Kepler's laws of planetary motion

Johannes Kepler's primary contributions to astronomy/astrophysics were his three laws of planetary motion. Kepler, a nearly blind though brilliant German mathematician, derived these laws, in part, by studying the observations of the keen-sighted Danish astronomer Tycho Brahe. The article on Johannes Kepler gives a less mathematical description of the laws, as well as a treatment of their historical and intellectual context.

Connection to Newton's laws and conservation laws

Kepler did not understand why his laws were correct; it was Isaac Newton who discovered the answer to this more than fifty years later. 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.

Related Topics:
Isaac Newton - Angular momentum

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Kepler's first law

Newton proposed 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."

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The following assumes that acceleration is of the form

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:rac{d^2mathbf{r}}{dt^2} = f(r)widehat{mathbf{r}}.

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Recall that in polar coordinates:

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:rac{dmathbf{r}}{dt} = dot rwidehat{mathbf{r}} + rdot hetawidehat{ heta}

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:rac{d^2mathbf{r}}{dt^2} = (ddot r - rdot heta^2)widehat{mathbf{r}} + (rddot heta + 2dot r dot heta)widehat{ heta}

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In component form, we have:

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:ddot r - rdot heta^2 = f(r)

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:rddot heta + 2dot rdot heta = 0

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Since dot r = dr/dt and ddot heta={ddot heta}/{dt}, the latter equation is equivalent to

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:rac{ddot heta}{dot heta} = -2rac{dr}{r}.

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When integrated, this yields

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:logdot heta = -2log r + logell,

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:ell = r^2dot heta,

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for some constant ell, which can be shown to be the specific angular momentum. Now we substitute. Let:

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:r = rac{1}{u}

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:dot r = -rac{1}{u^2}dot u = -rac{1}{u^2}rac{d heta}{dt}rac{du}{d heta}= -ellrac{du}{d heta}

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:ddot r = -ellrac{d}{dt}rac{du}{d heta} = -elldot hetarac{d^2u}{d heta^2}= -ell^2u^2rac{d^2u}{d heta^2}

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The equation of motion in the hat{mathbf{r}} direction becomes:

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:rac{d^2u}{d heta^2} + u = - rac{1}{ell^2u^2}fleft(rac{1}{u} ight)

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If f(r)=-k/r^2, as Newton's law of gravitation claims, then:

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:rac{d^2u}{d heta^2} + u = rac{k}{ell^2}

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where -k is our proportionality constant.

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This differential equation has the general solution:

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:u = Acos( heta- heta_0) + rac{k}{ell^2}.

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Replacing u with r and letting θ0=0:

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:r = rac{1}{Acos heta + rac{k}{ell^2}}.

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This is indeed the equation of a conic section with 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.

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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 mathbf{L} of a point mass with mass m and velocity mathbf{v} is :

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:mathbf{L} equiv m mathbf{r} imes mathbf{v}.

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where mathbf{r} is the position vector of the particle.

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Since mathbf{v} = rac{dmathbf{r}}{dt} , we have:

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:mathbf{L} = mathbf{r} imes mrac{dmathbf{r}}{dt}

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taking the time derivative of both sides:

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:rac{dmathbf{L}}{dt} = mathbf{r} imes mathbf{F} = 0

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since the cross product of parallel vectors is 0. We can now say that |mathbf{L}| is constant.

Related Topics:
Cross product - Vectors

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The area swept out by the line joining the planet and the sun, is half the area of the parallelogram formed by mathbf{r} and dmathbf{r}.

Related Topics:
Planet - Sun - Area - Parallelogram

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:dA = egin{matrix}rac{1}{2}end{matrix} |mathbf{r} imes dmathbf{r}| = egin{matrix}rac{1}{2}end{matrix} left|mathbf{r} imes rac{dmathbf{r}}{dt}dt ight| = rac{mathbf{|L|}}{2m}dt

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Since |mathbf{L}| is constant, the area swept out by is also constant. Q.E.D.

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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 F=mv^2/r for the sun's force on the planet. The velocity is proportional to r/P, 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.) Reversing the direction of reasoning, we can consider this as a proof of Kepler's third law based on Newton's of gravity, and taking care of the proportionality factors that were neglected in the argument above, we have:

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:P^2 = rac{4pi^2}{GM} cdot r^3

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where:

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• P = 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:

Related Topics:
Mass - Semimajor

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:P^2 = rac{4pi^2}{G(M + m)} cdot a^3

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where:

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• P = 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 becomes 4π2, and with m<<M, and we have simply P^2=a^3.

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