Gravity, rosettes and inertia.        modified 3-9-10
Orbits: Gravity is so intertwined with centrifugal force that they are usually inseparable. When something moves in a circular orbit, satellites around the earth, planets around the sun, electrons around protons, light around black holes, the centrifugal force equals the gravitational force or the centrifugal force equals the electrostatic force of attraction. These can be considered two body binary systems, in orbit across from each other around a common center of mass with a common orbital period. Each planet imposes an additional force, center of mass and orbital period on the sun. Their x, y and z forces can be summed. The center of mass, or barycenter, is always someplace between their centers. Both masses always move. The centrifugal forces are with respect to the center of mass not the distance between the masses. The gravitational force is between the masses at the distance between the masses. The centrifugal forces are equal to each other and to the gravitational force. The accelerations are proportional to the masses.

Centrifugal force: When gravity exerts a center seeking centripetal force, inertia opposes this deviation from straight line motion with a center fleeing centrifugal force. The centripetal gravitational force equals the inertial centrifugal force along the orbital path. For every action there is an equal but opposite reaction. Slinging a rock on a rope, around in a circle, demonstrates this centrifugal force which can easily be measured with a spring scale used by fishermen. You and the rock are a binary system. Your centrifugal force, at your distance from the center of mass, equals the centrifugal force of the rock, at its distance from the common center of mass, equals the tension in the rope. If the rope is cut or released both the centripetal and centrifugal forces become zero. The rock continues on its inertial path. You continue on your inertial path determined by your momentum prior to release, that is, tangent to the circle at the point of release. The two paths are in opposite directions. Reality is defined by simple experiments like this one. I suspect Newton would have done something like this when he was in his twenties working on gravitation around 1660. He would have known this is equivalent to Kepler's third law.

Kepler's third law:
m*vt2/r = G*m*M/r2, The centrifugal force equals the gravitational force. This is just another way of writing Kepler's third law. M and m are mass. vt is tangent velocity. G is the gravitational constant. r is the radius apart. The period of a circular orbit is p.
p = 2*pi*r/vt, therefore vt2 = {4*pi2*r2/p2}
m*{4*pi2*r2/p2}/r = G*m*M/r2,
substituted for vt2
4*pi2*r3 = p2*G*M, collected terms. The cube of the radius is proportional to the square of the period. This is Kepler's third law but we will usually use it in the “centrifugal force equals the gravitational force” form. This equation and the idea of conservation of energy are both indubitably correct and are central to our arguments.

Sun and earth forces:
centrifugal forceearth = centrifugal forcesun = gravitational forceearth&sun
mearth*vearth2/dbcearth = msun*vsun2/dbcsun = G*mearth*msun/(dbcearth+dbcsun)2 = 3.547E22_kg*m/s2, G is the gravitational constant. mearth and msun are the masses. dbcearth and dbcsun are the distances to the barycenter, the center of mass. vearth and vsun are the orbital velocities around the center of mass. There are three equal forces. The centrifugal force of the earth at its distance from the common center of mass equals the centrifugal force of the sun at its distance from the common center of mass equals the gravitational force between the earth and sun. Their mass, velocity and distance to the center of mass are related by ratios.
The left term is the centrifugal force of the earth around the earth-sun center of mass.
The middle term is the centrifugal force of the sun around the earth-sun center of mass. The sun also moves. This term is always neglected in over simplications. Where is it not neglected? This term is necessary to understand inertia, force = mass*acceleration.
The right term is the gravitational force between between the earth and sun across the center of mass. They have the same orbital period. The earth and sun both orbit at different distances about the same point. That point, the barycenter, is the center of mass of the earth-sun system. The three equal forces are along a line through the center of mass. If the forces had a propogation delay then the forces would pull on where the objects used to be and not be along this line. The forces would form a triangle. Energy would not be conserved. The orbits would not endure. See Van Flandern.

Sun and earth accelerations:
There are two different accelerations which are proportional to the masses.
vearth2/dbcearth = G*Msun/(dbcearth+dbcsun)2 = 5.93E-3_m/s2, the acceleration is the force divided by mearth and the force per unit mass of the Earth. The earths gravitational acceleration is caused by the sun.
vsun2/dbcsun = G*mearth/(dbcearth+dbcsun)2 = 1.78E-8_m/s2, the acceleration is the force divided by Msun and the force per unit mass of the sun. The suns gravitational acceleration is caused by the earth.

