Kepler's laws of planetary motion

Illustration of Kepler's three laws with two planetary orbits. (1) The orbits are ellipses, with focal points ƒ₁ and ƒ₂ for the first planet and ƒ₁ and ƒ₃ for the second planet. The sun is placed in focal point ƒ₁. (2) The two shaded sectors A₁ and A₂ have the same surface area and the time for planet 1 to cover segment A₁ is equal to the time to cover segment A₂. (3) The total orbit times for planet 1 and planet 2 have a ratio a₁^3/2 : a₂^3/2.

In astronomy, Kepler's laws of planetary motion are three mathematical laws that describe the motion of planets in the Solar System. German mathematician and astronomer Johannes Kepler (1571–1630) discovered them.

Kepler studied the observations (the Rudolphine tables) of Tycho Brahe. Around 1605, Kepler found that Brahe's observations of the planets' positions followed three relatively simple mathematical laws.

Kepler's laws challenged Aristotelean and Ptolemaic astronomy and physics. His assertion that the Earth moved, his use of ellipses rather than epicycles, and his proof that the planets' speeds varied, changed astronomy and physics. Almost a century later Isaac Newton was able to deduce Kepler's laws from Newton's own laws of motion and his law of universal gravitation, using classical Euclidean geometry.

In modern times, Kepler's laws are used to calculate approximate orbits for artificial satellites, and bodies orbiting the Sun of which Kepler was unaware (such as the outer planets and smaller asteroids). They apply where any relatively small body is orbiting a larger, relatively massive body, though the effects of atmospheric drag (e.g. in a low orbit), relativity (e.g. Perihelion precession of Mercury), and other nearby bodies can make the results insufficiently accurate for a specific purpose.

Introduction to the three laws

First law

Kepler's first law placing the Sun at the focus of an elliptical orbit

"The orbit of every planet is an ellipse with the sun at a focus."

At the time, this was a radical claim; the prevailing belief (particularly in epicycle-based theories) was that orbits should be based on perfect circles. This observation was very significant at the time as it supported the Copernican view of the Universe. This does not mean it loses relevance in a more modern context. Although technically an ellipse is not the same as a circle, most of the planets follow an orbit of low eccentricity, meaning that they can be crudely approximated as circles. So it is not very evident from the orbit of the planets that the orbits are indeed elliptic. However, Kepler's calculations proved they were, which also allowed for other heavenly bodies farther away from the Sun with highly eccentric orbits (like really long stretched out circles). These other heavenly bodies indeed have been identified as the numerous comets or asteroids by astronomers after Kepler's time. The dwarf planet Pluto was discovered as late as 1930, the delay mostly due to its small size and its highly elongated orbit compared to the other planets.

Second law

Illustration of Kepler's second law. The planet moves faster near the Sun, so the same area is swept out in a given time as at larger distances, where the planet moves more slowly.

"A line joining a planet and the sun sweeps out equal areas during equal intervals of time."

This is also known as the law of equal areas. To understand this let us suppose a planet takes one day to travel from point A to point B. The lines from the Sun to points A and B, together with the planet orbit, will define a (roughly triangular) area. This same area will be covered every day regardless of where in its orbit the planet is. Now as the first law states that the planet follows an ellipse, the planet is at different distances from the Sun at different parts in its orbit. This leads to the conclusion that the planet has to move faster when it is closer to the sun so that it sweeps an equal area.

About a century later, Newton showed using his theory of gravitation that this is a direct consequence of conservation of angular momentum. It can be shown that the angular momentum is proportional to the area swept by the orbit of the planet. This can also be understood as the sun's gravity accelerating the planet as it comes closer to the sun, and decelerating it on the way out. However since Kepler predates Newtonian gravity he arrived at his conclusions empirically.

Third law

Planets distant from the sun have longer orbital periods than close planets. Kepler's third law describes this fact quantitatively.

"The square of the orbital period of a planet is directly proportional to the third power of the semi-major axis of its orbit. Moreover, the constant of proportionality has the same value for all planets."

