Kepler’s laws

1. From an inverse-square law to planetary motion

Kepler's laws describe orbit shape, the rate at which an orbit is swept out, and the relation between size and period. Newtonian inverse-square attraction explains all three within an isolated two-body model. Real planetary systems have additional perturbations, but they are not needed for these derivations.

Let $k=Gm_1m_2$, $\mu=m_1m_2/(m_1+m_2)$, and $\mathbf r=\mathbf r_1-\mathbf r_2$. The orbital elements in this chapter describe the relative orbit. When one mass is much larger, its motion is small and the relative orbit is approximately the lighter body's orbit about the heavier body.

2. First law: the ellipse and its focus

Statement. A bounded non-collision orbit under inverse-square attraction is an ellipse with the force center at a focus; a circle is the zero-eccentricity case.

The derivation begins with the radial Newton equation and constant angular momentum. With $u=1/r$ they give $u''+u=\mu k/\ell^2$. Solving yields

Energy determines $e^2=1+2E\ell^2/(\mu k^2)$. For bounded regular inverse-square motion, $E<0$, so $0\leq e<1$. The Cartesian completion of the square in Chapter 20 gives an ellipse centered at $(-ae,0)$, with $a=p/(1-e^2)$ and $b=a\sqrt{1-e^2}$. Its focal distance satisfies $c^2=a^2-b^2=a^2e^2$, hence $c=ae$. One focus is therefore at $x=-ae+c=0$, exactly the force center.

This focus result is dynamical. It does not follow from calling the path an ellipse: an isotropic spring also generates ellipses, but centered on the force origin. Under two-body gravity, each body orbits the center of mass on a scaled ellipse, and the center of mass is a focus of each body’s ellipse in that frame.

3. Second law: equal areas in equal times

Statement. The line of relative separation sweeps equal areas in equal time intervals.

For a short angular change $d\theta$, the swept area is $dS=r^2d\theta/2$. One way to see this is to take a triangle with sides $r$ and the small tangential displacement $r\,d\theta$; the radial displacement affects area only at higher order. Thus

The central force exerts zero torque, so $\ell=\mu r^2\dot\theta$ is constant. Therefore

The integral form directly establishes equal areas for finite equal time intervals. It is not merely a statement about infinitesimal sectors. Near periapsis, the radius is smaller, so the angular speed must be larger. Equal time intervals do not correspond to equal arc lengths or equal angle changes.

At the apsides the velocity is tangential, so $\ell=\mu r_pv_p=\mu r_av_a$. Consequently $v_p/v_a=r_a/r_p=(1+e)/(1-e)$. Away from an apsis, angular momentum involves the tangential component of velocity, not the full speed.

This law needs a central force, but does not need the inverse-square magnitude. Its derivation is therefore more general than the first and third laws.

4. Third law: period and semimajor axis

Statement. For relative Kepler ellipses with fixed total gravitating mass, the square of the orbital period is proportional to the cube of the relative semimajor axis.

Write the period as $\tau$ to avoid confusion with kinetic energy $T$. The area of an ellipse is $\pi ab$: scaling a unit disk by factors $a$ and $b$ scales its area by $ab$. Divide the complete swept area by the constant areal rate:

Squaring and substituting $b^2=a^2(1-e^2)$ gives

But $p=a(1-e^2)=\ell^2/(\mu k)$, so $\ell^2=\mu ka(1-e^2)$. Cancel the common eccentricity factor:

For gravity,

and hence

When $m_2\ll m_1$, replace $m_1+m_2$ by $m_1$ to recover the familiar planetary form. Comparing planets as if the proportionality constant were exactly identical neglects their differing masses; the approximation is usually excellent in the intended model.

The result is independent of eccentricity at fixed $a$. More eccentric motion is slower near a more distant apoapsis and faster near a closer periapsis; the area and angular-momentum factors compensate exactly in the full period.

Dimensions check: $G(m_1+m_2)$ has units of length cubed divided by time squared, so $a^3/[G(m_1+m_2)]$ has units of time squared. For a circle $a=r$, combining $v^2=G(m_1+m_2)/r$ with $\tau=2\pi r/v$ reproduces the same expression.

