orbital mechanics – Elite Math Program

Why do planets move faster near the Sun? Why do they sweep out equal areas in equal times? And why does a planet farther from its star take longer to complete an orbit? These questions connect geometry, derivatives, Newton’s laws, and a remarkably simple relationship between orbital size and orbital period.

Model and notation. We use Newtonian gravity and first treat a planet whose mass is negligible compared with its star’s mass M. Write G for the gravitational constant and μ = GM for the gravitational parameter. For two bodies of comparable mass, replace GM with G(M + m) in the relative-orbit formulas. We assume an ideal isolated two-body system unless stated otherwise.

1. Three laws, one mathematical story

First law: A planet follows an ellipse with the Sun at one focus. Second law: The line from the Sun to the planet sweeps out equal areas in equal times. Third law: For planets orbiting the same star, the square of the orbital period is proportional to the cube of the semimajor axis.

Figure 1. The Sun lies at a focus of the elliptical orbit, not at its center. The semimajor axis is a and the semiminor axis is b.

Let the focal distance be c. The eccentricity is e = c/a, with 0 ≤ e < 1 for an ellipse. The familiar circle is the special case e = 0. We will derive all three laws from a central inverse-square force.

2. Newton’s inverse-square force and the orbit equation

The acceleration of a planet at position vector r relative to the star is directed toward the star:

d^{2} 𝐫 / d t^{2} = - \frac{G M}{r^{3}} 𝐫

Here r without boldface is the distance to the star. In plane polar coordinates, the radial and transverse components of acceleration are

a_{r} = r^{″} - r {θ^{'}}^{2}, a_{θ} = r θ^{″} + 2 r^{'} θ^{'}

Primes in this section mean derivatives with respect to time. Since gravity has no transverse component,

d / dt (r^{2} θ^{'}) = 0

Therefore h = r²θ′ is constant. To derive the orbit shape, set u(θ) = 1/r. The radial equation becomes the Binet equation:

\frac{d^{2}}{d θ^{2}} u + u = \frac{μ}{h^{2}}

To see this, note that r′ = −h u_θ and r″ = −h²u²u_θθ, while rθ′² = h²u³. Substituting into the radial equation and dividing by −h²u² gives the displayed equation. Its general solution is

u (θ) = \frac{μ}{h^{2}} [1 + e \cos (θ - θ_{0})]

Choose θ = 0 in the perihelion direction, so θ₀ = 0. Inverting gives the polar conic equation:

r = \frac{p}{1 + e \cos θ}

p = \frac{h^{2}}{μ}

For 0 ≤ e < 1, this is an ellipse. Thus Newton’s inverse-square force yields Kepler’s First Law for bound noncircular and circular orbits. The same equation also describes parabolic or hyperbolic paths when e ≥ 1.

3. Kepler’s Second Law: why equal areas take equal times

Let v = dr/dt. Because gravitational acceleration is parallel to r,

\begin{aligned} \frac{d}{d t} (r \times v) & = \frac{d r}{d t} \times v + r \times \frac{d v}{d t} \\ = v \times v + r \times a \\ = 0 . \end{aligned}

The vector r × v is constant. Its magnitude h equals r² dθ/dt, the specific angular momentum. A narrow sector of angle dθ has area dA ≈ ½r²dθ. Taking the limit,

\frac{dA}{dt} = \frac{1}{2} r^{2} \frac{dθ}{dt} = \frac{h}{2}

Therefore the area swept out per unit time is constant. Near perihelion, r is small and the angular speed dθ/dt = h/r² is large; near aphelion, the angular speed is smaller. This is Kepler’s Second Law.

Figure 2. Near the Sun, the planet moves faster. The colored sectors are schematic representations of equal swept areas and remain inside the elliptical orbit.

4. Kepler’s Third Law from the area of an ellipse

An ellipse with semiaxes a and b has area πab. Since the radius vector sweeps out this entire area in one period T and dA/dt = h/2,

T = \frac{2 π a b}{h}

The ellipse geometry gives b² = a²(1 − e²). The polar conic parameter p satisfies p = a(1 − e²) = b²/a. From Section 2, h² = μp, so

h^{2} = μ \frac{b^{2}}{a}

Squaring the period formula and substituting yields

T^{2} = \frac{4 π^{2} a^{3}}{μ}

This is Kepler’s Third Law in its Newtonian form. In a two-body system with planet mass m not negligible, the more general relation is

T^{2} = \frac{4 π^{2} a^{3}}{G (M + m)}

Figure 3. The semiaxes and focal distance form a right triangle: a² = b² + c². Hence b² = a²(1 − e²).

5. Worked example: the orbital period of Mars

Use a new example: Mars has semimajor axis approximately 1.524 AU. Earth’s semimajor axis is approximately 1 AU, and its orbital period is approximately one year. Since both orbit the Sun, Kepler’s Third Law gives

\frac{{T_{Mars}}^{2}}{{T_{Earth}}^{2}} = {\frac{a_{Mars}}{a_{Earth}}}^{3}

T_{Mars} = {1.524}^{\frac{3}{2}} years \approx 1.881 years

Multiplying by 365.25 days per year gives approximately 687 days, consistent with Mars’s orbital period. Notice that the eccentricity does not appear: the semimajor axis, not the semiminor axis, controls the period in the ideal two-body model.

