# Kepler's laws of planetary motion

(Redirected from Kepler's third law)

In astronomy, Kepler's laws of planetary motion are three scientific laws describing the motion of planets around the Sun. Figure 1: Illustration of Kepler's three laws with two planetary orbits.
1. The orbits are ellipses, with focal points F1 and F2 for the first planet and F1 and F3 for the second planet. The Sun is placed in focal point F1.
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 ${\textstyle \left({\frac {a_{1}}{a_{2}}}\right)^{\frac {3}{2}}}$ .
1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
3. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Most planetary orbits are nearly circular, and careful observation and calculation are required in order to establish that they are not perfectly circular. Calculations of the orbit of Mars indicated an elliptical orbit. From this, Johannes Kepler inferred that other bodies in the Solar System, including those farther away from the Sun, also have elliptical orbits.

Kepler's work (published between 1609 and 1619) improved the heliocentric theory of Nicolaus Copernicus, explaining how the planets' speeds varied, and using elliptical orbits rather than circular orbits with epicycles.

Isaac Newton showed in 1687 that relationships like Kepler's would apply in the Solar System to a good approximation, as a consequence of his own laws of motion and law of universal gravitation.

## Comparison to Copernicus

Kepler's laws improved the model of Copernicus. If the eccentricities of the planetary orbits are taken as zero, then Kepler basically agreed with Copernicus:

1. The planetary orbit is a circle
2. The Sun is at the center of the orbit
3. The speed of the planet in the orbit is constant

The eccentricities of the orbits of those planets known to Copernicus and Kepler are small, so the foregoing rules give fair approximations of planetary motion, but Kepler's laws fit the observations better than does the model proposed by Copernicus.

Kepler's corrections are not at all obvious:

1. The planetary orbit is not a circle, but an ellipse.
2. The Sun is not at the center but at a focal point of the elliptical orbit.
3. Neither the linear speed nor the angular speed of the planet in the orbit is constant, but the area speed (closely linked historically with the concept of angular momentum) is constant.

The eccentricity of the orbit of the Earth makes the time from the March equinox to the September equinox, around 186 days, unequal to the time from the September equinox to the March equinox, around 179 days. A diameter would cut the orbit into equal parts, but the plane through the Sun parallel to the equator of the Earth cuts the orbit into two parts with areas in a 186 to 179 ratio, so the eccentricity of the orbit of the Earth is approximately

$e\approx {\frac {\pi }{4}}{\frac {186-179}{186+179}}\approx 0.015,$

which is close to the correct value (0.016710219) (see Earth's orbit).

The calculation is correct when perihelion, the date the Earth is closest to the Sun, falls on a solstice. The current perihelion, near January 3, is fairly close to the solstice of December 21 or 22.

## Nomenclature

It took nearly two centuries for the current formulation of Kepler's work to take on its settled form. Voltaire's Eléments de la philosophie de Newton (Elements of Newton's Philosophy) of 1738 was the first publication to use the terminology of "laws". The Biographical Encyclopedia of Astronomers in its article on Kepler (p. 620) states that the terminology of scientific laws for these discoveries was current at least from the time of Joseph de Lalande. It was the exposition of Robert Small, in An account of the astronomical discoveries of Kepler (1814) that made up the set of three laws, by adding in the third. Small also claimed, against the history, that these were empirical laws, based on inductive reasoning.

Further, the current usage of "Kepler's Second Law" is something of a misnomer. Kepler had two versions, related in a qualitative sense: the "distance law" and the "area law". The "area law" is what became the Second Law in the set of three; but Kepler did himself not privilege it in that way.

## History

Johannes Kepler published his first two laws about planetary motion in 1609, having found them by analyzing the astronomical observations of Tycho Brahe. Kepler's third law was published in 1619. Kepler had believed in the Copernican model of the solar system, which called for circular orbits, but he could not reconcile Brahe's highly precise observations with a circular fit to Mars' orbit – Mars coincidentally having the highest eccentricity of all planets except Mercury. His first law reflected this discovery.

