Section 2.6: Kepler’s Laws

Kepler’s Laws describe the motion of planets around the Sun. They are fundamental to understanding orbital mechanics.

Kepler’s First Law (Law of Ellipses):

Planets move in elliptical orbits with the Sun at one focus.

Kepler’s Second Law (Law of Equal Areas):

A line joining a planet and the Sun sweeps out equal areas in equal times.

Kepler’s Third Law (Harmonic Law):

\( \frac{T^2}{r^3} = \text{constant} \), where \( T \) is orbital period and \( r \) is semi-major axis.

Example 1

A planet orbits a star at radius 2×10^11 m. Its orbital period is 1.5×10^8 s. Compute the constant for this system using Kepler’s Third Law.

\( \frac{T^2}{r^3} = \frac{(1.5\times10^8)^2}{(2\times10^{11})^3} \approx 2.81\times10^{-8} \text{ s²/m³} \)

Compute the orbital period of a satellite at 4×10^7 m from the Earth’s center using the same Kepler constant as Earth.

\( T = \sqrt{k r^3} = \sqrt{2.97\times10^{-19} * (4\times10^7)^3} \approx 2.0\times10^4 \text{ s} \)

Verify Kepler’s Third Law for Mars (radius 2.28×10^11 m, period 5.94×10^7 s).
Find the semi-major axis of a satellite with orbital period 1.2×10^5 s around Earth.
A planet’s orbital period is 8×10^7 s. Determine its orbital radius if the Sun’s constant is 2.97×10^-19 s²/m³.
Calculate the time to sweep out 1/12 of the area for a circular orbit of radius 1.5×10^11 m.
Compare the speed of a planet at perihelion and aphelion.
For a comet with elliptical orbit, determine the ratio of velocities at closest and farthest points.
Compute orbital speed of a satellite in low Earth orbit (radius 6.78×10^6 m).
Determine orbital period of a geostationary satellite (radius 4.23×10^7 m).
Plot areas swept in equal intervals for a planet in elliptical orbit.
Find the orbital radius for a satellite with period 24 hours around a planet of mass 5×10^24 kg.