Sun and earth as a typical binary system:
periodearth = periodsun, the orbital period of the earth and sun are equal. They are a binary system.
2*pi*dbcearth/vearth = 2*pi*dbcsun/vsun,
dbcearth/vearth = dbcsun/vsun, multiplied by 1/(2*pi)
vsun/vearth = dbcsun/dbcearth, this uses velocity and distance to the barycenter or
(1) vsun/dbcsun = vearth/dbcearth, Angular velocity is the orbital velocity of the object divided by the distance to the center of mass. The angular velocity of the sun and earth are the same. Their orbital periods are the same. In all binary systems, the distance, velocity and mass are related by ratios and products. This uses velocity and distance to the barycenter, the center of mass.

centrifugal forceearth = centrifugal forcesun
mearth*vearth2/dbcearth = msun*vsun2/dbcsun = 3.547E22_kg*m/s2
The centrifugal force of the earth at its distance from the common center of mass equals the centrifugal force of the sun at its distance from the common center of mass.
mearth*{vsun2*dbcearth2/dbcsun2}/dbcearth = msun*vsun2/dbcsun, substituted for vearth2
mearth*{dbcearth2/dbcsun}/dbcearth = msun
mearth*dbcearth = msun*dbcsun, collected terms. The mass-distance products are equal. This is the balance equation used in scales.
(2) mearth/msun = dbcsun/dbcearth, This uses masses and distance to the barycenter.

mearth*{dbcsun*vearth/vsun} = msun*dbcsun, substituted for dbcearth
msun*vsun = mearth*vearth, their momentums are equal.
(3) msun/mearth = vearth/vsun, This uses mass and orbital velocity.

This is a binary system so we can calculate the distance to the center of mass of each of the bodies if we know their masses and their total distance apart. The distance between the bodies is their center distance. The center of mass is between the centers of the sun and earth.
dbcearth + dbcsun = center distance = au, an au is the distance between the earth and sun. The sun and earth are opposite each other across the center of mass.
{dbcsun*msun/mearth} + dbcsun = au, substituted for dbcearth = dbcsun*msun/mearth.
dbcsun*(msun+mearth)/mearth = au,
dbcsun = au*mearth/(msun+mearth), the distance from the center of mass to the center of the sun. The sun is not stationary. The sun orbits around a point, the barycenter, at this distance from its center toward the earth. The planetary data shows this as "solar wobble distance".
dbcsun = 149.598E9_m *5.9722E24_kg /(1.98843E30_kg +5.9722E24_kg) = 449312_m, this is the offset from the solar center which the sun orbits around with respect to the earth.
dbcearth = au*msun/(msun+mearth) = 149.59755E9_m
vearth = 2*pi*dbcearth/period = 29786_m/s, the earth's orbital period around the center of mass is one year as is the sun's with respect to the earth.
vsun = vearth*dbcsun/dbcearth = 0.08955_m/s. Galileo was wrong, the sun does move with respect to the earth, but not fast.

binary2.png
Click image to animate! Wobbles and a donut:
When the barycenter, the center of mass of the system, is inside a member of a binary system, as it is in the animation, then its motion is a wobble. The earth appears to wobble because of the moon. The sun appears to wobble, because of its planets, when seen from its poles. We see the sun from its equator. There is a different center of mass and wobble in the sun for each planet. Galaxies wobble. The universe wobbles. The sun orbits about all of its different centers of mass of all the planets simultaneously. This is a complex pattern of movement. Only the sun-jupiter center of mass is outside the surface of the sun. The sun makes, offset from its center, orbits or wobbles, around its center of mass, for all the planets except for Jupiter. The wobbles have the orbital period of the planets. The sun makes a donut, not a wobble in the sky, in its orbit with Jupiter. This center of mass is outside the radius of the sun. The donut has the orbital period of Jupiter. The donut is seen from the solar poles against the background of the fixed stars. If the sun and Jupiter orbit around their common center of mass then in some real way the Earth orbits with Jupiter. In what way? This multiple binary pattern of movement is one way which can reveal planets around other stars. The sum of the x, y, and z components of the forces of the planets, when divided by the mass of the sun is an acceleration. The suns acceleration over time is a velocity. The direction, velocity and location of the sun with respect to the background stars varies as the sun loops around. These movements produce huge tidal forces in the sun. These tides may produce solar tsunamis. The tides suck energy from the suns rotation. As the sun slows in its rotation, the sun receeds in its orbit, the planets receed in their orbits. This is a solar system wide Hubble like expansion.

