## Elliptical equation of motion,what is the best inexpensive elliptical trainer india,best treadmills with cushioning,recumbent exercise bike kmart - PDF Books

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Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. An elliptical field has the equation of its boundary $x^2+3y^2=3$ with A at an end of its major axis.A tower stands vertically at A. WLOG, assume we are working on the right hand side of the ellipse, since working on one side translates to working on the other side. If the angles of elevation are to be equal, then the base lengths must be the same, since they share the same height.
The points of intersection between the ellipse, y=x and y=-x are $(\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2})$ and $(\frac{\sqrt{3}}{2},\frac{-\sqrt{3}}{2})$. Using the distance formula, the length of $AC$ turns out to be the same as the length of $OC$. Not the answer you're looking for?Browse other questions tagged conic-sections or ask your own question.
Polar equation of an ellipse given the origin coordinates and major and minor axis lengths? As was explained, and as can be seen in the diagram, the angle we are interested in is the difference between a and v, specifically (a-v). Lets first calculate the average angle the earth travels in its orbit over the course of a single day. An ellipse can range anywhere from a perfect circle to a line, and its eccentricity can range anywhere from 0 (a=b) to 1(b=0). It is known that Bohr model agrees with experimental results and Schrodinger equation in hydrogen atom, as shown on this site.
In Bohr model, the circumference of the orbit is just equal to an integer number of de Broglie wavelength.
Surprisingly, also in Schrodinger equation, both radial and angular wavefunctions satisfy an integer number of de Broglie wavelength, as shown in their nodes ( see this site (p.3).
Different from simple circular Bohr's orbit, elliptical orbit contains movement in "radial" directions. We can separate each momentum at each point into angular (= tangential ) and radial components, which are perpendicular to each other. Numbers (= n ) of de Broglie wavelength in each tangential and radial directions are given by dividing one orbital length in each direction by each de Broglie wavelength (= λ ).
When the circumference of elliptical orbit is n × de Broglie wavelemgth, this integer "n" can be expressed as the sum of angular and radial waves. As shown in Fig.5, when an electron is moving in circular orbit, it moves Only in tangential (= angular ) direction. So in circular orbit, the concept of radial momentum and de Broglie wavelength do NOT exist.
We don't need to think about the radial wave quantization, different from elliptical orbit.
This is a circular orbit, so all these "5" de Broglie waves are in tangential (= angular ) direction. Schrodinger equation also depends on an integer number of de Broglie wavelength in radial and angular directions. Comparing the energies of Bohr model and Schrodinger equation, they are just consistent with each other.
And the principal quantum number "n" is the sum of radial and angular ( wave ) numbers in both models.
In the latter part of this site, there are some examples of this radial amplitude of χ = rR. If you check hydrogen wavefunction based on Schrodinger equation in some websites or textbooks, you will find 1s radial wavefunction like Fig.10.
Radial quantization condition of Eq.2 also means an integer times de Broglie wavelength is contained in one orbit in radial direction. So Sommerfeld quantization conditions of Eq.2 require quantization in both radial and angular directions.
In elliptical orbit, the electron is moving both in angular and radial directions, as shown in Fig.13. Using Pythagorean theorem, we can divide total momentum "p" into radial (= pr ) and tangential (= pφ ) components at each point.

