Kepler's Laws of Planetary Motion

Johannes Kepler mathematically described the motion of planets using empirical data. His three laws apply to any system bound by gravity, including planets orbiting stars or satellites orbiting planets.

Kepler's First Law

Kepler's First Law states that the orbit of each planet is an ellipse with the central body at one of the two foci.

An ellipse is characterized by its semi-major axis $a$ (half the longest diameter) and its eccentricity $e$ (a measure of elongation, where $0 \le e < 1$ ).

Using these parameters, the distances of closest approach (perihelion/perigee) and farthest approach (aphelion/apogee) relative to the central body are given by:

r_{\text{min}} = a(1 - e)

r_{\text{max}} = a(1 + e)

Kepler's Second Law

Kepler's Second Law states that a line segment joining a planet and the central body sweeps out equal areas during equal intervals of time.

This law is a direct physical consequence of the conservation of angular momentum. Because the gravitational force acts radially toward the central body, it exerts zero torque ( $\vec{\tau} = 0$ ). Therefore, total orbital angular momentum ( $\vec{L}$ ) remains constant throughout the orbit:

L = m r v_\perp = \text{constant}

At the points of closest ( $r_{\text{min}}$ ) and farthest ( $r_{\text{max}}$ ) approach, the velocity vector is perpendicular to the radial position vector. This yields a simple relation between speed and distance at these two points:

r_{\text{min}} v_{\text{max}} = r_{\text{max}} v_{\text{min}}

(1)

As a result, an orbiting body moves fastest when closest to the central mass and slowest when farthest away.

Kepler's Third Law

Kepler's Third Law states that the square of the orbital period $T$ of a planet is directly proportional to the cube of the semi-major axis $a$ of its orbit.

Deriving Kepler's Third Law for a Circular Orbit

For a circular orbit where $a = r$ , the gravitational force provides the necessary centripetal acceleration:

\frac{G M_{\text{central}} m}{r^2} = m \frac{v^2}{r}

Solving for orbital velocity $v$ :

v^2 = \frac{G M_{\text{central}}}{r}

Substituting the relationship between velocity, distance, and period ( $v = \frac{2\pi r}{T}$ ):

\left(\frac{2\pi r}{T}\right)^2 = \frac{G M_{\text{central}}}{r}

\frac{4\pi^2 r^2}{T^2} = \frac{G M_{\text{central}}}{r}

Rearranging to solve for $T^2$ and generalizing the radius $r$ to the semi-major axis $a$ yields:

T^2 = \left( \frac{4\pi^2}{G M_{\text{central}}} \right) a^3

(2)

Notice that in Equation 2, the mass of the orbiting body $m$ cancels out completely. The period depends only on the mass of the central attracting body ( $M_{\text{central}}$ ) and the semi-major axis $a$ .

Example 1

Comet Halley has an elliptical orbit around the Sun ( $M_S = 1.99 \cdot 10^{30} \text{ kg}$ ) with a perihelion distance $r_{\text{min}} = 8.75 \cdot 10^{10} \text{ m}$ and an aphelion distance $r_{\text{max}} = 5.26 \cdot 10^{12} \text{ m}$ . If its speed at perihelion is $v_{\text{max}} = 54.6 \text{ km/s}$ , calculate its speed at aphelion ( $v_{\text{min}}$ ).

Hint Click to unblur

Apply Kepler's Second Law using Equation 1 at perihelion and aphelion.

Solution Click to unblur

Using Equation 1:

r_{\text{min}} v_{\text{max}} = r_{\text{max}} v_{\text{min}}

Isolate $v_{\text{min}}$ :

v_{\text{min}} = v_{\text{max}} \left( \frac{r_{\text{min}}}{r_{\text{max}}} \right)

Substitute known values:

v_{\text{min}} = (54,600 \text{ m/s}) \left( \frac{8.75 \cdot 10^{10} \text{ m}}{5.26 \cdot 10^{12} \text{ m}} \right)

v_{\text{min}} \approx 908 \text{ m/s} \quad (0.908 \text{ km/s})

Because of the highly eccentric orbit, Comet Halley moves roughly 60 times slower at aphelion than it does at perihelion.

Example 2

An astronomical body is observed orbiting a distant star with a semi-major axis of $a = 2.50 \cdot 10^{11} \text{ m}$ and an orbital period of $T = 4.10 \cdot 10^7 \text{ s}$ . What is the mass of the central star?

Solution Click to unblur

Start with Kepler's Third Law (Equation 2):

T^2 = \frac{4\pi^2}{G M_{\text{star}}} a^3

Rearrange to isolate the mass of the star $M_{\text{star}}$ :

M_{\text{star}} = \frac{4\pi^2 a^3}{G T^2}

Substitute the known values ( $a = 2.50 \cdot 10^{11} \text{ m}$ , $T = 4.10 \cdot 10^7 \text{ s}$ , $G = 6.674 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}$ ):

M_{\text{star}} = \frac{4\pi^2 (2.50 \cdot 10^{11} \text{ m})^3}{(6.674 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2})(4.10 \cdot 10^7 \text{ s})^2}

M_{\text{star}} \approx 5.50 \cdot 10^{30} \text{ kg}