Orbital Angular Momentum

Gravity from a spherical star pulls a planet along the line to the star's centre: F⃗ = −(GMm/r²) r̂, with r̂ the unit vector from the centre to the planet. Take the torque about that centre: τ⃗ = r⃗ × F⃗ = −(GMm/r)(r̂ × r̂) = 0, because any vector crossed with itself vanishes. Nothing in that line used the size of F, its 1/r² dependence or the planet's speed, so every central force gives the same zero, from a string tension pulling a puck towards a hole to the electric pull on an orbiting charge. The zero belongs to the origin, not to the planet. About a point Q at position s⃗ from the Sun the torque is (r⃗ − s⃗) × F⃗ = −s⃗ × F⃗. The Sun pulls on Earth with 3.54 × 10²² N; put Q only 0.10 AU from the Sun, perpendicular to the Earth–Sun line, and that force has a lever arm of 1.50 × 10¹⁰ m and a torque of 5.3 × 10³² N m about Q. Declare the origin before claiming conservation.

L⃗ = m r⃗ × v⃗ is perpendicular to both r⃗ and v⃗. If L⃗ never changes direction, the position vector is perpendicular to the same fixed vector at every instant, so r⃗ is confined to one plane through the force centre, and v⃗ lies in that plane too. The starting position and velocity pick the plane, and a purely central force can never tilt it. The one exception fits the rule: if v⃗ starts parallel to r⃗, then L⃗ = 0, no plane is defined, and the motion is a straight line through the centre, like a probe released from rest. Tilting a real orbit therefore needs a force with a sideways part. A satellite moving at 7.7 km s⁻¹ that must turn its orbital plane through 10° has to rotate its velocity through 10°, which costs Δv = 2v sin 5° = 1.34 km s⁻¹, about 17% of the speed it already has.

Split the velocity into a radial part vᵣ = dr/dt along r̂ and a sideways part v⊥ = r dθ/dt perpendicular to it. Only v⊥ survives the cross product, so L = m r v⊥ = m r² dθ/dt. With L fixed, v⊥ ∝ 1/r and the turning rate dθ/dt ∝ 1/r². Mercury shows both. Its perihelion is 4.600 × 10¹⁰ m and its aphelion 6.982 × 10¹⁰ m, a ratio of 1.518, and its speeds there are 58.98 and 38.86 km s⁻¹, the same ratio, as rₚ vₚ = rₐ vₐ demands. The angle Mercury sweeps per day changes by 1.518² = 2.30 between the two points. Angular momentum says nothing about vᵣ: whether the planet is falling in or climbing out, and how fast, is set by energy. That division of labour is why the two conservation laws together, and neither alone, determine an orbit.

In a short time dt the radius line turns through dθ and sweeps a thin triangle with two sides of length r and an included angle dθ, so dA = ½ r (r dθ). Divide by dt: dA/dt = ½ r² dθ/dt = L/(2m). A constant L means a constant sweeping rate. Kepler extracted the rule from Tycho Brahe's observations of Mars and published it in 1609, nearly eighty years before Newton could say why it held. The derivation shows it holds for any central force: the inverse-square law is what makes the path an ellipse, a separate result. For Earth, L/(2m) = ½ × 1.496 × 10¹¹ m × 2.978 × 10⁴ m s⁻¹ = 2.23 × 10¹⁵ m² s⁻¹. Over one year, 3.156 × 10⁷ s, that sweeps 7.03 × 10²² m², exactly the area πab enclosed by Earth's orbit. Read backwards, T = πab/(dA/dt): an orbit's period is its area divided by its areal velocity.

At perihelion and aphelion the distance is momentarily neither growing nor shrinking, so v⃗ ⟂ r⃗ and L = m rₚ vₚ = m rₐ vₐ. That fixes the ratio of the two speeds but not their values; energy supplies the second equation. Write ½vₚ² − GM/rₚ = ½vₐ² − GM/rₐ, substitute vₐ = vₚ rₚ/rₐ, and solve: vₚ² = 2GM rₐ/[rₚ(rₚ + rₐ)]. Apply it to the transfer ellipse that touches Earth's orbit at 1.000 AU and Mars's at 1.524 AU, with GM = 1.327 × 10²⁰ m³ s⁻² for the Sun. The perihelion speed is 32.7 km s⁻¹ and the aphelion speed 32.7 × 1.000/1.524 = 21.5 km s⁻¹. Earth moves at 29.8 km s⁻¹, so a spacecraft needs about 2.9 km s⁻¹ more, along Earth's direction of motion, to start the transfer, and it reaches Mars's orbit slower than Mars's own 24.1 km s⁻¹, so it must speed up again to stay there.

The claim that L⃗ about the Sun is constant hides three conditions, and each can fail. The origin must be the force centre, as the torque about Q showed. The centre must not accelerate: Jupiter alone swings the Sun around a common centre of mass 7.42 × 10⁸ m from the Sun's centre, just outside its surface, so the exact law conserves total angular momentum about the centre of mass, and a fixed Sun is an approximation good to about m/M, 1 part in 1000 for Jupiter. And the force must be exactly central. Air drag on a low satellite points against v⃗, has a lever arm about Earth's centre and drains L, so the orbit shrinks, and the satellite speeds up as it sinks. Earth's equatorial bulge adds a slightly non-central pull that turns L⃗ itself, swinging the International Space Station's orbital plane round by about 5° a day. Tidal torques carry angular momentum from Earth's spin into the Moon's orbit, which is why the Moon recedes by 3.8 cm a year.