Kepler's laws of planetary motion
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Sir Isaac Newton's later discovery of the laws of motion and universal gravitation depended strongly on Kepler's work. Although from the modern point of view, Kepler's laws can be seen as a consequence of Newton's laws, historically, it was the other way around: Kepler provided a kinematic mathematical model of the empirical observations, which Newton then interpreted using calculus and his new physics.
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[edit] Kepler's first law
The revolution of every planet is an ellipse with a star at one focus.
There is no object at the other focus of a planet's orbit. The semimajor axis, a, is half the major axis of the ellipse. In some sense it can be regarded as the average distance between the planet and its star, but it is not the time average in a strict sense, as more time is spent near apocentre than near pericentre.
Strictly speaking the paths of both objects form ellipses around their common center of mass. As the mass of the star is so much greater than that of the planet, the centre of mass of the system is very close to that of the star and the orbit of the star will be very small.
[edit] Kepler's second law
A line joining a planet and its star sweeps out equal areas during equal intervals of time. This is also known as the law of equal areas.
Suppose a planet takes one day to travel from points A to B. During this time, an imaginary line, from the Sun to the planet, will form out a roughly triangular area. This same amount of area will be formed every day regardless of where in its orbit the planet is.
As a planet travels in its elliptical orbit, its distance from the Sun will vary. As an equal area is swept during any period of time and since the distance from a planet to its orbiting star varies, one can conclude that in order for the area being swept to remain constant, a planet must vary in speed. The physical meaning of the law is that the planet moves faster when it is closer to the sun. This is because the sun's gravity accelerates the planet as it falls toward the sun, and decelerates it on the way back out.
[edit] Kepler's third law
The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits.
- <math>T^2 \propto a^3</math>
- <math>T</math> = orbital period of planet
- <math>a</math> = semimajor axis of orbit
So the expression T2a–3 has the same value for all planets in the solar system as it has for Earth.
That value is (with T in seconds, a in meters) <math>3.00\times 10^{-19} \frac{s^{2}}{m^{3}} \pm \ 0.7%\, </math>.
The general form for this equation is as follows:
- <math>T^2 = {4 \pi^2 a^3 \over G (M+m)}. </math>
- <math>T</math> = orbital period of planet
- <math>a</math> = semimajor axis of orbit
- <math>G</math> = gravitational constant
- <math>M</math> = mass of star
- <math>m</math> = mass of planet
Thus, not only does the length of the orbit increase with distance, the orbital speed decreases, so that the increase of the sidereal period is more than proportional.
See the actual figures: attributes of major planets.
This law is also known as the harmonic law.
[edit] Connection to Newton's laws and conservation laws
The second law can also be seen as a statement of conservation of angular momentum, which is a logical consequence of Newton's laws in the special case of a force that acts along the line connecting two objects.
[edit] Kepler's first law
Newton said that "every object in the universe attracts every other object along a line of the centres of the objects, proportional to each object's mass, and inversely proportional to the square of the distance between the objects."
The following assumes that acceleration a is of the form
- <math> \mathbf{a} = \frac{d^2\mathbf{r}}{dt^2} = f(r)\hat{\mathbf{r}}.</math>
Recall that in polar coordinates
- <math>\frac{d\mathbf{r}}{dt} =\dot r\hat{\mathbf{r}} + r\dot\theta\hat{\boldsymbol\theta},</math>
- <math>\frac{d^2\mathbf{r}}{dt^2} = (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}.</math>
In component form, we have
- <math>\ddot r - r\dot\theta^2 = f(r),</math>
- <math>r\ddot\theta + 2\dot r\dot\theta = 0.</math>
Substituting for <math>\ddot \theta</math> and <math>\dot r</math> in the second equation, we have
- <math>r { d \dot\theta \over dt } + 2 {dr \over dt} \dot\theta = 0,</math>
which simplifies to
- <math>\frac{d\dot\theta}{\dot\theta} = -2\frac{dr}{r}.</math>
When integrated, this yields
- <math>\log\dot\theta = -2\log r + \log\ell,</math>
- <math> \log\ell = \log r^2 + \log\dot\theta, </math>
- <math>\ell = r^2\dot\theta,</math>
for some constant <math>\ell</math>, which can be shown to be the specific angular momentum. Now we substitute. Let
- <math>r = \frac{1}{u},</math>
- <math>\dot r = -\frac{1}{u^2}\dot u = -\frac{1}{u^2}\frac{d\theta}{dt}\frac{du}{d\theta}= -\ell\frac{du}{d\theta},</math>
- <math>\ddot r = -\ell\frac{d}{dt}\frac{du}{d\theta} = -\ell\dot\theta\frac{d^2u}{d\theta^2}= -\ell^2u^2\frac{d^2u}{d\theta^2}.</math>
The equation of motion in the <math>\hat{\mathbf{r}}</math> direction becomes
- <math>\frac{d^2u}{d\theta^2} + u = - \frac{1}{\ell^2u^2}f\left(\frac{1}{u}\right).</math>
Newton's law of gravitation relates the force per unit mass to the radial distance as
- <math> f \left( {1 \over u} \right) = f(r)= - \, { GM \over r^2 } = - GM u^2 </math>
where G is the constant of universal gravitation and M is the mass of the star.
