THE WORLD OF ENGINEER: kepler'S law of planetary motion

Wednesday, 14 December 2011

kepler'S law of planetary motion

Kepler's laws are:

The orbit of every planet is an ellipse with the Sun at one of the two foci.
A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.^[1]
The square of the orbital period of a planet is directly proportional to the cubeof the semi-major axis of its orbit.
First Law

See also: ellipse and orbital eccentricity

Figure 2: 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 one of the two foci."

An ellipse is a particular class of mathematical shapes that resemble a stretched out circle. (See the figure to the right.) Note as well that the Sun is not at the center of the ellipse but is at one of the focal points. The other focal point is marked with a lighter dot but is a point that has no physical significance for the orbit. Ellipses have two focal points neither of which is in the center of the ellipse (except for the one special case of the ellipse being a circle). Circles are a special case of an ellipse that are not stretched out and in which both focal points coincide at the center.

How stretched out that ellipse is from a perfect circle is known as its eccentricity; a parameter that varies from 0 (a simple circle) to 1 (an ellipse that is so stretched out that it is a straight line back and forth between the two focal points). The eccentricities of the planets known to Kepler varies from 0.007 (Venus) to 0.2 (Mercury). (See List of planetary objects in the Solar System for more detail.)

After Kepler, though, bodies with highly eccentric orbits have been identified, among them many comets and asteroids. The dwarf planet Pluto was discovered as late as 1929, the delay mostly due to its small size, far distance, and optical faintness. Heavenly bodies such as comets with parabolic or even hyperbolic orbits are possible under the Newtonian theory and have been observed.^[7]

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 = r_min and for θ = 180°, r = r_max.

Symbolically an ellipse can be represented in polar coordinates as:

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

where (r, θ) are the polar coordinates (from the focus) for the ellipse, p is the semi-latus rectum, and ε is the eccentricity of the ellipse. For a planet orbiting the Sun then r is the distance from the Sun to the planet and θ is the angle with its vertex at the Sun from the location where the planet is closest to the Sun.

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

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

At θ = 90° and at θ = 270°, the distance is $\, p.$

At θ = 180°, aphelion, the distance is maximum

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

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

$\,r_\max - a=a-r_\min$

so

$a=\frac{p}{1-\varepsilon^2}.$

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

$\frac{r_\max} b =\frac b{r_\min}$

so

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

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

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

so

$pa=r_\max r_\min=b^2\,.$

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

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

The area of the ellipse is

$A=\pi a b\,.$

The special case of a circle is ε = 0, resulting in r = p = r_min = r_max = a = b and A = π r².

[edit]Second law

Figure 3: 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. The green arrow represents the planet's velocity, and the purple arrows represents the force on the planet.

"A line joining a planet and the Sun sweeps out equal areas during equal intervals of time."^[1]

In a small time

$dt\,$

the planet sweeps out a small triangle having base line

$r\,$

and height

$r d\theta\,.$

The area of this triangle is

$dA=\tfrac 1 2\cdot r\cdot r d\theta$

and so the constant areal velocity is

$\frac{dA}{dt}=\tfrac{1}{2}r^2 \frac{d\theta}{dt}.$

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. So the planet has to move faster when it is closer to the Sun so that it sweeps equal areas in equal times.

The total area enclosed by the elliptical orbit is

$A=\pi ab\,$ .

Therefore the period

$P\,$

satisfies

$\pi ab=P\cdot \tfrac 12r^2 \dot\theta$

or

$r^2\dot \theta = nab$

where

$\dot\theta=\frac{d\theta}{dt}$

is the angular velocity, (using Newton notation for differentiation), and

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

is the mean motion of the planet around the sun.

[edit]Third law

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

The third law, published by Kepler in 1619 [1] captures the relationship between the distance of planets from the Sun, and their orbital periods. For example, suppose planet A is 4 times as far from the Sun as planet B. Then planet A must traverse 4 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 order to maintain equilibrium with the reduced gravitational centripetal force due to being 4 times further from the Sun. In total it takes 4×2=8 times as long for planet A to travel an orbit, in agreement with the law (8²=4³).

This third law used to be known as the harmonic law,^[8] because Kepler enunciated it 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.^[9]

This third law currently receives additional attention as it can be used to estimate the distance from an exoplanet to its central star, and help to decide if this distance is inside the habitable zone of that star.^[10]

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.

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

So the constant is 1 (sidereal year)²(astronomical unit)⁻³ or 2.97472505×10⁻¹⁹ s²m⁻³. See the actual figures: attributes of major planets.

[edit]Generality

These laws approximately describe the motion of any two bodies in orbit around each other. (The statement in the first law about the focus becomes closer to exactitude as one of the masses becomes closer to zero mass. Where there are more than two masses, all of the statements in the laws become closer to exactitude as all except one of the masses become closer to zero mass and as the perturbationsthen also tend towards zero).^[4] The masses of the two bodies can be nearly equal, e.g. Charon—Pluto (~1:10), in a small proportion, e.g.Moon—Earth (~1:100), or in a great proportion, e.g. Mercury—Sun (~1:10,000,000).

In all cases of two-body motion, rotation is about the barycenter of the two bodies, with neither one having its center of mass exactly at one focus of an ellipse. However, both orbits are ellipses with one focus at the barycenter. When the ratio of masses is large, the barycenter may be deep within the larger object, close to its center of mass. In such a case it may require sophisticated precision measurements to detect the separation of the barycenter from the center of mass of the larger object. But in the case of the planets orbiting the Sun, the largest of them are in mass as much as 1/1047.3486 (Jupiter) and 1/3497.898 (Saturn) of the solar mass,^[11] and so it has long been known that thesolar system barycenter can sometimes be outside the body of the Sun, up to about a solar diameter from its center.^[12] Thus Kepler's first law, though not far off as an approximation, does not quite accurately describe the orbits of the planets around the Sun under classical physics.