6. Why a geostationary satellite stays over one point

A geostationary satellite moves in a circular, equatorial, prograde orbit whose period equals Earth’s sidereal rotation period, approximately 86,164 seconds—not the 86,400 seconds of a mean solar day.

For a circular orbit of radius r, gravity supplies centripetal acceleration:

\frac{G M}{r^{2}} = \frac{v^{2}}{r}

Because v = 2πr/T, solving for r gives

r = {\frac{G M T^{2}}{4 π^{2}}}^{\frac{1}{3}}

Using Earth’s gravitational parameter GM ≈ 3.986004418 × 10¹⁴ m³/s² and equatorial radius approximately 6,378 km gives an orbital radius of about 42,164 km and an altitude of about 35,786 km above the equator. A satellite with the same period but an inclined or eccentric orbit is geosynchronous, not necessarily geostationary.

Figure 4. A geostationary satellite travels in a circular equatorial orbit, rotating with Earth. The orbit is schematic and not to scale.

7. Beyond Kepler: orbital energy and the limits of the model

The specific mechanical energy (energy per unit planet mass) is

ε = \frac{v^{2}}{2} - \frac{μ}{r}

For a bound Kepler ellipse, a standard consequence of the orbit equation and conservation laws is

ε = - \frac{μ}{2 a}

Equating these expressions gives the vis-viva equation:

v^{2} = μ (\frac{2}{r} - \frac{1}{a})

At perihelion r = a(1 − e), and at aphelion r = a(1 + e). Using h = rv at these turning points, where velocity is tangential,

\frac{v_{peri}}{v_{aph}} = \frac{1 + e}{1 - e}

Real planetary orbits experience perturbations from other planets, nonspherical gravity fields, and relativistic corrections. Kepler’s laws remain an extraordinarily accurate first approximation, not an exact description of every real orbit.

Figure 5. Two ellipses with equal semimajor axes but different eccentricities have the same ideal Keplerian period around the same central mass. They are drawn with a common center for shape comparison; their orbital foci differ.

Discovery problems

Try these before opening the solutions. Each problem extends one of the article’s central ideas.

Problem 1. How much faster at perihelion?

For an elliptical orbit of eccentricity e = 0.30, find the ratio of perihelion speed to aphelion speed. Explain your result without calculating either speed separately.

Problem 2. A planet around a heavier star

A planet has semimajor axis 3 AU and orbits a star of mass twice the Sun’s mass. Neglect the planet’s mass. Find its period in Earth years.

Problem 3. Escape velocity

Starting from conservation of mechanical energy, derive the minimum escape speed from distance R from a spherical planet of mass M. Estimate Earth’s surface escape speed using GM ≈ 3.986 × 10¹⁴ m³/s² and R ≈ 6.371 × 10⁶ m. Ignore the atmosphere and rotation.

Problem 4. Different shapes, same period

Two satellites orbit the same planet in ideal elliptical orbits with the same semimajor axis a but eccentricities e₁ = 0.1 and e₂ = 0.7. Prove that their periods agree, and explain why their speeds need not agree at corresponding locations.

Solutions (open only after trying the problems)

Solution to Problem 1

Angular momentum per unit mass is conserved. At perihelion and aphelion the velocity is perpendicular to the radius, so r_pv_p = r_av_a. Since r_p = a(1 − e) and r_a = a(1 + e),

\frac{v_{p}}{v_{a}} = \frac{1 + 0.30}{1 - 0.30} = \frac{13}{7} \approx 1.857

The planet moves about 1.86 times as fast at perihelion.

Solution to Problem 2

Relative to Earth’s orbit, the semimajor axis is three times as large and the central mass is twice as large. Therefore

\frac{T^{2}}{{1 year}^{2}} = \frac{3^{3}}{2}

Hence T = √(27/2) years ≈ 3.67 years.

Solution to Problem 3

At the threshold of escape, the spacecraft arrives infinitely far away with zero residual speed, so its specific mechanical energy is zero:

\frac{{v_{esc}}^{2}}{2} - \frac{G M}{R} = 0

v_{esc} = {\frac{2 G M}{R}}^{\frac{1}{2}}

Substituting Earth’s values gives approximately 11.2 km/s. This neglects drag and Earth’s rotation.

Solution to Problem 4

Kepler’s Third Law gives T² = 4π²a³/(GM) for both satellites. Their shared a and M imply identical periods. But the vis-viva equation v² = GM(2/r − 1/a) shows that speed depends on the instantaneous distance r. The more eccentric orbit has a greater variation in r and therefore a greater variation in speed.

Further connections

Kepler’s laws are an especially useful example of how parametric curves and geometry lead to physical predictions. For another application of parametric descriptions, explore Bézier Curves: How Four Points Create Beautiful Shapes.

Further reading: Newton, Philosophiæ Naturalis Principia Mathematica (1687); standard introductory treatments of Newtonian gravitation, central forces, and conic sections. Numerical constants above are rounded for illustration.

Newton’s laws explain not only planetary orbits but also the motion of projectiles near Earth’s surface. For another application of vector calculus, see Projectile Motion in Three Dimensions .

Calculus can also reveal surprising behavior in everyday mechanical systems. Explore another example in The Sliding Ladder: Does It Slide or Jump? .

Tag: orbital mechanics

Kepler’s Laws: How Calculus Explains Planetary Motion