As a result,
- <math>\frac{d^2u}{d\theta^2} + u = \frac{GM}{\ell^2}.</math>
This differential equation has the general solution:
- <math>u = \frac{GM}{\ell^2} \left[ 1 + e\cos(\theta-\theta_0) \right] .</math>
for arbitrary constants of integration e and θ0.
Replacing u with 1/r and letting θ0 = 0:
- <math>r = { 1 \over u } = \frac{ \ell^2 / GM }{ 1+ e\cos\theta}. </math>
This is indeed the equation of a conic section with eccentricity e and the origin at one focus. Thus, Kepler's first law is a direct result of Newton's law of gravitation and Newton's second law of motion.
[edit] Kepler's second law
Assuming Newton's laws of motion, we can show that Kepler's second law is consistent. By definition, the angular momentum <math>\mathbf{L}</math> of a point mass with mass <math>m</math> and velocity <math>\mathbf{v}</math> is :
- <math>\mathbf{L} \ \stackrel{\mathrm{def}}{=}\ \mathbf{r} \times \mathbf{p} = \mathbf{r} \times ( m \mathbf{v} )</math>.
where <math>\mathbf{r}</math> is the position vector of the particle and <math>\mathbf{p} = m \mathbf{v} </math> is the momentum of the particle.
By definition,
- <math>\mathbf{v} = \frac{d\mathbf{r}}{dt} </math>.
As a result, we have
- <math>\mathbf{L} = \mathbf{r} \times m\frac{d\mathbf{r}}{dt}</math>.
Taking the derivative of both sides with respect to time, we have
- <math>\frac{d\mathbf{L}}{dt} = (\mathbf{r} \times \mathbf{F}) + \left( \frac{d\mathbf{r}}{dt} \times m\frac{d\mathbf{r}}{dt} \right)
= ( \mathbf{r} \times \mathbf{F} ) + ( \mathbf{v} \times \mathbf{p} ) = 0 </math>
since the cross product of parallel vectors is 0. Note that F is always parallel to r, since the force is entirely radial, and p is always parallel to v by definition. Therefore, we can now say that <math>|\mathbf{L}|</math> is constant.
The area swept out by the line joining the planet and the sun is half the area of the parallelogram formed by <math>\mathbf{r}</math> and <math>d\mathbf{r}</math>.
- <math>dA = \begin{matrix}\frac{1}{2}\end{matrix} |\mathbf{r} \times d\mathbf{r}| = \begin{matrix}\frac{1}{2}\end{matrix} \left|\mathbf{r} \times \frac{d\mathbf{r}}{dt}dt\right| = \frac{\mathbf{|L|}}{2m}dt</math>
Since <math>|\mathbf{L}|</math> is constant, the area swept out by the planet in a time interval is also a constant.
[edit] Kepler's third law
Newton used the third law as one of the pieces of evidence used to build the conceptual and mathematical framework supporting his law of gravity. If we take Newton's laws of motion as given, and consider a hypothetical planet that happens to be in a perfectly circular orbit of radius r, then we have <math>F=mv^2/r</math> for the sun's force on the planet. The velocity is proportional to r/T, which by Kepler's third law varies as one over the square root of r. Substituting this into the equation for the force, we find that the gravitational force is proportional to one over r squared. (Newton's actual historical chain of reasoning is not known with certainty, because in his writing he tended to erase any traces of how he had reached his conclusions.[specify] ) Reversing the direction of reasoning, we can consider this as a proof of Kepler's third law based on Newton's law of gravity, and taking care of the proportionality factors that were neglected in the argument above, we have:
- <math>T^2 = \frac{4\pi^2}{GM} \cdot r^3</math>
where:
- T = planet's sidereal period
- r = radius of the planet's circular orbit
- G = the gravitational constant
- M = mass of the sun
The same arguments can be applied to any object orbiting any other object. This discussion implicitly assumed that the planet orbits around the stationary sun, although in reality both the planet and the sun revolve around their common center of mass. Newton recognized this, and modified this third law, noting that the period is also affected by the orbiting body's mass. However typically the central body is so much more massive that the orbiting body's mass may be ignored. Newton also proved that in the case of an elliptical orbit, the semimajor axis could be substituted for the radius. The most general result is:
- <math>T^2 = \frac{4\pi^2}{G(M + m)} \cdot a^3</math>
where:
- T = object's sidereal period
- a = object's semimajor axis
- G = 6.67 × 10−11 N • m²/kg2 = the gravitational constant
- M = mass of one object
- m = mass of the other object
For objects orbiting the sun, it can be convenient to use units of years, AU, and solar masses, so that G, 4π2 and the various conversion factors cancel out. Also with m<<M we can set m+M = M, so we have simply <math>T^2=a^3</math>. Note that the values of G and planetary masses are not known with good accuracy; however, the products GM (the Keplerian attraction) are known to extremely high precision.
