Orbital velocity: falling but never landing
A satellite in orbit is always falling toward the Earth. But it also moves sideways so fast that the Earth's surface curves away beneath it. So it never lands.
Deriving v₀
- For a circular orbit of radius r = R + h, gravity provides the centripetal force: GMm/r² = m v₀²/r.
- So v₀ = √(GM/r) = √(GM/(R + h)).
- Just above the surface (h ≪ R): v₀ = √(GM/R) = √(gR) = √(9.8 × 6.4 × 10⁶) ≈ 7.9 km/s.
v₀ does not depend on the satellite's mass. Higher orbit (bigger r) means lower speed.
Time period
T = 2πr / v₀ = 2π√(r³/GM). Near the surface T ≈ 84 minutes. So T² ∝ r³ — Kepler's third law again.
Escape speed: leaving for ever
Escape speed is the smallest speed with which a body must be thrown from the surface so that it never comes back (it just reaches infinity with zero speed).
Deriving vₑ from energy
- At the surface: KE = ½ m vₑ², PE = −GMm/R.
- At infinity, just escaping: KE = 0, PE = 0. Total = 0.
- Energy is conserved: ½ m vₑ² − GMm/R = 0.
- vₑ = √(2GM/R) = √(2gR) = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s.
- It does not depend on the mass of the body or on the direction of throw (if we ignore air).
- vₑ = √2 × v₀ (both near the surface): 11.2 ≈ 1.414 × 7.9.
- Moon: vₑ ≈ 2.4 km/s. That is why the Moon lost its air.
Energy of an orbiting satellite
For a satellite of mass m in a circular orbit of radius r:
- Kinetic energy: KE = ½ m v₀² = ½ m (GM/r) = GMm/2r (positive)
- Potential energy: U = −GMm/r (negative)
- Total energy: E = KE + U = −GMm/2r
So E = −KE and U = −2 KE. The total is negative: the satellite is bound to the Earth.
Binding energy
The energy you must add to free the satellite from its orbit = +GMm/2r. For an orbit at r, going to a higher orbit needs extra energy, even though the satellite ends up moving slower (KE falls, but U rises twice as much).
Kinds of satellites and weightlessness
- Geostationary satellite: T = 24 h, moves west to east above the equator, so it looks fixed in the sky. Radius ≈ 42,200 km, height ≈ 35,800 km. Used for TV, communication and weather.
- Polar satellite: low orbit (about 500–800 km) passing over the poles; Earth turns beneath it, so it can photograph every part of the Earth. Used for mapping and remote sensing.
Why do astronauts float?
The astronaut and the station fall together with the same acceleration. The floor does not push up on the astronaut, so the weighing machine reads zero. Gravity is still there; the apparent weight is zero. This is weightlessness.
Try it: the sideways throw
Throw a ball sideways gently, then harder, then harder again, from the same height. Each time it lands farther away. Newton imagined a cannon on a very tall mountain: throw fast enough and the ball lands 'beyond the curve' of the Earth — it orbits. In the 3D free play, try 7.0, 7.9, 9.0, 11.2 and 12 km/s and predict before you press Launch: fall back, circle, stretched ellipse or escape?
Key formulas and definitions
- v₀ = √(GM/r) = √(gR²/(R + h))
- v₀ (near surface) = √(gR) ≈ 7.9 km/s
- T = 2π√(r³/GM)
- vₑ = √(2GM/R) = √(2gR) ≈ 11.2 km/s
- vₑ = √2 v₀
- KE = GMm/2r, U = −GMm/r, E = −GMm/2r
- Binding energy = GMm/2r
Worked examples
1. Find the orbital speed just above Earth's surface (g = 9.8 m/s², R = 6.4 × 10⁶ m).
v₀ = √(gR) = √(6.27 × 10⁷) ≈ 7.92 × 10³ m/s ≈ 7.9 km/s.
2. Find the escape speed from Earth.
vₑ = √(2gR) = √(1.254 × 10⁸) ≈ 1.12 × 10⁴ m/s = 11.2 km/s.
3. Find the orbital speed at a height h = R.
r = 2R. v₀ = √(gR²/2R) = √(gR/2) = 7.9/√2 ≈ 5.6 km/s.
4. A planet has twice Earth's mass and twice its radius. Find its escape speed (Earth: 11.2 km/s).
vₑ ∝ √(M/R) = √(2/2) = 1. So vₑ = 11.2 km/s.
5. Find the period of a satellite orbiting close to Earth's surface.
T = 2π√(R/g) = 2π√(6.4 × 10⁶/9.8) = 2π × 808 ≈ 5077 s ≈ 84.6 min.
6. A 500 kg satellite orbits at r = 2R. Find its KE, PE and total energy (g = 9.8, R = 6.4 × 10⁶ m).
GMm/r = gR²m/2R = mgR/2 = 500 × 9.8 × 6.4 × 10⁶/2 = 1.57 × 10¹⁰ J. KE = 7.84 × 10⁹ J, U = −1.57 × 10¹⁰ J, E = −7.84 × 10⁹ J.
7. How much energy is needed to move that 500 kg satellite from r = 2R to r = 4R?
E = −mgR²/2r. E(2R) = −mgR/4, E(4R) = −mgR/8. ΔE = mgR/8 = 500 × 9.8 × 6.4 × 10⁶/8 = 3.92 × 10⁹ J.
8. Find the radius of a geostationary orbit (T = 86400 s, GM = 4 × 10¹⁴ m³/s²).
r³ = GMT²/4π² = 4 × 10¹⁴ × 7.46 × 10⁹ / 39.48 = 7.56 × 10²². r ≈ 4.23 × 10⁷ m ≈ 42,300 km, so height ≈ 35,900 km.
Common mistakes
- Thinking a heavier satellite needs a bigger orbital or escape speed. Neither depends on the satellite's mass.
- Thinking a higher satellite moves faster. It moves slower: v₀ = √(GM/r).
- Writing total energy of a satellite as positive. It is E = −GMm/2r, negative because it is bound.
- Using vₑ = √(2gR) at a height. At distance r from the centre, escape speed is √(2GM/r).