For example, suppose planet A is four times as far from the sun as planet B. Then planet A must traverse four times the distance of Planet B each orbit, and moreover it turns out that planet A travels at half the speed of planet B. In total it takes 4×2=8 times as long for planet A to travel an orbit, in agreement with the law (8²=4³).

Symbolically:

${P^{2}}\propto {a^{3}}$

where $P$ is the orbital period of planet and $a$ is the semimajor axis of the orbit.

The proportionality constant is the same for any planet around the sun.

${\frac {P_{planet}^{2}}{a_{planet}^{3}}}={\frac {P_{earth}^{2}}{a_{earth}^{3}}}.$

So the constant is 1 (sidereal year)²(astronomical unit)⁻³ or 2.97473×10⁻¹⁹ s²m⁻³. See the actual figures: attributes of major planets.

Newton determined that the proportionality constant was related to the mass of the sun according to the following general expression:

$\left({\frac {P}{2\pi }}\right)^{2}={a^{3} \over G(M+m)},$

where P is time per orbit and P/2π is time per radian. $G$ is the gravitational constant, $M$ is the mass of the sun, and $m$ is the mass of planet. Since the mass of a planet is much smaller than the mass of the sun, the mass of the planet can be neglected, so Kepler's third law works (that is the constant of proportionality is independent of which planet we choose to look at) despite the fact that different planets have different masses. For example, the mass of Jupiter is 318 Earth masses, while the mass of the Sun is 332,946 Earth masses (see data tabulated at Planet attributes). Thus, the discrepancy in Kepler's constant calculated neglecting the mass of Jupiter is approximately a tenth of a percent.

This law is also known as the harmonic law.

Position as a function of time

Kepler used these three laws for computing the position of a planet as a function of time. His method involves the solution of a transcendental equation called Kepler's equation.

The derivations in this section use heliocentric polar coordinates, i.e. polar coordinates with the sun as the origin. However, they can alternatively be formulated and derived using Cartesian coordinates.

Mathematics of the ellipse

Main article: Ellipse

Heliocentric coordinate system (r, θ) for ellipse. Also shown are: semi-major axis a, semi-minor axis b and semi-latus rectum p; center of ellipse and its two foci marked by large dots. For θ = 0°, r = r_min and for θ = 180°, r = r_max.

The mathematics of the ellipse is as follows. The equation is

$r={\frac {p}{1+\varepsilon \cdot \cos \theta }}$

where (r, θ) are heliocentrical polar coordinates for the planet (see figure), p is the semi-latus rectum, and ε is the eccentricity.

For θ = 0 the planet is at the perihelion at minimum distance:

$r_{\mathrm {min} }={\frac {p}{1+\varepsilon }}$

for θ = 90°: r = p, and for θ = 180° the planet is at the aphelion at maximum distance:

$r_{\mathrm {max} }={\frac {p}{1-\varepsilon }}.$

The semi-major axis is the arithmetic mean between r_min and r_max:

$a={\frac {r_{min}+r_{max}}{2}}={\frac {p}{1-\varepsilon ^{2}}}.$

The semi-minor axis is the geometric mean between r_min and r_max:

$b={\sqrt {r_{min}r_{max}}}={\frac {p}{\sqrt {1-\varepsilon ^{2}}}}.$

The area of the ellipse is

$A=\pi ab\,.$

The eccentricity of the ellipse is related to the semi-major axis a and semi-minor axis b as shown in the figure by:

$\varepsilon ={\sqrt {1-\left({\frac {b}{a}}\right)^{2}}}\ ,$

which is greater than or equal to zero, and less than one.

Summary

Using these ellipse-related terms, Kepler's procedure for calculating the heliocentric polar coordinates (r,θ) to a planetary position as a function of the time t since perihelion, and the orbital period P, is the following four steps.

1. Compute the mean anomaly M from the formula

$M={\frac {2\pi t}{P}}$

2. Compute the eccentric anomaly E by numerically solving Kepler's equation:

$\ M=E-\varepsilon \cdot \sin E$

3. Compute the true anomaly θ by the equation:

$\tan {\frac {\theta }{2}}={\sqrt {\frac {1+\varepsilon }{1-\varepsilon }}}\cdot \tan {\frac {E}{2}}$

4. Compute the heliocentric distance r from the first law:

$r={\frac {p}{1+\varepsilon \cdot \cos \theta }}$

The important special case of circular orbit, $\varepsilon =0$, gives simply $\theta =E=M$ and $r=p.$

The proof of this procedure is shown below.