Kepler in 1621 and Godefroy Wendelin in 1643 noted that Kepler's third law applies to the four brightest moons of Jupiter.[Nb 1] The second law, in the "area law" form, was contested by Nicolaus Mercator in a book from 1664, but by 1670 his Philosophical Transactions were in its favour. As the century proceeded it became more widely accepted. The reception in Germany changed noticeably between 1688, the year in which Newton's Principia was published and was taken to be basically Copernican, and 1690, by which time work of Gottfried Leibniz on Kepler had been published.

Newton was credited with understanding that the second law is not special to the inverse square law of gravitation, being a consequence just of the radial nature of that law; while the other laws do depend on the inverse square form of the attraction. Carl Runge and Wilhelm Lenz much later identified a symmetry principle in the phase space of planetary motion (the orthogonal group O(4) acting) which accounts for the first and third laws in the case of Newtonian gravitation, as conservation of angular momentum does via rotational symmetry for the second law.

## Formulary

The mathematical model of the kinematics of a planet subject to the laws allows a large range of further calculations.

### First law of Kepler (law of orbits)

The orbit of every planet is an ellipse with the Sun at one of the two foci.

Figure 4: 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 = rmin and for θ = 180°, r = rmax.

Mathematically, an ellipse can be represented by the formula:

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

where $p$  is the semi-latus rectum, ε is the eccentricity of the ellipse, r is the distance from the Sun to the planet, and θ is the angle to the planet's current position from its closest approach, as seen from the Sun. So (rθ) are polar coordinates.

For an ellipse 0 < ε < 1 ; in the limiting case ε = 0, the orbit is a circle with the Sun at the centre (i.e. where there is zero eccentricity).

At θ = 0°, perihelion, the distance is minimum

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

At θ = 90° and at θ = 270° the distance is equal to $p$ .

At θ = 180°, aphelion, the distance is maximum (by definition, aphelion is – invariably – perihelion plus 180°)

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

The semi-major axis a is the arithmetic mean between rmin and rmax:

{\begin{aligned}r_{\max }-a&=a-r_{\min }\\[3pt]a&={\frac {p}{1-\varepsilon ^{2}}}\end{aligned}}

The semi-minor axis b is the geometric mean between rmin and rmax:

{\begin{aligned}{\frac {r_{\max }}{b}}&={\frac {b}{r_{\min }}}\\[3pt]b&={\frac {p}{\sqrt {1-\varepsilon ^{2}}}}\end{aligned}}

The semi-latus rectum p is the harmonic mean between rmin and rmax:

{\begin{aligned}{\frac {1}{r_{\min }}}-{\frac {1}{p}}&={\frac {1}{p}}-{\frac {1}{r_{\max }}}\\[3pt]pa&=r_{\max }r_{\min }=b^{2}\,\end{aligned}}

The eccentricity ε is the coefficient of variation between rmin and rmax:

$\varepsilon ={\frac {r_{\max }-r_{\min }}{r_{\max }+r_{\min }}}.$

The area of the ellipse is

$A=\pi ab\,.$

The special case of a circle is ε = 0, resulting in r = p = rmin = rmax = a = b and A = πr2.

### Second law of Kepler

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

The same (blue) area is swept out in a fixed time period. The green arrow is velocity. The purple arrow directed towards the Sun is the acceleration. The other two purple arrows are acceleration components parallel and perpendicular to the velocity.

The orbital radius and angular velocity of the planet in the elliptical orbit will vary. This is shown in the animation: the planet travels faster when closer to the Sun, then slower when farther from the Sun. Kepler's second law states that the blue sector has constant area.

In a small time $dt\,$  the planet sweeps out a small triangle having base line $r\,$  and height $r\,d\theta$  and area $dA={\frac {1}{2}}\cdot r\cdot r\,d\theta$  and so the constant areal velocity is ${\frac {dA}{dt}}={\frac {1}{2}}r^{2}{\frac {d\theta }{dt}}.$

The area enclosed by the elliptical orbit is $\pi ab.\,$  So the period $P\,$  satisfies

$P\cdot {\frac {1}{2}}r^{2}{\frac {d\theta }{dt}}=\pi ab$

and the mean motion of the planet around the Sun

$n={\frac {2\pi }{P}}$

satisfies

$r^{2}\,d\theta =abn\,dt.$

### Third law of Kepler

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

This captures the relationship between the distance of planets from the Sun, and their orbital periods.