5. Locating the body as a function of time

Deeper insight (optional). The area law can give an explicit time parameter along an ellipse. Introduce eccentric anomaly $\psi$ by $x=a(\cos\psi-e)$, $y=b\sin\psi$, measured relative to the focus at the origin. A direct cross product gives

Thus $dS=ab(1-e\cos\psi)d\psi/2$. Integrating from periapsis, where $\psi=0$, gives $S=ab(\psi-e\sin\psi)/2$. Equating this with $\ell(t-t_p)/(2\mu)$ and using $\tau=2\pi\mu ab/\ell$ yields

This is Kepler's equation. Once $\psi$ is found, the displayed Cartesian formulas give position. The true polar angle $\theta$ is generally not equal to $\psi$; the distinction is why uniform increase of an angular parameter cannot be assumed.

6. Applications

Example 1 · Period ratio

Two light satellites orbit the same dominant mass on ellipses with semimajor axes $a_2=4a_1$. Neglect their masses in the total. Then $\tau_2^2/\tau_1^2=(a_2/a_1)^3=64$, so $\tau_2/\tau_1=8$. If the first period is $3$ hours, the second is $24$ hours. No eccentricity values are needed.

Example 2 · Speeds at periapsis and apoapsis

For an ellipse of eccentricity $e=0.6$, $r_a/r_p=(1.6)/(0.4)=4$. Hence $v_p/v_a=4$. If $v_a=2\,\mathrm{km\,s^{-1}}$, $v_p=8\,\mathrm{km\,s^{-1}}$. The angular-speed ratio is $(r_a/r_p)^2=16$, since angular momentum is proportional to $r^2\dot\theta$. The speed and angular-speed ratios should not be interchanged.

Example 3 · Inferring a binary system's total mass

Suppose the relative orbit has semimajor axis $a=2\,\mathrm{AU}$ and period $\tau=2\,\mathrm{yr}$. Using the idealized normalization $G M_\odot=4\pi^2\,\mathrm{AU^3\,yr^{-2}}$, the third law gives

This is the total mass. To separate the masses, more information is needed, such as the ratio of their center-of-mass orbital semimajor axes. Using one body's semimajor axis instead of the relative $a=a_1+a_2$ would give the wrong total.

Example 4 · Time from swept area

An orbit has period $80$ days. A sector between two positions sweeps one eighth of the full ellipse area in the direction of motion. The elapsed time is $(1/8)\times80=10$ days. If the same two points are connected by the complementary forward arc, the time is $70$ days. Sector area, direction, and the chosen arc specify the result; the central angle alone does not.

7. A connected derivation

The first law follows from the inverse-square orbit equation and its conic geometry. The second follows from torque-free angular momentum and swept area. The third combines the ellipse area from the first with the area rate from the second. Keeping this order makes the eccentricity cancellation in the period law physically meaningful rather than an isolated algebraic trick.

Exercises

Hints and solutions are collected separately.

1. 21.1 Derive the second law from angular momentum.

2. 21.2 Which Kepler law follows from centrality alone?

3. 21.3 Derive the third law for the relative two-body ellipse.

4. 21.4 If semimajor axis increases by a factor of $9$ around the same dominant mass, how does period change?

5. 21.5 For eccentricity $e=1/3$, find the speed ratio at periapsis and apoapsis.

6. 21.6 Two ellipses have the same semimajor axis and central masses but different eccentricities. Compare their periods and energies.

7. 21.7 Why must a binary-star mass calculation use the relative semimajor axis?

8. 21.8 A relative orbit has $a=3\,\mathrm{AU}$ and $\tau=3\,\mathrm{yr}$. Find total mass using the solar normalization.

9. 21.9 Does the line from ellipse center to the body sweep equal areas?

10. 21.10 An orbit has period $120$ days. How long does sweeping $30\%$ of its area take?

11. 21.11 In the optional anomaly parametrization, derive Kepler’s equation.

12. 21.12 Give a connected account of how the three Kepler laws follow from central-force mechanics.