Orbital Velocity — Circular, Escape, and Transfer Orbits
Why does the ISS travel at 7.66 km/s? Why does escape from Earth require exactly √2 times that speed? How does a spacecraft move from a low parking orbit to geostationary without continuous thrust? All of these questions are answered by four equations — the visionary simplicity of Kepler, Newton, and Hohmann.
1. Circular Orbital Velocity
For a circular orbit of radius r around a body of mass M, gravity provides exactly the centripetal acceleration needed:
v_c = √(GM/r)
For Earth (GM = 3.986 × 10¹⁴ m³/s²):
r = 6 771 km (ISS altitude ~400 km + R_Earth 6371 km)
v_c = √(3.986×10¹⁴ / 6.771×10⁶) ≈ 7.66 km/s
Note that v_c depends only on the central mass and the
orbital radius — not on the satellite's mass. A marble in the same
orbit as the ISS travels at identical speed.
2. Escape Velocity
Escape velocity is the minimum speed needed to escape a gravitational well with zero energy at infinity (i.e. kinetic + potential = 0):
v_esc = √(2GM/r) = √2 · v_c
From Earth's surface: v_esc = √(2 × 3.986×10¹⁴ / 6.371×10⁶)
≈ 11.19 km/s
The factor √2 between circular and escape velocity is universal — it depends only on energy conservation, not on Earth's specific properties. For the Moon (much smaller GM): v_esc ≈ 2.38 km/s.
3. Vis-Viva Equation
For any conic-section orbit (circle, ellipse, parabola, hyperbola), the speed at any point depends on the current radius r and the semimajor axis a:
Circular orbit: a = r → v² = GM/r ✓
Escape orbit: a = ∞ → v² = 2GM/r ✓
At apoapsis of ellipse (r = r_a = a(1+e)):
v_a = √(GM/a · (1−e)/(1+e))
The vis-viva equation is the single most useful formula in orbital mechanics — it directly gives speed at any orbital position without integrating equations of motion.
4. Hohmann Transfer Orbit
A Hohmann transfer is the most fuel-efficient two-burn manoeuvre to move between two circular coplanar orbits. The transfer orbit is an ellipse whose periapsis is the inner orbit and apoapsis the outer orbit:
Burn 1 (at r₁, increase speed to enter transfer ellipse):
v_t1 = √(GM(2/r₁ − 1/a_t))
Δv₁ = v_t1 − v_c1 = v_t1 − √(GM/r₁)
Burn 2 (at r₂, circularise):
v_t2 = √(GM(2/r₂ − 1/a_t))
Δv₂ = √(GM/r₂) − v_t2
Transfer time: t = π √(a_t³/GM) (half the ellipse period)
5. Geostationary Orbit
A satellite with an orbital period equal to Earth's sidereal rotation period (23 h 56 min 4 s) appears stationary from the ground. Solving Kepler's third law for this period:
r_GEO = (GM·T²/(4π²))^(1/3)
= (3.986×10¹⁴ × (86164)² / (4π²))^(1/3)
≈ 42 164 km from Earth's centre
≈ 35 786 km altitude above surface
v_GEO = √(GM/r_GEO) ≈ 3.07 km/s
All geostationary satellites share the same altitude and always orbit in the equatorial plane (inclination = 0°). Communication satellites, weather satellites (GOES/Meteosat), and GPS relay stations are typical occupants.
6. Kepler's Third Law
Kepler (1619) empirically found that the square of the orbital period T is proportional to the cube of the semimajor axis a. Newton showed why:
Normalised to Earth-Sun system (AU and years):
T² = a³ (T in years, a in AU)
| Body | Semimajor axis (AU) | Period (years) | T²/a³ |
|---|---|---|---|
| Mercury | 0.387 | 0.241 | 0.998 |
| Earth | 1.000 | 1.000 | 1.000 |
| Mars | 1.524 | 1.881 | 1.000 |
| Jupiter | 5.203 | 11.86 | 0.999 |
7. Lagrange Points
In a two-body system (e.g. Sun-Earth) there are five special points where a small body can remain stationary relative to both primary masses (in the rotating frame):
- L1: Between the two bodies — Earth-Sun L1 is ~1.5 million km from Earth toward the Sun. DSCOVR solar observatory, SOHO.
- L2: On the far side from the smaller body — Sun-Earth L2 is ~1.5 million km from Earth away from the Sun. James Webb Space Telescope, Gaia, Planck.
- L3: On the opposite side of the Sun — unstable and inaccessible for practical use.
- L4, L5: 60° ahead and behind the smaller body on its orbit. Stable if the mass ratio M₁/M₂ > 24.96. Jupiter's Trojan asteroids cloud L4 and L5.
8. Delta-v Budget
Delta-v (Δv) is the total change in velocity a spacecraft must execute to complete a mission. The Tsiolkovsky rocket equation relates Δv to propellant fraction:
v_e = effective exhaust velocity (= I_sp × g₀)
m₀ = initial (wet) mass
m_f = final (dry) mass
Example: Δv = 6 km/s, v_e = 3 km/s (kerosene engine)
m₀/m_f = e^(6/3) = e² ≈ 7.4
→ 86% of launch mass is propellant
Typical Δv budget for ISS (LEO): ~9.5 km/s from Earth's surface. GEO communication satellite: ~12 km/s from surface. Mars landing: ~16–18 km/s round trip via Hohmann, not including re-entry.