Kepler's Laws of Planetary Motion
Kepler's Laws of Planetary Motion
Between 1609 and 1619, Johannes Kepler published three laws of planetary motion based on meticulous analysis of Tycho Brahe's astronomical observations. These empirical laws, later derived from Newton's gravitational theory, describe the motion of any body orbiting another under gravity.
Kepler's first law states that every orbit is an ellipse with the central body at one focus. A circle is just a special case of an ellipse with zero eccentricity. Most planetary orbits in our solar system are nearly circular — Earth's eccentricity is only 0.0167 — but comets can have eccentricities close to 1, tracing long, narrow ellipses.
Kepler's second law, the law of equal areas, states that a line from the central body to the orbiting body sweeps out equal areas in equal times. This means a satellite moves faster when closer to the body (near periapsis) and slower when farther away (near apoapsis). This is a direct consequence of conservation of angular momentum.
Kepler's third law establishes the relationship between orbital period and semi-major axis: T squared is proportional to a cubed. For orbits around the same central body, the ratio T squared over a cubed is the same constant for all orbiting objects. This law allows us to determine the semi-major axis of any orbit from its period alone, and vice versa.
These three laws, combined with Newton's mechanics, form the complete foundation for understanding satellite orbits, planetary trajectories, and interplanetary navigation. Every space mission is designed using these relationships.
Every orbit is an ellipse with the central body at one focus.
The orbit's shape is characterized by semi-major axis $a$ and eccentricity $e$ ($0 \le e < 1$ for ellipses).
$$r(\theta) = \frac{a(1-e^2)}{1 + e\cos\theta}$$
A radius vector sweeps out equal areas in equal times:
$$\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$$
where $L$ is the angular momentum. The satellite moves faster at periapsis and slower at apoapsis.
$$T^2 = \frac{4\pi^2}{GM} a^3 = \frac{4\pi^2}{\mu} a^3$$
For all orbits around the same body: $T^2/a^3 = \text{constant}$.
Periapsis ($r_p$): closest approach = $a(1-e)$.
Apoapsis ($r_a$): farthest distance = $a(1+e)$.
Semi-major axis: $a = (r_p + r_a)/2$.
A satellite orbits Earth with periapsis altitude 300 km and apoapsis altitude 1500 km. Find $a$, $e$, and $T$.
Step 1: Convert to orbital radii.
$r_p = 6371 + 300 = 6671$ km, $r_a = 6371 + 1500 = 7871$ km.
Step 2: Semi-major axis.
$$a = \frac{r_p + r_a}{2} = \frac{6671 + 7871}{2} = 7271 \text{ km}$$
Step 3: Eccentricity.
$$e = \frac{r_a - r_p}{r_a + r_p} = \frac{7871 - 6671}{7871 + 6671} = \frac{1200}{14542} = 0.0825$$
Step 4: Period.
$$T = 2\pi\sqrt{\frac{a^3}{\mu}} = 2\pi\sqrt{\frac{(7.271 \times 10^6)^3}{3.986 \times 10^{14}}} = 6168 \text{ s} = 102.8 \text{ min}$$
Earth orbits the Sun with $a = 1$ AU $= 1.496 \times 10^{11}$ m and $T = 365.25$ days. Mars has $a = 1.524$ AU. Find Mars's orbital period.
Step 1: Use $T^2/a^3 = $ constant for both planets.
$$\frac{T_M^2}{a_M^3} = \frac{T_E^2}{a_E^3}$$
Step 2: $T_M = T_E \left(\frac{a_M}{a_E}\right)^{3/2} = 365.25 \times 1.524^{3/2} = 365.25 \times 1.883 = 687.8$ days.
Mars's year is about 687 Earth days — close to the observed value of 687.0 days. ✓
Practice Problems
Show Answer Key
1. $a = (7000+12000)/2 = 9500$ km. $e = (12000-7000)/(12000+7000) = 5000/19000 = 0.263$.
2. $r_p = a(1-e) = 17.83(0.033) = 0.586$ AU. $r_a = a(1+e) = 17.83(1.967) = 35.07$ AU.
3. $T = 1 \text{ yr} \times 5.203^{3/2} = 11.86$ years.
4. A circle (special case of ellipse with zero eccentricity).
5. Fastest — Kepler's second law (equal areas in equal times) requires faster speed at closer distance to sweep equal area.
6. $T = 10800$ s. $a = (\mu T^2/4\pi^2)^{1/3} = (2.953 \times 10^{22})^{1/3} = 3.088 \times 10^7$ m. $a = 30{,}880$ km. (MED orbit)
7. $a = (T/T_E)^{2/3}$ AU $= (88/365.25)^{2/3} = 0.2408^{2/3} = 0.387$ AU. ✓
8. $T^2/a^3 = (5553)^2/(6.779 \times 10^6)^3 = 3.083 \times 10^7/3.114 \times 10^{20} = 9.90 \times 10^{-14}$. $4\pi^2/\mu = 39.48/3.986 \times 10^{14} = 9.90 \times 10^{-14}$. ✓
9. $r_a = a(1+e) = 3r_p = 3a(1-e)$. $1+e = 3-3e \Rightarrow 4e = 2 \Rightarrow e = 0.5$.
10. $a = r_p/(1-e) = 0.5/0.005 = 100$ AU. $r_a = a(1+e) = 100(1.995) = 199.5$ AU.