# Hyperbola mathematical relationship between energy

### Eccentricity (mathematics) - Wikipedia

How to look at a data set and decide what sort of mathematical relationship it shows. This shape is called a rectangular hyperbola — a hyperbola since it has. conics is hyperbola for = 0, but kinetic energy never be imaginary, only be . From the relation between angular velocity and time (see [1],[2]), we have. e > 1 the trajectory is open (hyperbola), and E > 0. Equations 1, 2, and 3, together with the energy integral 7, provide most of relationships necessary to solve.

They are emitted from certain heavy nuclei, such as Polonium, as they strive to a more stable state, with energies in the MeV mega-electron-volt range. They knock electrons out of any atoms near their paths, creating densely ionized paths that can be observed in cloud chambers where they trigger condensation. They exhaust their energy in a few centimeters in air, and in a very short distance in solid materials.

They cause ZnS crystals to give a flash of luminescence if they hit them, so they can be observed and counted in a spinthariscope. Modern instrumentation makes their observation and counting much more convenient, but this is all that was available in the early 's.

Ernest Rutherford and his students noted in that alpha particles passing through very thin gold foils were occasionally scattered through large angles. This is an extraordinary effect, like firing a rifle through a wheat field and having the bullet come back at you. What would be expected were numerous slight deflections by the positive charges distributed through the matter. Electrons were known to be light, and could not produce large deflections, just slight wiggles in the paths which are observed.

To cause large deflections, the positive charge and the mass must be concentrated in very small volumes. Rutherford showed that although atoms have a radius of the order of cm, the mass and positive charge are concentrated within a radius of about cm. If an atom were the size of the earth, then its nucleus would be a few meters in diameter. Only for small b would there be a considerable deflection, and in this case the electrons could be considered distant and diffuse, with the full nuclear charge effective.

The impact parameter b is just the minor axis of the hyperbola verify this by drawing a right trianglewhile a is determined by the total energy. We note that the two branches of a hyperbola correspond to attractive and repulsive orbits.

Knowing a and b, we know the parabola, and can find the angle between the asymptotes, and thus the deflection D.

This is the relation between the impact parameter b and the deflection D. This is how Rutherford determined an upper limit on the size of the nucleus, from the maximum observed deflection of the alpha particles. This is the famous Rutherford scattering differential cross section, proportional to the inverse fourth power of half the angle of deflection.

Rutherford's discovery of the nucleus led soon after to Bohr's atom, and from there to quantum mechanics, revealing our modern view of matter. Rutherford received a peerage and a Nobel Prize, which were richly deserved. Discoveries in mathematics and physics may lead to understanding; discoveries in most other sciences lead only to knowledge.

Hyperbolas and the Dew Bow Light that enters a sphere of water and is reflected from the far side to re-emerge in the direction from which it entered is responsible for the lovely phenomenon of the rainbow. The colors arise near a caustic surface created by the fact that there is an angle of minimum deviation.

Near this limit there are two beams that interfere to produce a maximum of intensity. Since the angle differs slightly due to the variation of the index of refraction with wavelength, bright colors are seen. Red is on the outside of the rainbow, and blue is on the inside. There may be a secondary rainbow outside the primary one, with a reversed order of colors, which corresponds to two internal reflections.

The space between the primary and secondary rainbows is darker than the space outside, since there is no scattered light in this area. Beyond the blue edge of the rainbow supernumerary arcs are often seen, which alternate green and pink.

This is just a brief review of the properties of a variable phenomenon.

For more information, see a reliable source such as R. This holds for raindrops a and b, as well as for dewdrop c. The rainbow is familiar and is often seen, especially on summer afternoons, but the dewbow is less often noticed. It appears when looking westward over a lawn on a misty morning. The dewdrops give a brilliant reflection in the direction of the antisolar point, where your head casts a shadow, so you can recognize the axis of the cone clearly. This is the heiligenschein, a different, cat's eye, effect that is not related to the dewbow.