[edit]Zero eccentricity

Kepler's laws refine the model of Copernicus. If the eccentricity of a planetary orbit is zero, then Kepler's laws state:
1. The planetary orbit is a circle
2. The Sun is in the center
3. The speed of the planet in the orbit is constant
4. The square of the sidereal period is proportionate to the cube of the distance from the Sun.
Actually the eccentricities of the orbits of the six planets known to Copernicus and Kepler are quite small, so this gives excellent approximations to the planetary motions, but Kepler's laws give even better fit to the observations.

Kepler's corrections to the Copernican model 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
3. Neither the linear speed nor the angular speed of the planet in the orbit is constant, but the area speed is constant.
4. The square of the sidereal period is proportionate to the cube of the mean between the maximum and minimum distances from the Sun.
The nonzero 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. The equator cuts the orbit into two parts having areas in the proportion 186 to 179, while a diameter cuts the orbit into equal parts. So the eccentricity of the orbit of the Earth is approximately

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

close to the correct value (0.016710219). (See Earth's orbit). The calculation is correct when the perihelion, the date that the Earth is closest to the Sun, is on a solstice. The current perihelion, near January 4, is fairly close to the solstice on December 21 or 22.

[edit]Relation to Newton's laws

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 in inverse proportion to the square of the distance from the Sun.
This suggests that the Sun may be the physical cause of the acceleration of planets.

Newton defined the force 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 on a planet is in direct proportion to the mass of the planet and in inverse proportion to the square of the distance from the Sun.
Here the Sun plays an unsymmetrical part which is unjustified. So he assumed 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 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 to Kepler's laws approximately. Newton's model improves Kepler's model and gives better fit to the observations. See two-body problem.

A deviation of the motion of a planet from Kepler's laws due to attraction from other planets is called a perturbation.

[edit]Computing position as a function of time

Kepler used his two first 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 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 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, ε = 0, gives simply 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.

[edit]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 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,

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

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.

[edit]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 θfrom 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

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

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.

[edit]True anomaly, θ

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

$\tan\frac \theta2=\sqrt\frac{1+\varepsilon}{1-\varepsilon}\cdot\tan\frac E2.$

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. [2]

[edit]Distance, r

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

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

[edit]Computing the planetary acceleration

In his Principia Mathematica Philosophiae Naturalis, Newton showed that Kepler's laws imply that the acceleration of the planets are directed towards the sun and depend on the distance from the sun by the inverse square law. However, The geometrical method used by Newton to prove the result is quite complicated. The demonstration below is based on calculus.^[13]

[edit]Acceleration vector

See also: Polar coordinate#Vector calculus and Mechanics of planar particle motion

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 the direction $\hat {\mathbf{r}}$ is aunit vector. When the planet moves the direction vector $\hat {\mathbf{r}}$ changes:

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

where $\scriptstyle \hat{\boldsymbol\theta}$ is the unit vector orthogonal to $\scriptstyle \hat{\mathbf{r}}$ and pointing in the direction of rotation, and $\scriptstyle \theta$ is the polar angle, and where a dot on top of the variable signifies differentiation with respect to time.

So differentiating the position vector twice to obtain the velocity and the acceleration vectors:

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

So

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

where the radial acceleration is

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

and the tangential acceleration is

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

[edit]The inverse square law

Kepler's second law implies that the areal velocity $\tfrac 1 2 r^2 \dot \theta$ is a constant of motion. The tangential acceleration $a θ$ is zero by Kepler's second law:

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

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

Kepler's first law implies that the area enclosed by the orbit is $π a b$ , where $a$ is the semi-major axis and $b$ is the semi-minor axis of the ellipse. Therefore the period $P$ satisfies $\pi ab=\tfrac 1 2 r^2\dot \theta P$ or

$r^2\dot \theta = nab$

where

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

is the mean motion of the planet around the sun.

The radial acceleration $a r$ is

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

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

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

The radial acceleration $a r$ satisfies

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

Substituting the equation of the ellipse gives

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

The relation $b 2 = p a$ gives the simple final result

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

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

$\alpha = n^2 a^3=\frac{4\pi^2 a^3}{P^2}\,$

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, $α$ 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 includes 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.

[edit]Newton's law of gravitation

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

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

where $α$ only depends on the property of the Sun. According to Newton's third Law, the Sun is also attracted by the planet with a force of the same magnitude. Now that 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. So the form of the gravitational force should be

$\mathbf{F} = - \frac{GMm}{r^2}\hat{\mathbf{r}}$

where $G$ is a universal constant. This is Newton's law of universal gravitation.

The acceleration of solar system body no i is, according to Newton's laws:

$\mathbf{\ddot r_i} = G\sum_{j\ne i} \frac{m_j}{r_{ij}^2}\hat{\mathbf{r}}_{ij}$

where $m j$ is the mass of body no j, and $r i j$ is the distance between body i and body j, and $\hat{\mathbf{r}}_{ij}$ is the unit vector from body i pointing towards body j, and the vector summation is over all bodies in the world, besides no i itself. In the special case where there are only two bodies in the world, Planet and Sun, the acceleration becomes

$\mathbf{\ddot r}_{Planet} = G\frac{m_{Sun}}{r_{{Planet},{Sun}}^2}\hat{\mathbf{r}}_{{Planet},{Sun}}$

which is the acceleration of the Kepler motion.

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