Details and proof

Geometric construction for Kepler's calculation of θ. The sun (located at the focus) is labeled S and the planet P. The auxiliary circle is an aid to calculation. Line xd is perpendicular to the base and through the planet P. The shaded sectors are arranged to have equal areas by positioning of point y.

The Keplerian problem assumes an elliptical orbit and the four points:

s the sun (at one focus of ellipse);

z the perihelion

c the center of the ellipse

p the planet

and

$\ a=|cz|,$ distance between center and perihelion, the semimajor axis,

$\ \varepsilon ={|cs| \over a},$ the eccentricity,

$\ b=a{\sqrt {1-\varepsilon ^{2}}},$ the semiminor axis,

$\ r=|sp|,$ the distance between sun and planet.

$\theta =\angle zsp,$ the direction to the planet as seen from the sun, the true anomaly.

The problem is to compute the polar coordinates (r,θ) of the planet from the time since perihelion, t.

It is solved in steps. Kepler considered the circle with the major axis as a diameter, and

$\ x,$ the projection of the planet to the auxiliary circle

$\ y,$ the point on the circle such that the sector areas |zcy| and |zsx| are equal,

$M=\angle zcy,$ the mean anomaly.

The sector areas are related by $|zsp|={\frac {b}{a}}\cdot |zsx|.$

The circular sector area $\ |zcy|={\frac {a^{2}M}{2}}.$

The area swept since perihelion,

$|zsp|={\frac {b}{a}}\cdot |zsx|={\frac {b}{a}}\cdot |zcy|={\frac {b}{a}}\cdot {\frac {a^{2}M}{2}}$ $={\frac {abM}{2}}$ ,

is by Kepler's second law proportional to time since perihelion. So the mean anomaly, M, is proportional to time since perihelion, t.

$M={2\pi t \over P},$

where P is the orbital period.

The mean anomaly M is first computed. The goal is to compute the true anomaly θ. The function θ=f(M) is, however, not elementary. Kepler's solution is to use

$E=\angle zcx$, x as seen from the centre, the eccentric anomaly

as an intermediate variable, and first compute E as a function of M by solving Kepler's equation below, and then compute the true anomaly θ from the eccentric anomaly E. Here are the details.

$\ |zcy|=|zsx|=|zcx|-|scx|$

${\frac {a^{2}M}{2}}={\frac {a^{2}E}{2}}-{\frac {a\varepsilon \cdot a\sin E}{2}}$

Division by a²/2 gives Kepler's equation

$M=E-\varepsilon \cdot \sin E.$

The catch is that Kepler's equation cannot be rearranged to isolate E. The function E = f(M) is not an elementary formula, but Kepler's equation is solved either iteratively by a root-finding algorithm or, as derived in the article on eccentric anomaly, by an infinite series.

Having computed the eccentric anomaly E from Kepler's equation, the next step is to calculate the true anomaly θ from the eccentric anomaly E.

Note from the figure that

${\overrightarrow {cd}}={\overrightarrow {cs}}+{\overrightarrow {sd}}$

so that

$a\cdot \cos E=a\cdot \varepsilon +r\cdot \cos \theta .$

Dividing by $a$ and inserting from Kepler's first law

$\ {\frac {r}{a}}={\frac {1-\varepsilon ^{2}}{1+\varepsilon \cdot \cos \theta }}$

to get

$\cos E=\varepsilon +{\frac {1-\varepsilon ^{2}}{1+\varepsilon \cdot \cos \theta }}\cdot \cos \theta$ $={\frac {\varepsilon \cdot (1+\varepsilon \cdot \cos \theta )+(1-\varepsilon ^{2})\cdot \cos \theta }{1+\varepsilon \cdot \cos \theta }}$ $={\frac {\varepsilon +\cos \theta }{1+\varepsilon \cdot \cos \theta }}.$

The result is a usable relationship between the eccentric anomaly E and the true anomaly θ.