There may even be a colored glory if there is a mist. The dewbow is seen between the antisolar point and your station, stretching right and left in a curve along the ground. It is the section of the rainbow cone by the earth, and is, therefore, a hyperbola.

The secondary rainbow and supernumerary arcs have not been reported in the dewbow, but they certainly exist under the proper conditions, and are something to look for specially.

Dew occurs when the surface has cooled by radiation below the temperature of the air, and below the dew point at which the air is saturated by water.

### Equations for Elliptical, Parabolic, Hyperbolic Orbits

Upon reaching its destination, the spacecraft must decelerate so that the planet's gravity can capture it into a planetary orbit. To send a spacecraft to an inner planet, such as Venus, the spacecraft is launched and accelerated in the direction opposite of Earth's revolution around the sun i. It should be noted that the spacecraft continues to move in the same direction as Earth, only more slowly.

To reach a planet requires that the spacecraft be inserted into an interplanetary trajectory at the correct time so that the spacecraft arrives at the planet's orbit when the planet will be at the point where the spacecraft will intercept it.

This task is comparable to a quarterback "leading" his receiver so that the football and receiver arrive at the same point at the same time. The interval of time in which a spacecraft must be launched in order to complete its mission is called a launch window. Newton's Laws of Motion and Universal Gravitation Newton's laws of motion describe the relationship between the motion of a particle and the forces acting on it.

The first law states that if no forces are acting, a body at rest will remain at rest, and a body in motion will remain in motion in a straight line. Thus, if no forces are acting, the velocity both magnitude and direction will remain constant.

The second law tells us that if a force is applied there will be a change in velocity, i. This law may be summarized by the equation where F is the force, m is the mass of the particle, and a is the acceleration. The third law states that if body 1 exerts a force on body 2, then body 2 will exert a force of equal strength, but opposite in direction, on body 1.

This law is commonly stated, "for every action there is an equal and opposite reaction". In his law of universal gravitation, Newton states that two particles having masses m1 and m2 and separated by a distance r are attracted to each other with equal and opposite forces directed along the line joining the particles.

The common magnitude F of the two forces is where G is an universal constant, called the constant of gravitation, and has the value 6.

Let's now look at the force that the Earth exerts on an object. If we drop the object, the Earth's gravity will cause it to accelerate toward the center of the Earth. Many of the upcoming computations will be somewhat simplified if we express the product GM as a constant, which for Earth has the value 3. The product GM is often represented by the Greek letter.

For additional useful constants please see the appendix Basic Constants. For a refresher on SI versus U. Uniform Circular Motion In the simple case of free fall, a particle accelerates toward the center of the Earth while moving in a straight line. The velocity of the particle changes in magnitude, but not in direction.

In the case of uniform circular motion a particle moves in a circle with constant speed. The velocity of the particle changes continuously in direction, but not in magnitude. From Newton's laws we see that since the direction of the velocity is changing, there is an acceleration.

This acceleration, called centripetal acceleration is directed inward toward the center of the circle and is given by where v is the speed of the particle and r is the radius of the circle. Thus, a particle undergoing uniform circular motion is under the influence of a force, called centripetal force, whose magnitude is given by The direction of F at any instant must be in the direction of a at the same instant, that is radially inward.

A satellite in orbit is acted on only by the forces of gravity. The inward acceleration which causes the satellite to move in a circular orbit is the gravitational acceleration caused by the body around which the satellite orbits. Therefore, by setting these equations equal to one another we find that, for a circular orbit, Click here for example problem 4. Using the data compiled by his mentor Tycho BraheKepler found the following regularities after years of laborious calculations: These laws can be deduced from Newton's laws of motion and law of universal gravitation.

Indeed, Newton used Kepler's work as basic information in the formulation of his gravitational theory. As Kepler pointed out, all planets move in elliptical orbits, however, we can learn much about planetary motion by considering the special case of circular orbits. We shall neglect the forces between planets, considering only a planet's interaction with the sun.

