one sidereal day ≈ 86164 s for precision)
Solve for r:
T = 2\pi\sqrt{\frac{r^3}{\mu}}
r^3 = \frac{\mu T^2}{4\pi^2}
r = \left(\frac{\mu T^2}{4\pi^2}\right)^{1/3}
Substitute T = 86164 s:
r = \left(\frac{3.986 \times 10^{14} \times (86164)^2}{4\pi^2}\right)^{1/3}
r = (7.496 \times 10^{22})^{1/3} = 4.214 \times 10^7 \text{ m}
Altitude:
h = r - R_{\oplus} = 42140 - 6371 = 35769 \text{ km}
Geostationary altitude: 35,786 km (often rounded to 36,000 km)
Orbital velocity:
v = \sqrt{\frac{\mu}{r}} = \sqrt{\frac{3.986 \times 10^{14}}{4.214 \times 10^7}} = 3075 \text{ m/s} \approx 3.1 \text{ km/s}
Applications: Communication satellites, weather satellites (GOES, Meteosat), broadcast TV
Goal: The orbital plane precesses at the same rate Earth orbits the Sun (~1°/day), so the satellite crosses the equator at the same local solar time each day.
Typical altitude: 600–800 km
Inclination: 98°–99° (near-polar, retrograde)
Why it matters: Consistent lighting for optical imagery (always 10:30 AM local time, for example).
Examples: Landsat, Sentinel-2, most polar-orbiting weather satellites
Below is an interactive orbit calculator.
<label>
Altitude (km):
<input type="range" id="altitude-slider" min="200" max="40000" step="100" value="500">
<span id="altitude-value">500</span> km
</label>
<div class="button-group">
<button id="set-leo">ISS (400 km)</button>
<button id="set-geo">GEO (35,786 km)</button>
<button id="set-sso">SSO (800 km)</button>
</div><p><strong>Orbital radius:</strong> <span id="radius-display"></span> km</p>
<p><strong>Orbital velocity:</strong> <span id="velocity-display"></span> km/s</p>
<p><strong>Orbital period:</strong> <span id="period-display"></span></p>
<p><strong>Orbits per day:</strong> <span id="orbits-day-display"></span></p>Try this: - Slide to LEO altitudes (200–800 km): High velocity, short periods, many orbits/day - Click “GEO”: The magic altitude where period = 24 hours - Notice: Velocity decreases with altitude (gravity weakens faster than radius increases) - The dashed gray circle shows GEO for reference
Key insight: Lower orbits are faster but have shorter periods. Higher orbits are slower but take longer to complete.
Once in orbit, a satellite coasts. Gravity provides exactly the centripetal force needed for circular motion.
No fuel required to maintain altitude (in the absence of drag).
Fuel is needed for: - Reaching orbit initially (launch) - Correcting for atmospheric drag (LEO only) - Changing orbits (maneuvers) - De-orbiting at end of life
Below ~800 km, Earth’s atmosphere still has measurable density.
Drag force:
F_d \propto \rho v^2 A
Where \rho is atmospheric density, v is velocity, A is cross-sectional area.
Effect: Satellite slowly spirals inward, eventually re-entering and burning up.
ISS altitude maintenance: Requires periodic “reboost” maneuvers (small thruster burns to raise orbit). Without these, it would de-orbit in ~2 years.
Problem: Debris from dead satellites and collisions creates more collisions in a cascading effect.
LEO is already crowded: - ~5000 active satellites - ~30,000 tracked debris objects >10 cm - ~1 million objects 1–10 cm (too small to track, large enough to damage)
Mitigation: Satellites are designed to de-orbit within 25 years after end of mission (drag in LEO, controlled burns for higher orbits).
Orbital radius r is measured from Earth’s center.
Altitude h is measured from Earth’s surface.
r = R_{\oplus} + h
Don’t plug altitude into the orbital velocity formula!
In the derivation, m appears on both sides and cancels. Orbital parameters don’t depend on satellite mass (for circular orbits around a much more massive body).
A 1 kg CubeSat and a 400,000 kg ISS at the same altitude have the same orbital velocity and period.
Real orbits are ellipses (Kepler’s First Law). Circular orbits are a special case where eccentricity e = 0.
Transfer orbits (e.g., Hohmann transfer) are elliptical and use the difference in velocity at periapsis and apoapsis to change altitude.
Earth’s equatorial bulge causes orbital precession — the orbital plane slowly rotates.
This effect is intentionally used for sun-synchronous orbits (the precession matches Earth’s orbit around the Sun).
For precision calculations, use the J2 perturbation term.
Question: How fast must you go to escape Earth’s gravity entirely?
At escape velocity, the satellite has enough kinetic energy to reach infinity with zero residual velocity.
Energy balance:
\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0
(Kinetic energy equals the magnitude of gravitational potential energy)
v_e = \sqrt{\frac{2GM}{r}} = \sqrt{2} \cdot v_{\text{orbital}}
At Earth’s surface:
v_e = \sqrt{\frac{2 \times 3.986 \times 10^{14}}{6.371 \times 10^6}} = 11,180 \text{ m/s} \approx 11.2 \text{ km/s}
Escape velocity is \sqrt{2} \approx 1.41 times the orbital velocity at the same altitude.
An object moving at constant speed v in a circle of radius r has acceleration toward the center:
a_c = \frac{v^2}{r} = \omega^2 r
Where \omega = v/r is the angular velocity (rad/s).
Why? The velocity vector is constantly changing direction. The rate of change of velocity is acceleration.
Period T: Time for one complete revolution (seconds)
Frequency f: Revolutions per unit time (Hz = 1/s)
f = \frac{1}{T}
Angular velocity:
\omega = \frac{2\pi}{T} = 2\pi f
Linear velocity:
v = \frac{2\pi r}{T} = \omega r