A computationally more convenient form follows by substituting into the trigonometric identity:

$\tan ^{2}{\frac {x}{2}}={\frac {1-\cos x}{1+\cos x}}.$

Get

$\tan ^{2}{\frac {E}{2}}={\frac {1-\cos E}{1+\cos E}}$ $={\frac {1-{\frac {\varepsilon +\cos \theta }{1+\varepsilon \cdot \cos \theta }}}{1+{\frac {\varepsilon +\cos \theta }{1+\varepsilon \cdot \cos \theta }}}}$ $={\frac {(1+\varepsilon \cdot \cos \theta )-(\varepsilon +\cos \theta )}{(1+\varepsilon \cdot \cos \theta )+(\varepsilon +\cos \theta )}}$ $={\frac {1-\varepsilon }{1+\varepsilon }}\cdot {\frac {1-\cos \theta }{1+\cos \theta }}={\frac {1-\varepsilon }{1+\varepsilon }}\cdot \tan ^{2}{\frac {\theta }{2}}.$

Multiplying by (1+ε)/(1−ε) and taking the square root gives the result

$\tan {\frac {\theta }{2}}={\sqrt {\frac {1+\varepsilon }{1-\varepsilon }}}\cdot \tan {\frac {E}{2}}.$

We have now completed the third step in the connection between time and position in the orbit.

One could even develop a series computing θ directly from M. [1]

The fourth step is to compute the heliocentric distance r from the true anomaly θ by Kepler's first law:

$\ r=a\cdot {\frac {1-\varepsilon ^{2}}{1+\varepsilon \cdot \cos \theta }}.$

Derivation from Newton's laws of motion and Newton's law of gravitation

Kepler's laws are concerned with the motion of the planets around the sun. Newton's laws of motion in general are concerned with the motion of objects subject to impressed forces. Newton's law of universal gravitation describes how masses attract each other through the force of gravity. Using the law of gravitation to determine the impressed forces in Newton's laws of motion enables the calculation of planetary orbits, as discussed below.

In the special case where there are only two particles, the motion of the bodies is the exactly soluble two-body problem, of which an approximate example is the motion of a planet around the Sun according to Kepler's laws, as shown below. The trajectory of the lighter particle may also be a parabola or a hyperbola or a straight line.

In the case of a single planet orbiting its sun, Newton's laws imply elliptical motion. The focus of the ellipse is at the center of mass of the sun and the planet (the barycenter), rather than located at the center of the sun itself. The period of the orbit depends a little on the mass of the planet. In the realistic case of many planets, the interaction from other planets modifies the orbit of any one planet. Even in this more complex situation, the language of Kepler's laws applies as the complicated orbits are described as simple Kepler orbits with slowly varying orbital elements. See also Kepler problem in general relativity.

While Kepler's laws are expressed either in geometrical language, or as equations connecting the coordinates of the planet and the time variable with the orbital elements, Newton's second law is a differential equation. So the derivations below involve the art of solving differential equations. Kepler's second law is derived first, as the derivation of the first law depends on the derivation of the second law.

Equations of motion

See also: Mechanics of planar particle motion § Polar coordinates in an inertial frame of reference, and Centrifugal_force (rotating reference frame) § Planetary motion

Assume that the planet is so much lighter than the sun that the acceleration of the sun can be neglected. In other words, the barycenter is approximated as the center of the sun. Introduce the polar coordinate system in the plane of the orbit, with radial coordinate from the sun's center, r and angle from some arbitrary starting direction θ.

Newton's law of gravitation says that "every object in the universe attracts every other object along a line of the centers of the objects, proportional to each object's mass, and inversely proportional to the square of the distance between the objects," and his second law of motion says that "the mass times the acceleration is equal to the force." So the mass of the planet times the acceleration vector of the planet equals the mass of the sun times the mass of the planet, divided by the square of the distance, times minus the radial unit vector ${\hat {\mathbf {r} }}$, times a constant of proportionality. This is written:

$m\cdot {\ddot {\mathbf {r} }}={\frac {M\cdot m}{r^{2}}}\cdot (-{\hat {\mathbf {r} }})\cdot G$

where a dot on top of the variable signifies differentiation with respect to time, and the second dot indicates the second derivative.

