Kepler's laws of planetary motion

First law

Kepler's first law states that:

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

Mathematically, an ellipse can be represented by the formula:

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

where ${\displaystyle p}$ 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 (r, θ) 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

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

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

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

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

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

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

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

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

The semi-latus rectum p is the harmonic mean between r_min and r_max:

${\displaystyle {\begin{aligned}p&=\left({\frac {r_{\max }^{-1}+r_{\min }^{-1}}{2}}\right)^{-1}\\pa&=r_{\max }r_{\min }=b^{2}\,\end{aligned}}}$

The eccentricity ε is the coefficient of variation between r_min and r_max:

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

The area of the ellipse is

${\displaystyle A=\pi ab\,.}$

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

Second law

Kepler's second law states that:

A line joining a planet and the Sun sweeps out equal areas during equal intervals of time. ^[23]

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 terms of elliptical parameters

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

The area enclosed by the elliptical orbit is ${\displaystyle \pi ab}$. So the period ${\displaystyle T}$ satisfies

${\displaystyle T\cdot {\frac {r^{2}}{2}}{\frac {d\theta }{dt}}=\pi ab}$

and the mean motion of the planet around the Sun

${\displaystyle n={\frac {2\pi }{T}}}$

satisfies

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

And so, $${\displaystyle {\frac {dA}{dt}}={\frac {abn}{2}}={\frac {\pi ab}{T}}.}$$

Orbits of planets with varying eccentricities.

Low	High
Planet orbiting the Sun in a circular orbit (e=0.0)	Planet orbiting the Sun in an orbit with e=0.5
Planet orbiting the Sun in an orbit with e=0.2	Planet orbiting the Sun in an orbit with e=0.8
The red ray rotates at a constant angular velocity and with the same orbital time period as the planet, ${\displaystyle T=1}$. S: Sun at the primary focus, C: Centre of ellipse, S': The secondary focus. In each case, the area of all sectors depicted is identical.

Third law

Kepler's third law states that:

The ratio of the square of an object's orbital period with the cube of the semi-major axis of its orbit is the same for all objects orbiting the same primary.

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

Kepler enunciated in 1619 ^[16] 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. ^[25] It was therefore known as the harmonic law. ^[26] The original form of this law (referring to not the semi-major axis, but rather a "mean distance") holds true only for planets with small eccentricities near zero. ^[27]

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:

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

Then, expressing the angular velocity ω in terms of the orbital period ${\displaystyle {T}}$ and then rearranging, results in Kepler's Third Law:

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

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, ${\displaystyle r}$, with the semi-major axis, ${\displaystyle a}$, of the elliptical relative motion of one mass relative to the other, as well as replacing the large mass ${\displaystyle M}$ with ${\displaystyle M+m}$. However, with planet masses being so much smaller than the Sun, this correction is often ignored. The full corresponding formula is:

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

where ${\displaystyle M}$ is the mass of the Sun, ${\displaystyle m}$ is the mass of the planet, ${\displaystyle G}$ is the gravitational constant, ${\displaystyle T}$ is the orbital period and ${\displaystyle a}$ is the elliptical semi-major axis, and ${\displaystyle {\text{AU}}}$ is the astronomical unit, the average distance from earth to the sun.

Table

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	5.20	4332.62	7.49
Saturn	9.510	10759.2	7.43

Kepler became aware of John Napier's recent invention of logarithms and log-log graphs before he discovered the pattern. ^[28]

Upon finding this pattern Kepler wrote: ^[29]

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:^{[ citation needed ]}

Modern data

Planet	Semi-major axis (AU)	Period (days)	${\textstyle {\frac {a^{3}}{T^{2}}}}$ (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

Acceleration vector

From the heliocentric point of view consider the vector to the planet ${\displaystyle \mathbf {r} =r{\hat {\mathbf {r} }}}$ where ${\displaystyle r}$ is the distance to the planet and ${\displaystyle {\hat {\mathbf {r} }}}$ is a unit vector pointing towards the planet. $${\displaystyle {\frac {d{\hat {\mathbf {r} }}}{dt}}={\dot {\hat {\mathbf {r} }}}={\dot {\theta }}{\hat {\boldsymbol {\theta }}},\qquad {\frac {d{\hat {\boldsymbol {\theta }}}}{dt}}={\dot {\hat {\boldsymbol {\theta }}}}=-{\dot {\theta }}{\hat {\mathbf {r} }}}$$