[edit] Proving Kepler's third law
Define point A to be the periapsis, and point B as the apoapsis of the planet when orbiting the sun.
Kepler's second law states that the orbiting body will sweep out equal areas in equal quantities of time. If we now look at a very small periods of time at the moments when the planet is at points A and B, then we can approximate the area swept out as a triangle with an altitude equal to the distance between the planet and the sun, and the base equal to the time times the speed of the planet.
- <math>\begin{matrix}\frac{1}{2}\end{matrix} \cdot(1-\epsilon)a\cdot V_A\,dt= \begin{matrix}\frac{1}{2}\end{matrix} \cdot(1+\epsilon)a\cdot V_B\,dt</math>
- <math>(1-\epsilon)\cdot V_A=(1+\epsilon)\cdot V_B</math>
- <math>V_A=V_B\cdot\frac{1+\epsilon}{1-\epsilon}</math>
Using the law of conservation of energy for the total energy of the planet at points A and B,
- <math>\frac{mV_A^2}{2}-\frac{GmM}{(1-\epsilon)a} =\frac{mV_B^2}{2}-\frac{GmM}{(1+\epsilon)a}</math>
- <math>\frac{V_A^2}{2}-\frac{V_B^2}{2} =\frac{GM}{(1-\epsilon)a}-\frac{GM}{(1+\epsilon)a}</math>
- <math>\frac{V_A^2-V_B^2}{2}=\frac{GM}{a}\cdot \left ( \frac{1}{(1-\epsilon)}-\frac{1}{(1+\epsilon)} \right ) </math>
- <math>\frac{\left ( V_B\cdot\frac{1+\epsilon}{1-\epsilon}\right ) ^2-V_B^2}{2}=\frac{GM}{a}\cdot \left ( \frac{1+\epsilon-1+\epsilon}{(1-\epsilon)(1+\epsilon)} \right ) </math>
- <math>V_B^2 \cdot \left ( \frac{1+\epsilon}{1-\epsilon}\right ) ^2-V_B^2=\frac{2GM}{a}\cdot \left ( \frac{2\epsilon}{(1-\epsilon)(1+\epsilon)} \right ) </math>
- <math>V_B^2 \cdot \left ( \frac{(1+\epsilon)^2-(1-\epsilon)^2}{(1-\epsilon)^2}\right )=\frac{4GM\epsilon}{a\cdot(1-\epsilon)(1+\epsilon)} </math>
- <math>V_B^2 \cdot \left ( \frac{1+2\epsilon+\epsilon^2-1+2\epsilon-\epsilon^2}{(1-\epsilon)^2} \right) =\frac{4GM\epsilon}{a\cdot(1-\epsilon)(1+\epsilon)} </math>
- <math>V_B^2 \cdot 4\epsilon =\frac{4GM\epsilon\cdot (1-\epsilon)^2}{a\cdot(1-\epsilon)(1+\epsilon)} </math>
- <math>V_B =\sqrt{\frac{GM\cdot(1-\epsilon)}{a\cdot(1+\epsilon)}}.</math>
Now that we have <math>V_B</math>, we can find the rate at which the planet is sweeping out area in the ellipse. This rate remains constant, so we can derive it from any point we want, specifically from point B.
