Conic Sections
Gravitational Orbits

and more
Time Reversal A symmetry?

here is something here that has always bothered me. It doesn't really mean all that much but someone in the physics or mathematics profession should be a stickler for accuracy and complain.

Parabolic Canonball Trajectories

How many times have I seen this?

The trajectory of a projectile fired from the earth's surface would
follow a parabolic trajectory in the absence of friction with the air.

This is clearly false as it would trace out a portion of an ellipse before striking the surface. Where did the "parabolic" idea come from? Were parabolic mirrors on everyone's mind at the time?

There are three conic sections.

Conic sections - 5kb They are formed by the intersection of a plane with the volume of two inverted cones. If the plane cuts one cone in two sections one of which is finite in volume, you get an ellipse of varying eccentricity. If it cuts both cones into two infinite volumes, you get hyperbolas. If it cuts one cone into two infinite volumes, you get a parabola.

There are infinitely many ellipses of varying eccentricity. There are infinitely many hyperbolas of varying asymptotic angles. There is only one parabola, i.e. all parabolas have the same shape and can lie on top of each other congruently if you just globally make them bigger or smaller. The asymptotes of a parabola are two parallel lines.

Each conic section represents a possible path of one small body in the presence of the gravitational field of a large one. Though, in practice, a parabolic trajectory has never been observed and never will be in the future.

The reason it won't be observed is that a parabolic trajectory is equivalent to standing a pin on its pointy end and having it stay up unassisted. (No, you can't stick it ;o) Trajectories - 8kb

Here are the only possible types of trajectories:

As you can see, the hyperbolic trajectories occur when an object passes a massive body with too great a velocity to go into orbit at its nearest approach. If it has just the right velocity it would go into orbit as a circle. However, if it is "approaching the earth" (as in "getting nearer") it can't, in principle, go into a perfect circular orbit because it is physically impossible ... you need some course adjustments to go from hyperbolic to circular or elliptical orbits ... namely a retro-burn at perigee perpendicular to the body's gravitational field ... that keeps it from departing. Too much and you're in an elliptical orbit ... just right and its a circle.
Time reversal symmetry requires that a film run backwards be equally physically possible. But if you could insert yourself without a retro-fire, the backward film would show an object in a stable orbit being flung out of orbit for no reason ... that is, with no force applied. Hmmmm ... let me cogitate on this a moment ... meanwhile on with the show.
Let's use the Earth as a test bed, but we'll shrink its surface to a point to get it out of the way. Now, shoot any projectile in any direction such that it does not have escape velocity. Then ...

It must be true that it will reach a "high point" in the G-field and begin falling back to the Earth (now a point field source). That is, it will have no vertical component to its velocity. However, it will have some horizontal component (unless you shot it straight up in the gravitational field).

That horizontal component is what will keep the projectile from passing through the point source on the way down. Rather, it will go into an elliptical orbit with the Earth at one focus, ala Kepler's laws of planetary motion. Fine. When it comes around again it will pass through the point from which it was fired again ... and again ... and again ... It is   in orbit ... elliptical. This is true for any projectile fired in any direction (except perfectly vertical) in the absence of air friction.

Its path is not parabolic. It is elliptical.
So why do people insist on saying parabolic?

Some additional notes

The orbit of the moon 240,000 miles distant requires a velocity of about 2000 miles per hour. That of a satellite just four thousand miles from the center of the Earth requires a velocity of 18,000 miles per hour. If we have to go faster to get to a higher orbit ... why is the velocity required at that orbit so much less?

Each orbit represents an "energy level" ... the higher the orbit, the greater the energy. But that energy is composed of two types: kinetic (velocity) and potential (height). The potential plus the kinetic is greater in the higher orbit even though the velocity is less.

To raise an object to a higher level, you increase its velocity to say 19,000 miles an hour. It then rises to a higher orbit but as it does so it loses velocity by way of angular momentum conservation ... m v r is constant (after the "rocket burn" to increase velocity). The object continues to lose velocity and rise until its velocity matches that required for the highest obtainable new orbit. See?

If you twirl a rock on a string and pull the string in ... its speed doubles if you make the string half as long. If you let the string out, its speed is cut in half if you double the length of the string. Easy. (The mass of the rock remains constant.)

Hmmmm ... Time Reversal Symmetry again ...

conic3.gif - 30kb Suppose you are in a perfectly circular orbit (2). You shoot a cannonball in the direction of your orbit. This causes the craft to slow down. You drop to a lower orbit (1). The cannonball goes to a higher stable orbit (3). They do this ... at NASA ... but with rockets ... gradually.

Then there is this problem.

Since the reduction in velocity caused by the cannonball is almost instantaneous ... and ... the craft you are in does not fall instantaneously (nothing falls instantaneously) ... you drop for some finite time, unaccelerated by any rocket, before achieving a stable lower orbit ... what happens in the film of the event when run backwards?

You see what I'm getting at?

The backwards film shows a craft in a stable orbit which ... all of a sudden ... gains velocity and goes up to a higher orbit causelessly ... then ... gets hit with a cannonball which supplies it with the energy to get to where it already is !?

After the fact?

This is reminiscent of the uncertainty principle. Borrow energy then pay it back before anyone catches you.

Here is the solution ...

Both craft and cannonball must go into slightly elliptical orbits and therefore meet up every so many orbits. The higher orbit has the firing point as its perigee while the craft has its apogee at that same point.

This is what I like about proofs using symmetry principles.

You can draw very broad generalizations from them without any mathematical proof whatsoever. Here is what we can deduce from T-invariance in the gravitational interaction.

A test object in any state of motion in the gravitational field of a large body, fired in any direction at less than escape velocity, will result in an elliptical orbit around the center of that gravitational field ranging from zero eccentricity (a circle) to infinite eccentricity (straight up and down) which will return that object to the point of fire on completion of one full orbit.
There are two obvious exclusions. If the gravitational field is a planet, any orbit intersecting the planet's surface will result in a crash of course ... and ... all bets are off in the case of a "black hole" which extreme does not appear to be T-invariant (as given in the previous section).
There are also a relativistic modifications as in the case of the advance of the perihelion of Mercury.

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