where ${\displaystyle {\hat {\boldsymbol {\theta }}}}$ is the unit vector whose direction is 90 degrees counterclockwise of ${\displaystyle {\hat {\mathbf {r} }}}$, and ${\displaystyle \theta }$ 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: $${\displaystyle {\begin{aligned}{\dot {\mathbf {r} }}&={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\hat {\mathbf {r} }}}={\dot {r}}{\hat {\mathbf {r} }}+r{\dot {\theta }}{\hat {\boldsymbol {\theta }}},\\{\ddot {\mathbf {r} }}&=\left({\ddot {r}}{\hat {\mathbf {r} }}+{\dot {r}}{\dot {\hat {\mathbf {r} }}}\right)+\left({\dot {r}}{\dot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\ddot {\theta }}{\hat {\boldsymbol {\theta }}}+r{\dot {\theta }}{\dot {\hat {\boldsymbol {\theta }}}}\right)=\left({\ddot {r}}-r{\dot {\theta }}^{2}\right){\hat {\mathbf {r} }}+\left(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}\right){\hat {\boldsymbol {\theta }}}.\end{aligned}}}$$

So $${\displaystyle {\ddot {\mathbf {r} }}=a_{r}{\hat {\boldsymbol {r}}}+a_{\theta }{\hat {\boldsymbol {\theta }}}}$$ where the radial acceleration is $${\displaystyle a_{r}={\ddot {r}}-r{\dot {\theta }}^{2}}$$ and the transversal acceleration is $${\displaystyle a_{\theta }=r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }}.}$$

Inverse square law

Kepler's second law says that $${\displaystyle r^{2}{\dot {\theta }}=nab}$$ is constant.

The transversal acceleration ${\displaystyle a_{\theta }}$ is zero: $${\displaystyle {\frac {d\left(r^{2}{\dot {\theta }}\right)}{dt}}=r\left(2{\dot {r}}{\dot {\theta }}+r{\ddot {\theta }}\right)=ra_{\theta }=0.}$$

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

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

Kepler's first law states that the orbit is described by the equation: $${\displaystyle {\frac {p}{r}}=1+\varepsilon \cos(\theta ).}$$

Differentiating with respect to time $${\displaystyle -{\frac {p{\dot {r}}}{r^{2}}}=-\varepsilon \sin(\theta )\,{\dot {\theta }}}$$ or $${\displaystyle p{\dot {r}}=nab\,\varepsilon \sin(\theta ).}$$

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

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

Substituting the equation of the ellipse gives $${\displaystyle pa_{\text{r}}={\frac {n^{2}a^{2}b^{2}}{r^{2}}}\left({\frac {p}{r}}-1-{\frac {p}{r}}\right)=-{\frac {n^{2}a^{2}}{r^{2}}}b^{2}.}$$

The relation ${\displaystyle b^{2}=pa}$ gives the simple final result $${\displaystyle a_{\text{r}}=-{\frac {n^{2}a^{3}}{r^{2}}}.}$$

This means that the acceleration vector ${\displaystyle \mathbf {\ddot {r}} }$ of any planet obeying Kepler's first and second law satisfies the inverse square law $${\displaystyle \mathbf {\ddot {r}} =-{\frac {\alpha }{r^{2}}}{\hat {\mathbf {r} }}}$$ where $${\displaystyle \alpha =n^{2}a^{3}}$$ is a constant, and ${\displaystyle {\hat {\mathbf {r} }}}$ is the unit vector pointing from the Sun towards the planet, and ${\displaystyle r\,}$ is the distance between the planet and the Sun.

Since mean motion ${\displaystyle n={\frac {2\pi }{T}}}$ where ${\displaystyle T}$ is the period, according to Kepler's third law, ${\displaystyle \alpha }$ 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: $${\displaystyle \mathbf {F} =m_{\text{planet}}\mathbf {\ddot {r}} =-m_{\text{planet}}\alpha r^{-2}{\hat {\mathbf {r} }}}$$

where ${\displaystyle m_{\text{planet}}}$ is the mass of the planet and ${\displaystyle \alpha }$ 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, ${\displaystyle m_{\text{Sun}}}$. So $${\displaystyle \alpha =Gm_{\text{Sun}}}$$ where ${\displaystyle G}$ is the gravitational constant.