## Proof of the hyperbola foci formula

These considerations apply equally well to the motion of a satellite about a planet. Let's examine the case of two bodies of masses M and m moving in circular orbits under the influence of each other's gravitational attraction. The large body of mass M moves in an orbit of constant radius R and the small body of mass m in an orbit of constant radius r, both having the same angular velocity.

For this to happen, the gravitational force acting on each body must provide the necessary centripetal acceleration.

Since these gravitational forces are a simple action-reaction pair, the centripetal forces must be equal but opposite in direction.

That is, m 2r must equal M 2R. The specific requirement, then, is that the gravitational force acting on either body must equal the centripetal force needed to keep it moving in its circular orbit, that is If one body has a much greater mass than the other, as is the case of the sun and a planet or the Earth and a satellite, its distance from the center of mass is much smaller than that of the other body.

If we assume that m is negligible compared to M, then R is negligible compared to r. This is a basic equation of planetary and satellite motion. So, that's one and that's the other asymptote. It looks something like that. It's going to intersect at a comma 0, right there. This is going to be a comma 0. And the intersect at minus a comma 0. We saw this in the previous video.

And then the focus points are going to sit out here someplace. And the focal length this a squared plus b square. That's just this distance right here. That distance is the focal length. So this is going to be the point f,0 and this is going to be the point minus f,0. Now we learned in the last video that one of the definitions of a hyperbola is the locus of all points, or the set of all points, where if I take the difference of the distances to the two foci, that difference will be a constant number.

## Eccentricity (mathematics)

So if this is the point x comma y, and it could be any point that satisfies this equation, it's any point on the byperbola, we know, or we are told, that if we take this distance right here-- --let's call that d1 --and subtract from that the distance to the high other foci-- call that d2 --that that number is a constant regardless of where we are on the hyperbola. In fact the locus of all points are-- the hyperbole in fact is all of the points that satisfy that condition.

And we learned in last video just by taking the difference of the distance, we picked this point and we said, OK, what's that distance minus that distance. And we figured out that it's 2a. So d1 minus d2 is equal to-- I'm going off the video screen --d1 minus d2 is equal to 2a. So let's use this fact right here that d1 minus d2 is equal to 2a, to try to prove this. So the first thing to do is figure out what d1 and d2.

Just using the distance formula. So what you do is we just use the distance formula, which is really just Pythagorean theorem. So it's the difference of the x's. It's the x distance. So it's x minus f squared plus the y distances.

### Hyperbola - Wikipedia

Take the square root of that. So that's d1, right there. And we want to subtract from that d2. Or you could take the absolute values if you didn't want to worry about that. And so here, we get the square root of x minus f, x minus f squared, plus y squared. What does that equal to? Well, we said that equals to 2a, that equals this distance right here.

So that is equal to 2a. Now let's see if we can simplify this at all. Well an interesting thing to do my just to be the put this on the other side of the equation. And this just can get hairy, so I really hope I don't make any careless mistakes. So this becomes-- and I might write small to save space --this becomes x plus f, right, minus minus, squared plus y squared is equal to 2a plus the square root of x minus f squared plus y squared.

- Curve Fitting

Now, to get rid of these radicals, let's square both sides of this equation. The left hand side, if you were to square it just becomes x plus f squared plus y squared. And then to square this we have to square the first term, which is 4a squared. Then we multiply the two terms and multiply that by 2, right? We're just taking this whole thing and squaring it, so that's-- and this is just a review of kind of binomial algebra --so this is equal to plus 2a times this times 2 is 4a times the square root of x minus f squared plus y.

And then we square this term. And this is just multiplying a binomial. So that's equal to-- you just get rid of the radical sign, and I'm just going to be staying in that color for now --that's equal to x minus f squared plus y squared. And already it looks like there's some cancellation that we can do. We can cancel out-- there's a y squared on both sides of this equation --so let's just cancel those out.