- <math>\frac{dA}{dt}=\frac{\frac{1}{2}\cdot(1+\epsilon)a\cdot V_B \,dt}{dt}= \begin{matrix}\frac{1}{2}\end{matrix} \cdot(1+\epsilon)a\cdot V_B </math>
- <math>= \begin{matrix}\frac{1}{2}\end{matrix} \cdot(1+\epsilon)a\cdot \sqrt{\frac{GM\cdot(1-\epsilon)}{a\cdot(1+\epsilon)}} =
\begin{matrix}\frac{1}{2}\end{matrix} \cdot\sqrt{GMa\cdot(1-\epsilon)(1+\epsilon)}</math>
However, the total area of the ellipse is equal to <math>\pi a \sqrt{(1-\epsilon^2)}a</math>. (That's the same as <math>\pi a b</math>, because <math>b=\sqrt{(1-\epsilon^2)}a</math>). The time the planet take out to sweep out the entire area of the ellipse equals the ellipse's area, so,
- <math>T\cdot \frac{dA}{dt}=\pi a \sqrt{(1-\epsilon^2)}a</math>
- <math>T\cdot \begin{matrix}\frac{1}{2}\end{matrix} \cdot\sqrt{GMa\cdot(1-\epsilon)(1+\epsilon)}=\pi \sqrt{(1-\epsilon^2)}a^2</math>
- <math>T=\frac{2\pi \sqrt{(1-\epsilon^2)}a^2}{\sqrt{GMa\cdot(1-\epsilon)(1+\epsilon)}} =\frac{2\pi a^2}{\sqrt{GMa}}=
\frac{2\pi}{\sqrt{GM}}\sqrt{a^3}</math>
- <math>T^2=\frac{4\pi^2}{GM}a^3.</math>
However, if the mass m is not negligible in relation to M, then the planet will orbit the sun with the exact same velocity and position as a very small body orbiting an object of mass <math>M+m</math> (see reduced mass). To integrate that in the above formula, M must be replaced with <math>M+m</math>, to give
- <math>T^2=\frac{4\pi^2}{G(M+m)}a^3.</math>
Q.E.D.
[edit] Solution for the motion as a function of time
The Keplerian problem assumes an orbit with semimajor axis a, semiminor axis b, and eccentricity ε. To convert the laws into predictions, Kepler began by adding the orbit's auxiliary circle (that with the major axis as a diameter) and defined these points:
- c center of auxiliary circle and ellipse
- s sun (at one focus of ellipse); <math>\mbox{length }cs=a\varepsilon</math>
- p the planet
- z perihelion
- x is the projection of the planet to the auxiliary circle; then <math>\mbox{area }sxz=\frac ab\mbox{area }spz</math>
- y is a point on the not true circle such that <math>\mbox{area }cyz=\mbox{area }sxz=\frac ab\mbox{area }spz</math>
and three angles measured from perihelion:
- true anomaly <math>T=\angle zsp</math>, the planet as seen from the sun
- eccentric anomaly <math>E=\angle zcx</math>, x as seen from the centre
- mean anomaly <math>M=\angle zcy</math>, y as seen from the centre
Then
- <math>\mbox{area }cxz=\mbox{area }cxs+\mbox{area }sxz</math><math>=\mbox{area }cxs+\mbox{area }cyz</math>
- <math>\frac{a^2}2E=a\varepsilon\frac a2\sin E+\frac{a^2}2M</math>
giving Kepler's equation
- <math>M=E-\varepsilon\sin E</math>.
To connect E and T, assume <math>r=\mbox{length }sp</math> then
- <math>a\varepsilon+r\cos T=a\cos E</math> and <math>r\sin T=b\sin E</math>
- <math>r=\frac{a\cos E-a\varepsilon}{\cos T}=\frac{b\sin E}{\sin T}</math>
- <math>\tan T=\frac{\sin T}{\cos T}=\frac ba\frac{\sin E}{\cos E-\varepsilon}=\frac{\sqrt{1-\varepsilon^2}\sin E}{\cos E-\varepsilon}</math>
which is ambiguous but useable. A better form follows by some trickery with trigonometric identities:
- <math>\tan\frac T2=\sqrt\frac{1+\varepsilon}{1-\varepsilon}\tan\frac E2</math>
(So far only laws of geometry have been used.)
Note that <math>\mbox{area }spz</math> is the area swept since perihelion; by the second law, that is proportional to time since perihelion. But we defined <math>\mbox{area }spz=\frac ba\mbox{area }cyz=\frac ba\frac{a^2}2M</math> and so M is also proportional to time since perihelion—this is why it was introduced.
We now have a connection between time and position in the orbit. The catch is that Kepler's equation cannot be rearranged to isolate E; going in the time-to-position direction requires an iteration (such as Newton's method) or an approximate expression, such as
- <math>E\approx M+\left(\varepsilon-\frac18\varepsilon^3\right)\sin M+\frac12\varepsilon^2\sin 2M+\frac38\varepsilon^3\sin 3M</math>
via the Lagrange reversion theorem. For the small ε typical of the planets (except Pluto) such series are quite accurate with only a few terms; one could even develop a series computing T directly from M.[1]
[edit] See also
[edit] External links
- Crowell, Benjamin, Conservation Laws, http://www.lightandmatter.com/area1book2.html, an online book that gives a proof of the first law without the use of calculus.ar:قوانين كبلر
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