Kepler enunciated in 1619 this third law in a laborious attempt to determine what he viewed as the "music of the spheres" according to precise laws, and express it in terms of musical notation. So it was known as the harmonic law.

Using Newton's Law of gravitation (published 1687), this relation can be found in the case of a circular orbit by setting the centripetal force equal to the gravitational force:

$mr\omega ^{2}=G{\frac {mM}{r^{2}}}$

Then, expressing the angular velocity in terms of the orbital period and then rearranging, we find Kepler's Third Law:

$mr\left({\frac {2\pi }{T}}\right)^{2}=G{\frac {mM}{r^{2}}}\rightarrow T^{2}=\left({\frac {4\pi ^{2}}{GM}}\right)r^{3}\rightarrow T^{2}\propto r^{3}$

A more detailed derivation can be done with general elliptical orbits, instead of circles, as well as orbiting the center of mass, instead of just the large mass. This results in replacing a circular radius, $r$ , with the elliptical semi-major axis, $a$ , as well as replacing the large mass $M$  with $M+m$ . However, with planet masses being so much smaller than the Sun, this correction is often ignored. The full corresponding formula is:

${\frac {a^{3}}{T^{2}}}={\frac {G(M+m)}{4\pi ^{2}}}\approx {\frac {GM}{4\pi ^{2}}}\approx 7.496\cdot 10^{-6}\left({\frac {{\text{AU}}^{3}}{{\text{days}}^{2}}}\right){\text{ is constant}}$

where $M$  is the mass of the Sun, $m$  is the mass of the planet, and $G$  is the gravitational constant, $T$  is the orbital period and $a$  is the elliptical semi-major axis.

The following table shows the data used by Kepler to empirically derive his law:

Data used by Kepler (1618)
Planet Mean distance
to sun (AU)
Period
(days)
${\textstyle {\frac {R^{3}}{T^{2}}}}$  (10-6 AU3/day2)
Mercury 0.389 87.77 7.64
Venus 0.724 224.70 7.52
Earth 1 365.25 7.50
Mars 1.524 686.95 7.50
Jupiter 5.2 4332.62 7.49
Saturn 9.510 10759.2 7.43

Upon finding this pattern Kepler wrote:

"I first believed I was dreaming… But it is absolutely certain and exact that the ratio which exists between the period times of any two planets is precisely the ratio of the 3/2th power of the mean distance."

translated from Harmonies of the World by Kepler (1619)

For comparison, here are modern estimates:

Modern data (Wolfram Alpha Knowledgebase 2018)
Planet Semi-major axis (AU) Period (days) ${\textstyle {\frac {R^{3}}{T^{2}}}}$  (10-6 AU3/day2)
Mercury 0.38710 87.9693 7.496
Venus 0.72333 224.7008 7.496
Earth 1 365.2564 7.496
Mars 1.52366 686.9796 7.495
Jupiter 5.20336 4332.8201 7.504
Saturn 9.53707 10775.599 7.498
Uranus 19.1913 30687.153 7.506
Neptune 30.0690 60190.03 7.504

## Planetary acceleration

Isaac Newton computed in his Philosophiæ Naturalis Principia Mathematica the acceleration of a planet moving according to Kepler's first and second law.

1. The direction of the acceleration is towards the Sun.
2. The magnitude of the acceleration is inversely proportional to the square of the planet's distance from the Sun (the inverse square law).

This implies that the Sun may be the physical cause of the acceleration of planets. However, Newton states in his Principia that he considers forces from a mathematical point of view, not a physical, thereby taking an instrumentalist view. Moreover, he does not assign a cause to gravity.

Newton defined the force acting on a planet to be the product of its mass and the acceleration (see Newton's laws of motion). So:

1. Every planet is attracted towards the Sun.
2. The force acting on a planet is directly proportional to the mass of the planet and is inversely proportional to the square of its distance from the Sun.