${\dot {\hat {\mathbf {r} }}}={\dot {\theta }}{\hat {\boldsymbol {\theta }}}$

where ${\hat {\boldsymbol {\theta }}}$ is the tangential (azimuthal) unit vector orthogonal to ${\hat {\mathbf {r} }}$ and pointing in the direction of rotation, and

${\dot {\hat {\boldsymbol {\theta }}}}=-{\dot {\theta }}{\hat {\mathbf {r} }}.$

So the position vector

$\mathbf {r} =r{\hat {\mathbf {r} }}$

is differentiated twice to give the velocity vector and the acceleration vector

${\dot {\mathbf {r} }}={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\hat {\mathbf {r} }}}={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\theta }}{\hat {\boldsymbol {\theta }}},$

${\ddot {\mathbf {r} }}=({\ddot {r}}{\hat {\mathbf {r} }}+{\dot {r}}{\dot {\hat {\mathbf {r} }}})+({\dot {r}}{\dot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\ddot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\dot {\theta }}{\dot {\hat {\boldsymbol {\theta }}}})=({\ddot {r}}-r{\dot {\theta }}^{2}){\hat {\mathbf {r} }}+(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}){\hat {\boldsymbol {\theta }}}.$

Note that for constant distance, $\ r$, the planet is subject to the centripetal acceleration, $r{\dot {\theta }}^{2}$, and for constant angular speed, ${\dot {\theta }}$, the planet is subject to the Coriolis acceleration, $2{\dot {r}}{\dot {\theta }}$.

Inserting the acceleration vector into Newton's laws, and dividing by m, gives the vector equation of motion

$({\ddot {r}}-r{\dot {\theta }}^{2}){\hat {\mathbf {r} }}+(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}){\hat {\boldsymbol {\theta }}}=-GMr^{-2}{\hat {\mathbf {r} }}$

Equating components, we get the two ordinary differential equations of motion, one for the acceleration in the ${\hat {\mathbf {r} }}$ direction, the radial acceleration

${\ddot {r}}-r{\dot {\theta }}^{2}=-GMr^{-2},$

and one for the acceleration in the ${\hat {\boldsymbol {\theta }}}$ direction, the tangential or azimuthal acceleration:

$r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}=0.$

Deriving Kepler's second law

In order to derive Kepler's second law only the tangential acceleration equation is needed.

The magnitude of the specific angular momentum

$\ell =r^{2}{\dot {\theta }}$

is a constant of motion, even if both the distance $\ r$, and the angular speed ${\dot {\theta }}$, and the tangential velocity $r{\dot {\theta }}$, vary, because

${\frac {d\ell }{dt}}={\frac {d(r^{2}{\dot {\theta }})}{dt}}=r^{2}{\ddot {\theta }}+2r{\dot {r}}{\dot {\theta }}=r(r{\ddot {\theta }}+2{\dot {r}}\theta )=0$

where the expression in the last parentheses vanishes due to the tangential acceleration equation.

The area swept out from time t₁ to time t₂,

$\ \int _{t_{1}}^{t_{2}}{\frac {1}{2}}\cdot base\cdot d(height)=\int _{t_{1}}^{t_{2}}{\frac {1}{2}}\cdot r\cdot r{\dot {\theta }}dt={\frac {1}{2}}\cdot \ell \cdot (t_{2}-t_{1})$

depends only on the duration t₂−t₁. This is Kepler's second law.

Deriving Kepler's first law

In order to derive Kepler's first law define

$\ u=pr^{-1}\,$

where the constant

$p=\ell ^{2}G^{-1}M^{-1}\,$

has the dimension of length. Then

$GMr^{-2}=\ell ^{2}p^{-3}u^{2}$

and

$\ {\dot {\theta }}=\ell r^{-2}=\ell p^{-2}u^{2}.$

Differentiation with respect to time is transformed into differentiation with respect to angle:

$\ {\dot {X}}={\frac {dX}{dt}}={\frac {dX}{d\theta }}\cdot {\frac {d\theta }{dt}}={\frac {dX}{d\theta }}\cdot {\dot {\theta }}={\frac {dX}{d\theta }}\cdot \ell p^{-2}u^{2}.$

