Jupiter2U School
Chapter 13

13. Orbital Mechanics — Kepler & Newton

College

From Newton's law of gravitation alone, Kepler's three empirical laws of planetary motion can be derived rigorously — and the full machinery of orbits opens up.

At a glance
Core ideaKepler's three laws all follow from Newton's inverse-square gravity.
Key termVis-viva — v² = GM(2/r − 1/a), speed anywhere on an orbit.
You can…Find a comet's perihelion speed or a planet's period.
Watch outTotal energy sets the shape: bound ellipse, parabola, or hyperbola.
01 · Theory

Deriving Kepler's laws from Newtonian gravity

Kepler's first law states orbits are ellipses with the Sun at one focus — a consequence of solving the two-body equation of motion under an inverse-square force, which yields a general conic section (ellipse, parabola, or hyperbola) depending on total energy. Kepler's second law — equal areas in equal times — follows directly from conservation of angular momentum, since gravity is a central force exerting zero torque about the focus: L = m v r sinφ = constant.

Kepler's third law, T² ∝ a³, comes from equating gravitational and centripetal force for a circular case and generalizes to ellipses with semi-major axis a:

T² = 4π²a³ / G(M+m)

For elliptical orbits, speed varies with position according to the vis-viva equation, derived from conservation of energy:

v² = GM (2/r − 1/a)
02 · Explanation

Energy, eccentricity and orbit shape

The total mechanical energy of an orbit — kinetic plus gravitational potential — determines its shape entirely: negative total energy gives a bound elliptical (or circular) orbit; exactly zero energy gives a parabolic escape trajectory; positive energy gives a hyperbolic flyby that never returns. The eccentricity e (0 = circle, 0<e<1 = ellipse, e=1 = parabola, e>1 = hyperbola) sets how elongated the orbit is, with rperi = a(1−e) at closest approach and rapo = a(1+e) at farthest.

Key ideaA spacecraft moves fastest at perihelion (closest to the Sun) and slowest at aphelion (farthest) — this is exactly Kepler's second law in action: to sweep equal areas in equal time, the orbit's tighter, closer arc must be traced faster than its wide, distant arc.
03 · Practical

Worked example — vis-viva for a comet

A comet has a semi-major axis a = 4.0 AU and reaches perihelion at r = 0.50 AU. Find its speed at perihelion. Use GM☉ = 1.327×10²&sup0; m³s⁻², and 1 AU = 1.496×10¹¹ m.

Solution
  1. Convert to metres: a = 5.98×10¹¹ m, r = 7.48×10¹&sup0; m.
  2. Apply vis-viva: v² = GM(2/r − 1/a).
  3. Compute 2/r = 2.674×10⁻¹¹ and 1/a = 1.672×10⁻¹²; the difference is 2.507×10⁻¹¹.
  4. Multiply by GM: v² = 1.327×10²&sup0; × 2.507×10⁻¹¹ = 3.33×10&sup9;, so v ≈ 57 700 m/s ≈ 57.7 km/s.

Answer: about 57.7 km/s at perihelion — far faster than Earth's own 29.8 km/s orbital speed, exactly as expected for a much closer, highly eccentric pass.

04 · Q&A

Test your understanding

Derive, in words, why Kepler's second law follows from angular momentum conservation.

Gravity always points directly from the orbiting body toward the central mass, so it exerts zero torque about that centre. With no torque, angular momentum L = mvr sinφ is conserved throughout the orbit. Since the rate at which the orbit's radius vector sweeps out area is proportional to L/2m — a constant — equal areas must be swept in equal times.

What determines whether an orbit is elliptical, parabolic, or hyperbolic?

The total mechanical energy of the orbit. Negative total energy binds the object into an ellipse; zero total energy gives exactly the escape trajectory (a parabola); positive total energy gives an unbound hyperbolic path that escapes with excess speed.

Using T² ∝ a³, if Mars's semi-major axis is 1.52 AU, estimate its orbital period in years.

With T in years and a in AU for solar orbits, T² = a³ = 1.52³ = 3.51, so T = √3.51 ≈ 1.87 years — matching Mars's real orbital period of about 687 days almost exactly.

Why does a spacecraft need more energy to reach the Sun than to leave the Solar System entirely?

Earth already orbits the Sun at about 29.8 km/s tangentially. To "fall into" the Sun, a spacecraft must cancel nearly all of that tangential velocity — an enormous energy cost — whereas escaping the Solar System only requires adding enough energy to reach solar escape velocity, roughly 42 km/s from Earth's orbit, which is comparatively far cheaper than cancelling existing orbital motion.

Concept mind map

How the ideas connect

Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.

Newton gravityKepler's 3 lawsEllipse orbitsEccentricityOrbital energyVis-vivaOrbital Mechanics
Infographic

The key facts, visualised

Kepler 1
Orbits are ellipses with the Sun at one focus
Kepler 2
A line to the Sun sweeps equal areas in equal times
Kepler 3
Period squared is proportional to semi-major axis cubed
Vis-viva
v^2 = GM(2/r - 1/a) links speed, distance, orbit size
Solved examples

Worked problems, step by step

Follow each solution line by line, then try to reproduce it on paper before moving on.

Example 1A comet has semi-major axis a = 17.8 AU. Using Kepler's third law (T^2 = a^3 in years and AU), find its period.

  1. T^2 = a^3 = 17.8^3 = ~5,640
  2. T = sqrt(5,640) = ~75 years

Example 2Use vis-viva to find a comet's speed at perihelion. Given GM_sun, r = 0.6 AU, a = 17.8 AU.

  1. v^2 = GM(2/r - 1/a)
  2. Since r is much smaller than a, 2/r dominates
  3. The small distance r makes v large at perihelion
Practice problem set

Now you try

Work each one out first, then tap to reveal the worked answer.

1State Kepler's first law.
Planets move in ellipses with the Sun at one focus.
2What does Kepler's second law imply about orbital speed?
A body moves faster near the Sun (perihelion) and slower when far (aphelion).
3What does the sign of an orbit's total energy tell you?
Negative energy is a bound ellipse, zero is a parabola (escape), positive is an unbound hyperbola.
4What does eccentricity describe?
How elongated an orbit is: 0 is a circle, closer to 1 is a long ellipse.
5What does the vis-viva equation relate?
Orbital speed to current distance r and semi-major axis a: v^2 = GM(2/r - 1/a).
6How does Kepler's third law connect period and orbit size?
The period squared is proportional to the semi-major axis cubed.