The Sun plays an unsymmetrical part, which is unjustified. So he assumed, in Newton's law of universal gravitation:

1. All bodies in the Solar System attract one another.
2. The force between two bodies is in direct proportion to the product of their masses and in inverse proportion to the square of the distance between them.

As the planets have small masses compared to that of the Sun, the orbits conform approximately to Kepler's laws. Newton's model improves upon Kepler's model, and fits actual observations more accurately (see two-body problem).

Below comes the detailed calculation of the acceleration of a planet moving according to Kepler's first and second laws.

### Acceleration vector

From the heliocentric point of view consider the vector to the planet $\mathbf {r} =r{\hat {\mathbf {r} }}$  where $r$  is the distance to the planet and ${\hat {\mathbf {r} }}$  is a unit vector pointing towards the planet.

${\frac {d{\hat {\mathbf {r} }}}{dt}}={\dot {\hat {\mathbf {r} }}}={\dot {\theta }}{\hat {\boldsymbol {\theta }}},\qquad {\frac {d{\hat {\boldsymbol {\theta }}}}{dt}}={\dot {\hat {\boldsymbol {\theta }}}}=-{\dot {\theta }}{\hat {\mathbf {r} }}$

where ${\hat {\boldsymbol {\theta }}}$  is the unit vector whose direction is 90 degrees counterclockwise of ${\hat {\mathbf {r} }}$ , and $\theta$  is the polar angle, and where a dot on top of the variable signifies differentiation with respect to time.

Differentiate the position vector twice to obtain the velocity vector and the acceleration vector:

{\begin{aligned}{\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} }}&=\left({\ddot {r}}{\hat {\mathbf {r} }}+{\dot {r}}{\dot {\hat {\mathbf {r} }}}\right)+\left({\dot {r}}{\dot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\ddot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\dot {\theta }}{\dot {\hat {\boldsymbol {\theta }}}}\right)=\left({\ddot {r}}-r{\dot {\theta }}^{2}\right){\hat {\mathbf {r} }}+\left(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}\right){\hat {\boldsymbol {\theta }}}.\end{aligned}}

So

${\ddot {\mathbf {r} }}=a_{r}{\hat {\boldsymbol {r}}}+a_{\theta }{\hat {\boldsymbol {\theta }}}$

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

and the transversal acceleration is

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

### Inverse square law

Kepler's second law says that

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

is constant.

The transversal acceleration $a_{\theta }$  is zero:

${\frac {d\left(r^{2}{\dot {\theta }}\right)}{dt}}=r\left(2{\dot {r}}{\dot {\theta }}+r{\ddot {\theta }}\right)=ra_{\theta }=0.$

So the acceleration of a planet obeying Kepler's second law is directed towards the Sun.

The radial acceleration $a_{\text{r}}$  is

$a_{\text{r}}={\ddot {r}}-r{\dot {\theta }}^{2}={\ddot {r}}-r\left({\frac {nab}{r^{2}}}\right)^{2}={\ddot {r}}-{\frac {n^{2}a^{2}b^{2}}{r^{3}}}.$

Kepler's first law states that the orbit is described by the equation:

${\frac {p}{r}}=1+\varepsilon \cos(\theta ).$

Differentiating with respect to time

$-{\frac {p{\dot {r}}}{r^{2}}}=-\varepsilon \sin(\theta )\,{\dot {\theta }}$

or

$p{\dot {r}}=nab\,\varepsilon \sin(\theta ).$

Differentiating once more

$p{\ddot {r}}=nab\varepsilon \cos(\theta )\,{\dot {\theta }}=nab\varepsilon \cos(\theta )\,{\frac {nab}{r^{2}}}={\frac {n^{2}a^{2}b^{2}}{r^{2}}}\varepsilon \cos(\theta ).$

