Orbital Mechanics

You’ll be able to

Apply Newton's gravitation to circular orbits, with gravity supplying the centripetal force
Derive orbital speed and relate the period to radius through Kepler's third law
Use conservation of angular momentum to relate speeds at perihelion and aphelion

Gravity provides the centripetal force

A satellite in a circular orbit is in free fall, continuously "missing" the planet. Gravity supplies exactly the centripetal force needed: GMm/r² = mv²/r. The satellite mass cancels, leaving the orbital speed v = √(GM/r). Larger orbits are slower — a distant satellite both feels weaker gravity and needs less centripetal force.

Circular orbit

GMm/r² = mv²/r → v = √(GM/r) T² = (4π²/GM) r³

Orbital speed falls as 1/√r. The period relation T² ∝ r³ is Kepler's third law, obtained by combining v = √(GM/r) with T = 2πr/v.

Angular momentum in an elliptical orbit

Gravity always points toward the central body, so it is a central force and exerts no torque about that center — which means the orbiting body's angular momentum is conserved even in a non-circular orbit. This is Kepler's second law (equal areas in equal times): at perihelion (closest) the planet moves fastest, at aphelion (farthest) slowest, with v_p r_p = v_a r_a since the velocity is perpendicular to the radius at both apses.

Angular momentum at the apses

L = mvr = constant → v_p r_p = v_a r_a

At perihelion and aphelion the velocity is perpendicular to the radius, so L = mvr directly. The closer approach forces the faster speed.

Worked example

A comet at perihelion is a distance r_p from the Sun moving at 60 km/s. At aphelion it is 3 r_p from the Sun. Find its speed at aphelion.

1.Gravity is central, so angular momentum is conserved: m v_p r_p = m v_a r_a.
2.At both apses the velocity is perpendicular to the radius, so the simple form L = mvr applies.
3.Solve for the aphelion speed: v_a = v_p (r_p/r_a) = 60 × (r_p / 3r_p) = 60/3.
4.So v_a = 20 km/s — the comet crawls when it is far from the Sun.

Answer: v_a = v_p (r_p/r_a) = 60 × (1/3) = 20 km/s

Checkpoint

A satellite in a circular orbit has speed v = √(GM/r). If its orbital radius is quadrupled, the orbital speed becomes:

Checkpoint

A comet moves in an elliptical orbit. At perihelion its distance is r_p and its speed is v_p; at aphelion its distance is 3r_p. What is its speed at aphelion?

On the exam

For circular orbits set gravity equal to the centripetal force (GMm/r² = mv²/r). For elliptical orbits, use conservation of angular momentum (v_p r_p = v_a r_a) at the apses and conservation of energy between them — energy and angular momentum together pin down the whole orbit.

Answer the 2 checkpoints as you read.