Mars has a = 1.52 AU. What is its orbital period?

T² = a³ → T = a^(3/2) = 1.52^1.5 ≈ 1.87 years.

At perihelion or aphelion — where is the planet moving fastest?

Kepler's Second Law: faster near the Sun (perihelion).

Prove Kepler's Third Law from Newton's gravity for circular orbits.

You can work it out this way: set F_gravity = F_centripetal: GMm/r² = mv²/r; v = 2πr/T; solve for T².

Pro 🔒~25 min

Kepler's Laws

Explore planetary motion with three fundamental laws

How it works

Kepler's First Law: planets orbit in ellipses with the Sun at one focus. Second Law: a line joining a planet and the Sun sweeps equal areas in equal times — meaning planets move faster near perihelion. Third Law: T² = (4π²/GM)a³, so the orbital period depends only on the semi-major axis. These laws were derived empirically by Kepler and later derived from Newton's law of gravity.

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Step-by-step

Set eccentricity and semi-major axis.
Watch the planet orbit and observe speed variation.
The shaded area sectors should all be equal for equal time intervals (Second Law).
Record period T for different values of a to verify T² ∝ a³.

Key formulas

$T^2 = \frac{4\pi^2}{GM}a^3$ Kepler's Third Law (a = semi-major axis)
$\frac{dA}{dt} = \text{const}$ Kepler's Second Law (equal areas)
$r_{perihelion} = a(1-e),\; r_{aphelion} = a(1+e)$ Orbit extremes

Frequently asked questions

Mars has a = 1.52 AU. What is its orbital period?: T² = a³ → T = a^(3/2) = 1.52^1.5 ≈ 1.87 years.
At perihelion or aphelion — where is the planet moving fastest?: Kepler's Second Law: faster near the Sun (perihelion).
Prove Kepler's Third Law from Newton's gravity for circular orbits.: You can work it out this way: set F_gravity = F_centripetal: GMm/r² = mv²/r; v = 2πr/T; solve for T².