The radial acceleration $a_{\text{r}}$  satisfies

$pa_{\text{r}}={\frac {n^{2}a^{2}b^{2}}{r^{2}}}\varepsilon \cos(\theta )-p{\frac {n^{2}a^{2}b^{2}}{r^{3}}}={\frac {n^{2}a^{2}b^{2}}{r^{2}}}\left(\varepsilon \cos(\theta )-{\frac {p}{r}}\right).$

Substituting the equation of the ellipse gives

$pa_{\text{r}}={\frac {n^{2}a^{2}b^{2}}{r^{2}}}\left({\frac {p}{r}}-1-{\frac {p}{r}}\right)=-{\frac {n^{2}a^{2}}{r^{2}}}b^{2}.$

The relation $b^{2}=pa$  gives the simple final result

$a_{\text{r}}=-{\frac {n^{2}a^{3}}{r^{2}}}.$

This means that the acceleration vector $\mathbf {\ddot {r}}$  of any planet obeying Kepler's first and second law satisfies the inverse square law

$\mathbf {\ddot {r}} =-{\frac {\alpha }{r^{2}}}{\hat {\mathbf {r} }}$

where

$\alpha =n^{2}a^{3}\,$

is a constant, and ${\hat {\mathbf {r} }}$  is the unit vector pointing from the Sun towards the planet, and $r\,$  is the distance between the planet and the Sun.

According to Kepler's third law, $\alpha$  has the same value for all the planets. So the inverse square law for planetary accelerations applies throughout the entire Solar System.

The inverse square law is a differential equation. The solutions to this differential equation include the Keplerian motions, as shown, but they also include motions where the orbit is a hyperbola or parabola or a straight line. See Kepler orbit.

### Newton's law of gravitation

By Newton's second law, the gravitational force that acts on the planet is:

$\mathbf {F} =m_{\text{planet}}\mathbf {\ddot {r}} =-m_{\text{planet}}\alpha r^{-2}{\hat {\mathbf {r} }}$

where $m_{\text{planet}}$  is the mass of the planet and $\alpha$  has the same value for all planets in the Solar System. According to Newton's Third Law, the Sun is attracted to the planet by a force of the same magnitude. Since the force is proportional to the mass of the planet, under the symmetric consideration, it should also be proportional to the mass of the Sun, $m_{\text{Sun}}$ . So

$\alpha =Gm_{\text{Sun}}$

where $G$  is the gravitational constant.

The acceleration of solar system body number i is, according to Newton's laws:

$\mathbf {\ddot {r}} _{i}=G\sum _{j\neq i}m_{j}r_{ij}^{-2}{\hat {\mathbf {r} }}_{ij}$

where $m_{j}$  is the mass of body j, $r_{ij}$  is the distance between body i and body j, ${\hat {\mathbf {r} }}_{ij}$  is the unit vector from body i towards body j, and the vector summation is over all bodies in the Solar System, besides i itself.

In the special case where there are only two bodies in the Solar System, Earth and Sun, the acceleration becomes

$\mathbf {\ddot {r}} _{\text{Earth}}=Gm_{\text{Sun}}r_{{\text{Earth}},{\text{Sun}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Earth}},{\text{Sun}}}$

which is the acceleration of the Kepler motion. So this Earth moves around the Sun according to Kepler's laws.

If the two bodies in the Solar System are Moon and Earth the acceleration of the Moon becomes

$\mathbf {\ddot {r}} _{\text{Moon}}=Gm_{\text{Earth}}r_{{\text{Moon}},{\text{Earth}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Moon}},{\text{Earth}}}$

So in this approximation, the Moon moves around the Earth according to Kepler's laws.