The acceleration of Solar System body number i is, according to Newton's laws: $${\displaystyle \mathbf {\ddot {r}} _{i}=G\sum _{j\neq i}m_{j}r_{ij}^{-2}{\hat {\mathbf {r} }}_{ij}}$$ where ${\displaystyle m_{j}}$ is the mass of body j, ${\displaystyle r_{ij}}$ is the distance between body i and body j, ${\displaystyle {\hat {\mathbf {r} }}_{ij}}$ 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 $${\displaystyle \mathbf {\ddot {r}} _{\text{Earth}}=Gm_{\text{Sun}}r_{{\text{Earth}},{\text{Sun}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Earth}},{\text{Sun}}}}$$ 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 $${\displaystyle \mathbf {\ddot {r}} _{\text{Moon}}=Gm_{\text{Earth}}r_{{\text{Moon}},{\text{Earth}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Moon}},{\text{Earth}}}}$$

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

In the three-body case the accelerations are $${\displaystyle {\begin{aligned}\mathbf {\ddot {r}} _{\text{Sun}}&=Gm_{\text{Earth}}r_{{\text{Sun}},{\text{Earth}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Sun}},{\text{Earth}}}+Gm_{\text{Moon}}r_{{\text{Sun}},{\text{Moon}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Sun}},{\text{Moon}}}\\\mathbf {\ddot {r}} _{\text{Earth}}&=Gm_{\text{Sun}}r_{{\text{Earth}},{\text{Sun}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Earth}},{\text{Sun}}}+Gm_{\text{Moon}}r_{{\text{Earth}},{\text{Moon}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Earth}},{\text{Moon}}}\\\mathbf {\ddot {r}} _{\text{Moon}}&=Gm_{\text{Sun}}r_{{\text{Moon}},{\text{Sun}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Moon}},{\text{Sun}}}+Gm_{\text{Earth}}r_{{\text{Moon}},{\text{Earth}}}^{-2}{\hat {\mathbf {r} }}_{{\text{Moon}},{\text{Earth}}}\end{aligned}}}$$

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.)

Mean anomaly, M

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

• ${\displaystyle a=|cz|,}$ distance between center and perihelion, the semimajor axis,
• ${\displaystyle \varepsilon ={|cs| \over a},}$ the eccentricity,
• ${\displaystyle b=a{\sqrt {1-\varepsilon ^{2}}},}$ the semiminor axis,
• ${\displaystyle r=|sp|,}$ the distance between Sun and planet.
• ${\displaystyle \theta =\angle zsp,}$ 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

• ${\displaystyle x,}$ the projection of the planet to the auxiliary circle
• ${\displaystyle y,}$ the point on the circle such that the sector areas |zcy| and |zsx| are equal,
• ${\displaystyle M=\angle zcy,}$ the mean anomaly.

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

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

The area swept since perihelion, $${\displaystyle |zsp|={\frac {b}{a}}\cdot |zsx|={\frac {b}{a}}\cdot |zcy|={\frac {b}{a}}\cdot {\frac {a^{2}M}{2}}={\frac {abM}{2}},}$$ is by Kepler's second law proportional to time since perihelion. So the mean anomaly, M, is proportional to time since perihelion, t. $${\displaystyle M=nt,}$$ 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. ^[33] Kepler's solution is to use $${\displaystyle E=\angle zcx,}$$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. $${\displaystyle {\begin{aligned}|zcy|&=|zsx|=|zcx|-|scx|\\with|scx|&={\frac {|cs|.|dx|}{2}}\\{\frac {a^{2}M}{2}}&={\frac {a^{2}E}{2}}-{\frac {a\varepsilon \cdot a\sin E}{2}}\end{aligned}}}$$

Division by a²/2 gives Kepler's equation $${\displaystyle M=E-\varepsilon \sin E.}$$

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 with reference to the center of ellipse are (a cos E, b sin E)

With reference to the Sun (with coordinates (c,0) = (ae,0) ), r = (a cos E – ae, b sin E)

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

True anomaly, θ

Note from the figure that $${\displaystyle |cd|=|cs|+|sd|}$$ so that $${\displaystyle a\cos E=a\varepsilon +r\cos \theta .}$$

Dividing by ${\displaystyle a}$ and inserting from Kepler's first law $${\displaystyle {\frac {r}{a}}={\frac {1-\varepsilon ^{2}}{1+\varepsilon \cos \theta }}}$$ to get $${\displaystyle \cos E=\varepsilon +{\frac {1-\varepsilon ^{2}}{1+\varepsilon \cos \theta }}\cos \theta ={\frac {\varepsilon (1+\varepsilon \cos \theta )+\left(1-\varepsilon ^{2}\right)\cos \theta }{1+\varepsilon \cos \theta }}={\frac {\varepsilon +\cos \theta }{1+\varepsilon \cos \theta }}.}$$

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

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

Multiplying by 1 + ε gives the result $${\displaystyle (1-\varepsilon )\tan ^{2}{\frac {\theta }{2}}=(1+\varepsilon )\tan ^{2}{\frac {E}{2}}}$$

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

Using the relation above between θ and E the final equation for the distance r is: $${\displaystyle r=a(1-\varepsilon \cos E).}$$