Kepler’s Laws

The three laws of Kepler can be stated as follows: 

1. Law of orbits : All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse (Fig. 1a).

This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

Fig. 1: An ellipse traced out by a planet around the sun. The closest point is P and the farthest point is A, P is called the perihelion and A the aphelion. The semimajor axis is half the distance AP.

Select two points F1 and F2. Take a length of a string and fix its ends at F1 and F2 by pins. With the tip of a pencil stretch the string taut and then draw a curve by moving the pencil keeping the string taut throughout.(Fig. 8.1(b)) The closed curve you get is called an ellipse.

Fig.2: Drawing an ellipse. A string has its ends fixed at F1 and F2. The tip of a pencil holds the string taut and is moved around.

Clearly for any point T on the ellipse, the sum of the distances from F1 and F2 is a constant. F1, F2 are called the focii. Join the points F1 and F2 and extend the line to intersect the ellipse at points P and A as shown in Fig. 2. The midpoint of the line PA is the centre of the ellipse O and the length PO = AO is called the semi-major axis of the ellipse. For a circle, the two focii merge into one and the semi-major axis becomes the radius of the circle.

Fig. 3: The planet P moves around the sun in an elliptical orbit. The shaded area is the area ∆A swept out in a small interval of time ∆t.

2. Law of areas : The line that joins any planet to the sun sweeps equal areas in equal intervals of time (Fig. 2). 

This law comes from the observations that planets appear to move slower when they are farther from the sun than when they are nearer.

3. Law of periods : The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

Table 1 gives the approximate time periods of revolution of eight* planets around the sun along with values of their semi-major axes.

Table 1: Data from measurement of planetary motions given below confirm Kepler’s Law of Periods (a ≡ Semi-major axis in units of $10^{10}$ m. T ≡ Time period of revolution of the planet in years (y). Q ≡ The quotient ($T^2/a^3$) in units of $10^{-34} y^2 m^{-3}$.)

The law of areas can be understood as a consequence of conservation of angular momentum whch is valid for any central force. A central force is such that the force on the planet is along the vector joining the Sun and the planet. Let the Sun be at the origin and let the position and momentum of the planet be denoted by r and p respectively. Then the area swept out by the planet of mass m in time interval ∆t is (Fig. 3) ∆A given by

$\Delta \mathbf{A}=\frac{1}{2}(\mathbf{r} \times \mathbf{v}\Delta t)$     (1)

Hence

$\frac{\Delta \mathbf{A}}{\Delta t}=\frac{1}{2}\frac{(\mathbf{r} \times \mathbf{p})}{m}$   (since v = p/m)

$\frac{\Delta \mathbf{A}}{\Delta t}=\frac{\mathbf{L}}{2m}$       (2)

where v is the velocity, L is the angular momentum equal to (r × p). For a central force, which is directed along r, L is a constant as the planet goes around. Hence, ∆A/∆t is a constant according to the last equation. This is the law of areas. Gravitation is a central force and hence the law of areas follows.

Example 1:

Let the speed of the planet at the perihelion P in Fig. 1 be $v_P$ and the Sun-planet distance SP be $r_P$. Relate {$r_P$, $v_P$} to the corresponding quantities at the aphelion {$r_A$, $v_A$}. Will the planet take equal times to traverse BAC and CPB?

Answer 

The magnitude of the angular momentum at P is $L_p = m_pr_pv_p$, since inspection tells us that $\mathbf{r_p}$ and $\mathbf{v_p}$ are mutually perpendicular. Similarly, $L_A = m_pr_Av_A$. From angular momentum conservation

$m_pr_pv_p=m_pr_Av_A$

or $\frac{v_p}{v_A}=\frac{r_A}{r_p}$

Since $r_A$ > $r_p$, $v_p$ > $v_A$.

The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in Fig. 1. From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse BAC than CPB.