In the three-body case the accelerations are

{\begin{aligned}\mathbf {\ddot {r}} _{\text{Sun}}&=Gm_{\text{Earth}}r_{{\text{Sun}},{\text{Earth}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Sun}},{\text{Earth}}}+Gm_{\text{Moon}}r_{{\text{Sun}},{\text{Moon}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Sun}},{\text{Moon}}}\\\mathbf {\ddot {r}} _{\text{Earth}}&=Gm_{\text{Sun}}r_{{\text{Earth}},{\text{Sun}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Earth}},{\text{Sun}}}+Gm_{\text{Moon}}r_{{\text{Earth}},{\text{Moon}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Earth}},{\text{Moon}}}\\\mathbf {\ddot {r}} _{\text{Moon}}&=Gm_{\text{Sun}}r_{{\text{Moon}},{\text{Sun}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Moon}},{\text{Sun}}}+Gm_{\text{Earth}}r_{{\text{Moon}},{\text{Earth}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Moon}},{\text{Earth}}}\end{aligned}}

These accelerations are not those of Kepler orbits, and the three-body problem is complicated. But Keplerian approximation is the basis for perturbation calculations. See Lunar theory.

## Position as a function of time

Kepler used his two first laws to compute the position of a planet as a function of time. His method involves the solution of a transcendental equation called Kepler's equation.

The procedure for calculating the heliocentric polar coordinates (r,θ) of a planet as a function of the time t since perihelion, is the following four steps:

1. Compute the mean anomaly M = nt where n is the mean motion.
$n\cdot P=2\pi$  radians where P is the period.
2. Compute the eccentric anomaly E by solving Kepler's equation:
$M=E-\varepsilon \sin E$
3. Compute the true anomaly θ by the equation:
$(1-\varepsilon )\tan ^{2}{\frac {\theta }{2}}=(1+\varepsilon )\tan ^{2}{\frac {E}{2}}$
4. Compute the heliocentric distance
$r=a(1-\varepsilon \cos E).$

The Cartesian velocity vector can be trivially calculated as $\mathbf {v} ={\frac {\sqrt {\mu a}}{r}}\left\langle -\sin {E},{\sqrt {1-\varepsilon ^{2}}}\cos {E}\right\rangle$ .

The important special case of circular orbit, ε = 0, gives θ = E = M. Because the uniform circular motion was considered to be normal, a deviation from this motion was considered an anomaly.

The proof of this procedure is shown below.

### Mean anomaly, M

Figure 5: 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 periheliont.

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.

$M=nt,\,$

where n is the mean motion.

### Eccentric anomaly, E

When the mean anomaly M is 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.

{\begin{aligned}|zcy|&=|zsx|=|zcx|-|scx|\\{\frac {a^{2}M}{2}}&={\frac {a^{2}E}{2}}-{\frac {a\varepsilon \cdot a\sin E}{2}}\end{aligned}}

Division by a2/2 gives Kepler's equation

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

This equation gives M as a function of E. Determining E for a given M is the inverse problem. Iterative numerical algorithms are commonly used.

Having computed the eccentric anomaly E, the next step is to calculate the true anomaly θ.

But note: Cartesian position coordinates reference the center of ellipse are (a cos Eb sin E)

Reference the Sun (with coordinates (c,0) = (ae,0) ), r = (a cos Eae, b sin E)

True anomaly would be arctan(ry/rx), magnitude of r would be r · r.

### True anomaly, θ

Note from the figure that

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

so that

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

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

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

to get

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

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

{\begin{aligned}\tan ^{2}{\frac {E}{2}}&={\frac {1-\cos E}{1+\cos E}}={\frac {1-{\frac {\varepsilon +\cos \theta }{1+\varepsilon \cos \theta }}}{1+{\frac {\varepsilon +\cos \theta }{1+\varepsilon \cos \theta }}}}\\[8pt]&={\frac {(1+\varepsilon \cos \theta )-(\varepsilon +\cos \theta )}{(1+\varepsilon \cos \theta )+(\varepsilon +\cos \theta )}}={\frac {1-\varepsilon }{1+\varepsilon }}\cdot {\frac {1-\cos \theta }{1+\cos \theta }}={\frac {1-\varepsilon }{1+\varepsilon }}\tan ^{2}{\frac {\theta }{2}}.\end{aligned}}

Multiplying by 1 + ε gives the result

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

This is the third step in the connection between time and position in the orbit.

### Distance, r

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

$r(1+\varepsilon \cos \theta )=a\left(1-\varepsilon ^{2}\right)$

Using the relation above between θ and E the final equation for the distance r is:

$r=a(1-\varepsilon \cos E).$