As shown in this section, total number (= n ) of de Broglie waves can be expressed as the sum of angular (= nφ ) and radial (= nr ) de Broglie waves. In Fig.14 orbit, the principal quantum number n = 6, which also expresses total number of de Broglie wavelength. Angular and radial quantum numbers (= wave numbers in each direction ) are "4" and "2", respectively. Fig.15 shows the examples of Schrodinger's wavefunctions including one de Broglie wavelength in radial direction of one orbit.
For example, R32 wavefunction is "3" principal number ( n = 3 ) and "2" angular (= tangential ) momentum ( l = 2 ). These wave's numbers have the same meaning as Bohr-Sommerfeld model, which also has an integer de Broglie waves. In the upper line of Fig.16, the radial one-round orbits are just two de Broglie wavelength.
For example, in R31, the principal quantum number is "3" and the angular momentum (= tangential ) is "1". In Fig.17, the angular moementum of Spherical Harmonics in Schrodinger's hydrogen shows the tangential de Broglie wave's number in one orbit.
As a result, the total number of radial and tangential de Broglie waves means the principal quantum number ( n = energy levels ), in both Bohr-Sommerfeld and Schrodinger's models. Unrealistically, the radial region of all orbitals in Schrodinger's hydrogen must be from 0 to infinity. In the area of r < a1, potential energy V is lower than total energy E, so, tunnel effect has Nothing to do with this 2p orbital. Furthermore, it is impossible for the bound-state electron to go away to infinity ( and then, return ! Here we think about the classial elliptical orbit with angular momentum = 1, like 2p orbital.
In classical orbit, when the electron has angular momentum, it cannot come closer to nucleus than perihelion (= a1 ). As shown in Fig.20, this tangential kinetic energy diverges to infinity, as r is closer to zero.
So, to cancel this divergence of tangential kinetic energy, radial kientic energy must be negative ! Unfortunately, our Nature does NOT depend on this artificial rule, so Schrodinger equation is unreasonable. But it is impossible, because this wavefunction becomes discontinuous at a1 and a2 in this case. As a result, the discontinuous wavefunction of Fig.21 cannot hold in Schrodinger equation for hydrogen. Because "discontinuous" means the gradient (= derivative ) of this wavefunction becomes divergent to infiniy. So sudden U-turn at these points makes this radial wavefunction unrelated to the original equation.
As shown on this site and this site, they replace the radial parts by χ = rR to satisfy de Broglie relation.
Eq.7 shows number of ( radial ) de Broglie wavelength in the whole radial path becomes "nr".
This means Schrodinger's hydrogen also satisfies an integer times de Broglie wavelength like Bohr model ! So, 2p radial wavefunction consists of infinite kinds of de Broglie wavelengths, as shown in Fig.24. Within each small segment (= dr ), we can consider the de Broglie wavelength is a constant value. In this section, we show this quantized azimuthal wavefunction means an integer times ( tangential ) de Broglie wavelength. As a result, the azimuthal wavefunction of Eq.8 also means an integer number of de Broglie wavelength like Sommerfeld model. This section is based on Sommerfeld's original paper in 1916 ( Annalen der Physik [4] 51, 1-167 ). If one electron is moving around one central positive charge, this angular momentum ( = p ) is constant (= law of constant areal velocity).

Here we prove the equation of Eq.21 ( B=0 ) means an ellipse with the nucleus at its focus.
Here we confirm Bohr-Sommerfeld solution of Eq.49 is valid also in the Schrodinger's hydrogen. As shown on this page, the radial quantization number (= nr ) means the number of de Broglie waves included in the radial orbits. Fig.27 shows the energy levels of Schrodinger's wavefunctions just obey Bohr-Sommerfeld quantization rules. The most important difference is that Schrodinger's solutions are always from zero to infinity, which is unreal. Schrodinger's hydrogen eigenfunctions consist of "radial" and angular momentum (= "tangential" ) parts. As you know, Schrodinger's radial parts Always include "radial" momentum, which means there are NO circular orbitals in Schrodinger's hydrogen. Because if the circular orbitals are included in Schrodinger's hydrogen, the radial eigenfunction needs to be constant (= C ), as shown in Fig.28. This is the main reason why Schrodinger's hydrogen includes many unrealistic "S" states which have no angular momentum. As shown on this page, Schrodinger's wavefunctions also satisfy an integer number of de Broglie wavelength. So in s orbital without angular momentum ( L = 0 ), opposite wave phases in one orbit cancel each other due to destructive interference in Schrodinger wavefunction.
Also in multi electron atoms such as sodium and potassium, 3s or 4s electrons always penetrate the inner electrons and nucleus ? On this page, we have proved rigorously that Schrodinger's energies are equal to Bohr Sommerfeld's model. First, as shown on this page, Bohr's single electron does NOT fall into nucleus just by acceleration. The third equation means the circumference of one circular orbit is an integer (= n ) number of de Broglie wavelength.
We suppose the electron travels the distance "dq" during the infinitesimal time interval "dt".
Using Phythagorean theorem, the moving distance "dq" can be separated into radial and angular (= tangential ) directions.
We can explain the reason why de Broglie waves can be separated into angular and radial directins. As you see Ap.16, angular and radial de Broglie wavelengths are longer than the total wavelength. So in this case, "radial" direction also satisfies an integer (= nr ) × de Broglie wavelength. In this section, we solve the Schrodinger equation of hydrogen atom, like on this site and this site.
Total energy E is the sum of kinetic energy and Coulomb potential energy (= V ) both in Schrodinger and Bohr's hydrogens. As shown in this section, Ap.32 means Schrodinger's hydrogen also obeys an integer number of de Broglie wavelength in angular direction. In both of Schrodinger and Bohr's hydrogens, the sum of radial and angular ( wave ) numbers are important for energy levels (= n ). To discover images and hq pictures, type your search terms into our powerful search engine box or browse our different categories. If the bases were not equal, then one of the triangles would have a greater angle of elevation than the other since one would have a point further away, resulting in differing angles of depression. The $y$ co-ordinates are negatives of each other, since they lie on opposite sides of the ellipse.
Since the angle of elevation is $\frac{\pi}{2}$, the triangle $TCA$ (where $T$ is the top of the tower) is a right angled (by definition) isosceles triangle (by computing the remaining angle). Since $\angle BOC$ is defined to be a right angle, the angle between the x-axis and the points $B$, $C$, are $\frac{\pi}{2}$.

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