law

Kepler's second law

Kepler's second law states that a planet sweeps out equal areas in equal times as it orbits the Sun.
9 related articles · 1 formulas

Description

A planet moves around the Sun: the closer, the faster. The line connecting the planet and the Sun sweeps out equal areas in equal time intervals. A figure skater pulling in their arms spins faster — the same principle applies here.

History

Johannes Kepler obtained the second law during many years of attempts to describe the orbit of Mars using Tycho Brahe's data. He abandoned circular orbits and discovered that the planet's speed changes, but the area swept out by the Sun-planet line remains constant. The result was published in 1609. Later, Isaac Newton in 'Mathematical Principles of Natural Philosophy' (1687) showed that Kepler's second law is equivalent to the conservation of angular momentum. Emmy Noether in 1918 linked this conservation to the isotropy of space.

How it works

The law is applied in celestial mechanics to compute the position of bodies in orbit at any moment of time, as well as in astrophysics to analyze radial velocity curves of stars with exoplanets. Limits of applicability: in strong gravitational fields, for example near neutron stars, relativistic corrections violate the Keplerian picture; in many-body systems, orbits are perturbed, and the law holds only approximately. When the potential deviates from spherical symmetry (flattened galaxies), the areal velocity is not strictly conserved.

Fun fact

If the orbit were a perfect circle, the planet's speed would remain constant. But on an ellipse, Earth is 3% faster in January than in July.

Concept graph →

Formulas

$$P = 2\pi \sqrt{\frac{a^3}{GM}}$$
Orbital period squared is proportional to semi-major axis cubed.
What is in the formula
Pvariableorbital period — period second
avariablesemi-major axis of the orbit — length metre
Mvariablemass of the central body — mass kilogram
Gconstantconstant of universal gravitation — gravitational constant = 6.67430e-11 cubic metre per kilogram per second squared
sqrtoperatortakes the square root — square root
p\approx2π\sqrt{\frac{r^3}{gm}} · 2605.04631v1
ω_p=\sqrt{\frac{(1+e)gm_\bullet}{r_p^3}} · 2606.02692v1
\frac{a^3}{p^2}=\frac{g(m_s+m_p)}{4π^2} · 2606.23774
p^2=\frac{4π^2a^3}{g(m_{wd}+m_{don})} · 2606.28993

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