IB MYP 4-5 Physics- Kepler’s laws of planetary motion- Study Notes - New Syllabus
IB MYP 4-5 Physics-Kepler’s laws of planetary motion- Study Notes
Key Concepts
- Kepler’s laws of planetary motion
Kepler’s Laws of Planetary Motion
Kepler’s Laws of Planetary Motion
Introduction:
- Johannes Kepler (1571–1630) formulated three fundamental laws describing the motion of planets around the Sun based on Tycho Brahe’s observational data.
- These laws apply to planets, satellites, and other celestial objects orbiting a massive body under gravity.
- Kepler’s work helped replace the geocentric (Earth-centered) model with the heliocentric (Sun-centered) model.
Kepler’s First Law: Law of Ellipses
Statement: The orbit of a planet around the Sun is an ellipse with the Sun at one of the foci.
Key points:
- An ellipse has two foci; in planetary orbits, the Sun occupies one focus.
- The shape of an ellipse is described by its eccentricity \( e \), where \( 0 \le e < 1 \). A circle is a special case with \( e = 0 \).
Mathematical form (for an ellipse with semi-major axis \( a \) and semi-minor axis \( b \)):
\( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \)
Implications:
- Planet-Sun distance varies; closest point is perihelion, farthest point is aphelion.
- Orbital speed is not constant; faster near perihelion and slower near aphelion.
Kepler’s Second Law: Law of Equal Areas
Statement: A line joining a planet and the Sun sweeps out equal areas in equal times.
Implications:
- Planet moves faster when closer to the Sun and slower when farther away.
Angular momentum of the planet is conserved (no external torque acts on it). Mathematically:
\( m r^2 \dfrac{d\theta}{dt} = \text{constant} \)
Area swept \( \Delta A \) in a small time interval \( \Delta t \) is:
\( \dfrac{dA}{dt} = \dfrac{1}{2} r^2 \dfrac{d\theta}{dt} = \text{constant} \)
Kepler’s Third Law: Law of Harmonies
Statement: The square of the orbital period \( T \) of a planet is proportional to the cube of the semi-major axis \( a \) of its orbit:
\( T^2 \propto a^3 \)
Mathematical form:
\( \dfrac{T_1^2}{a_1^3} = \dfrac{T_2^2}{a_2^3} \) for any two planets orbiting the same star.
Implications:
- Allows calculation of orbital period if the average distance from the Sun is known.
- Shows that outer planets move more slowly and have longer periods than inner planets.
Connection with Newton’s Law of Gravitation:
Using \( F = \dfrac{G M m}{r^2} \) and centripetal force \( F = \dfrac{mv^2}{r} \), we can derive \( T^2 \propto a^3 \).
Additional Notes:
- Kepler’s laws are valid for any object orbiting under an inverse-square central force, not just planets.
- They laid the foundation for Newton’s law of universal gravitation and classical orbital mechanics.
- They explain why planets have elliptical orbits rather than perfect circles and why orbital speeds vary.
Example:
A planet moves around the Sun in an elliptical orbit. Its closest distance to the Sun (perihelion) is $1.0 \times 10^{11}$ m, and its farthest distance (aphelion) is $1.5\times 10^{11}$ m. Find the semi-major axis and eccentricity of the orbit.
▶️ Answer/Explanation
Step 1: Semi-major axis \( a \) is the average of perihelion \( r_\text{p} \) and aphelion \( r_\text{a} \):
\( a = \dfrac{r_\text{p} + r_\text{a}}{2} = \dfrac{1.0 \times 10^{11} + 1.5 \times 10^{11}}{2} = 1.25 \times 10^{11} \text{ m} \)
Step 2: Eccentricity \( e \) is given by:
\( e = \dfrac{r_\text{a} – r_\text{p}}{r_\text{a} + r_\text{p}} = \dfrac{1.5 \times 10^{11} – 1.0 \times 10^{11}}{1.5 \times 10^{11} + 1.0 \times 10^{11}} = 0.2 \)
\(\boxed{a = 1.25 \times 10^{11} \text{ m}, \ e = 0.2}\)
Example:
A planet sweeps out an area of $5 \times 10^{16}$m in 30 days. If the area swept is always proportional to time, how much area will it sweep in 45 days?
▶️ Answer/Explanation
Step 1: Using Kepler’s second law, area swept is proportional to time:
\( \dfrac{A_1}{t_1} = \dfrac{A_2}{t_2} \)
Step 2: Substitute values:
\( \dfrac{5 \times 10^{16}}{30} = \dfrac{A_2}{45} \)
\( A_2 = \dfrac{5 \times 10^{16} \times 45}{30} = 7.5 \times 10^{16} \text{ m²} \)
\(\boxed{A_2 = 7.5 \times 10^{16} \text{ m²}}\)
Example:
Planet X orbits a star at an average distance of $2.0 \times 10^{11}$ m. Planet Y orbits the same star at $8.0 \times 10^{11}$m. If Planet X takes 1 Earth year to complete one orbit, calculate the orbital period of Planet Y using Kepler’s third law.
▶️ Answer/Explanation
Step 1: Kepler’s third law: \( \dfrac{T_1^2}{a_1^3} = \dfrac{T_2^2}{a_2^3} \)
Step 2: Substitute known values (\( T_1 = 1 \text{ yr}, a_1 = 2 \times 10^{11} \text{ m}, a_2 = 8 \times 10^{11} \text{ m} \)):
\( \dfrac{1^2}{(2 \times 10^{11})^3} = \dfrac{T_2^2}{(8 \times 10^{11})^3} \)
\( T_2^2 = \dfrac{(8 \times 10^{11})^3}{(2 \times 10^{11})^3} = 4^3 = 64 \)
\( T_2 = \sqrt{64} = 8 \text{ years} \)
\(\boxed{T_2 = 8 \text{ years}}\)