Differentiate

$\ r=pu^{-1}$

twice:

${\dot {r}}={\frac {d(pu^{-1})}{d\theta }}\cdot \ell p^{-2}u^{2}=-pu^{-2}{\frac {du}{d\theta }}\cdot \ell p^{-2}u^{2}=-\ell p^{-1}{\frac {du}{d\theta }}$

${\ddot {r}}={\frac {d{\dot {r}}}{d\theta }}\cdot \ell p^{-2}u^{2}={\frac {d}{d\theta }}\left(-\ell p^{-1}{\frac {du}{d\theta }}\right)\cdot \ell p^{-2}u^{2}=-\ell ^{2}p^{-3}u^{2}{\frac {d^{2}u}{d\theta ^{2}}}$

Substitute into the radial equation of motion

${\ddot {r}}-r{\dot {\theta }}^{2}=-GMr^{-2}$

and get

$\left(-\ell ^{2}p^{-3}u^{2}{\frac {d^{2}u}{d\theta ^{2}}}\right)-(pu^{-1})(\ell p^{-2}u^{2})^{2}=-\ell ^{2}p^{-3}u^{2}$

Divide by $-\ell ^{2}p^{-3}u^{2}$ to get a simple non-homogeneous linear differential equation for the orbit of the planet:

${\frac {d^{2}u}{d\theta ^{2}}}+u=1.$

An obvious solution to this equation is the circular orbit

$\ u=1.$

Other solutions are obtained by adding solutions to the homogeneous linear differential equation with constant coefficients

${\frac {d^{2}u}{d\theta ^{2}}}+u=0$

These solutions are

$\ u=\varepsilon \cdot \cos(\theta -\theta _{0})$

where $\ \varepsilon$ and $\theta _{0}\,$ are arbitrary constants of integration. So the result is

$\ u=1+\varepsilon \cdot \cos(\theta -\theta _{0})$

Choosing the axis of the coordinate system such that $\ \theta _{0}=0$, and inserting $\ u=pr^{-1}$, gives:

$\ pr^{-1}=1+\varepsilon \cdot \cos \theta .$

If $\ \varepsilon <1,$ this is Kepler's first law.

Deriving Kepler's third law

In the special case of circular orbits, which are ellipses with zero eccentricity, the relation between the radius a of the orbit and its period P can be derived relatively easily. The centripetal force of circular motion is proportional to a/P², and it is provided by the gravitational force, which is proportional to 1/a². Hence,

$P^{2}\propto a^{3}$

which is Kepler's third law for the special case.

In the general case of elliptical orbits, the derivation is more complicated.

The area of the planetary orbit ellipse is

$A=\pi ab=\pi a(a{\sqrt {1-\varepsilon ^{2}}})=\pi a^{2}{\sqrt {1-\varepsilon ^{2}}}\,.$

The area speed of the radius vector sweeping the orbit area is

${\dot {A}}=\ell /2\,$

where

$\ell ^{2}=pGM=a(1-\varepsilon ^{2})GM\,.$

The period of the orbit is

$P={\frac {A}{\dot {A}}}={\frac {\pi a^{2}{\sqrt {1-\varepsilon ^{2}}}}{\ell /2}}=2\pi {\frac {a^{2}{\sqrt {1-\varepsilon ^{2}}}}{\ell }}\,$

satisfying

$\left({\frac {P}{2\pi }}\right)^{2}=\left({\frac {a^{2}{\sqrt {1-\varepsilon ^{2}}}}{\ell }}\right)^{2}={\frac {a^{4}(1-\varepsilon ^{2})}{\ell ^{2}}}={\frac {a^{4}(1-\varepsilon ^{2})}{a(1-\varepsilon ^{2})GM}}={\frac {a^{3}}{GM}}\,$

implying Kepler's third law

$P^{2}\propto a^{3}\ .$

See also

• Kepler orbit
• Kepler problem
• Circular motion
• Gravity
• Two-body problem
• Free-fall time
• Laplace-Runge-Lenz vector