The Appolo missions to the moon left behind a laser reflector so that the round trip of laser pulses may be timed and the distance to the moon calculated. The moon is receeding from the earth, due it is said to tidal drag, at 38_mm per year. Who knows? Friction loss due to tidal drag is hard to estimate. It is possible that 25_mm of this expansion may actually be due to a Hubble expansion in the solar system.
Click image to enlarge! First, on the left. This is a figure from Wiki which shows the sun. The black line is the path and location of the sun over time. See Barycenter in astrophysics and astronomy. See the "Barycenter of the solar system" at Gravity simulator. The suns movement is the vector sum of the forces of the solar system and galactic and other forces from outside our solar system. The sun moves with respect to the background stars, galaxy and quasars as well as the planets.
Second, This uses the radial angles and the forces of all the planets. Forces are vectors=(red). You can add the x, y and z components of vectors or forces. x=radial force*cosine(angle)=(green). y=radial force*sine(angle)=(blue). The summation of the x and y planetary forces is the resultant force that the sun experiences from the planets alone. This resultant force causes the sun to move=(large red arrow). The sun is also moved by many other forces from outside the solar system but none are included in these calculations. The rosettes which follow were calculated in this way. We use the fact that both the sun and planets move.
Rosettes
Click to enlarge! When planets move, the sun moves:
Each of these rosettes shows the position of the sun and planets against the background stars, looking down on the north pole of the sun. The movement of the sun and the radial movement of the planets are to scale. The radius of the orbits of the planets are not to scale. The depth of the lobes reflects the movement of the sun. The pedals or lobes are exaggerated when we put them on a page rather than draw them to scale with Plutos orbit being 8491 times the solar radius. The lobes disappear in the scale of the planetary orbit as the size of the suns movement becomes miniscule compared with size of the planetary orbits. This is over a period of 247.7 years, the orbital period of Pluto. Yep, I still count Pluto. We look at the position of the sun and planets 2, 7 and 14 times per Mercury year of 88.023 days. Mercury makes 1027.8008 orbits around the sun for every orbit of Pluto.
First, on the left, we see the position of the planets as a dot every 44.01 days for 2055 dots per planet. Each dot is the position of the planet at that moment. The many dots merge into lines. Mercury has a circular orbit around the sun but we capture our pictures of Mercury twice in its orbit when it happens to be in the two little dark circles. This gives the effect of a stobe light freezing periodic motion. Mercury has its orbital motion frozen since we have fixed our strobe-light-periodic-views on its orbital period. The position of Venus is alway someplace in the odd seven lobe ring at these times. In the figure, there is a hint of lobes on the inside of earths orbit and on the outside of Mars orbit. These are circular orbits. The planets appear somewhere in these patterns when the position of the planets is recorded every 44.01 days. They demonstrate a non-integer harmonic relationship.
Second, we see the position of the planets as a dot every 12.57 days for 7194 dots per planet. Mercury has the pattern of 7 little dark circles.
Third, we see the position of the planets as a dot every 6.28 days for 14389 dots for the sun and each of the nine planets or 143890 dots. The harmonic information, from the inner planets, is lost in a sea of dots.
Forth, this blowup shows the path of the solar disk at the center of the rosettes. The sun is shown as blue for a month per year from 1944 to 2020. This is a scale drawing of the solar disk and the movement of the sun.
Fifth, The same period and path of the sun is shown, not as a monthly solar disk, but as a series of monthly dots. These dots shows the path of the center of the sun. This duplicates the start of the the time period seen in the figure from Wiki for confirmation of our calculations. The movements of the sun are similar in both figures. The suns orbit appears to be an ellipse spinning around the average position of the sun in time with Jupiter. This is something like the motion of a hula hoop. This is also like the Sun-Jupiter binary system orbitting around the average position of the sun. This orbit is caused by the other planets. The sun must experience huge tidal forces and friction losses to its angular momentum, to move like this.
(The fuzzy edges of system fonts look bad in bitmapped graphics but you can make your own fonts with free Bitmap Font Writer which was used for this figure.)

How it works: The mass and orbital period of the planets is recorded. The orbital period of the planets is considered constant and their orbits circular. Position and distance from the sun for each of the planets was recorded for a certain day. The gravitational force between the sun and each planet was calculated. The centrifugal force equals the gravitational force so it may also be used for these calculations. The x and y components of the forces of all the planets were summed yielding the x and y forces on the sun. The sum of these forces when divided by the mass of the sun yield x and y accelerations of the sun. The accelerations were multiplied by the duration of the period of observation in seconds, 44.01 days is 380246 seconds, yielding an x and y solar velocity. This is the velocity added by the solar acceleration. The sun also has a previous x and y velocity from the prior calculation which must be added to this new velocity from the acceleration. The new acceleration changes the direction and velocity of the sun. The sun is considered to move with the sum of these velocities to a new position for each observation. The circular path of the sun only comes from the summation of the x and y components of the radial planetary force vectors as the planets orbit. The sun orbits as the planets orbit. This is surprising and there is a lot more to this. The central force of gravitation allows rotation in the solar system and the cosmos. The normal condition for gravitational and electrostatic systems is to orbit. Orbits can be forever. What you see depends on the metaphors you use.

Each positional dot is a tiny circle filled with a color and outlined by another color. The circles overlap leaving behind the outline colors. The sun is a red circle outlined in yellow which leaves behind a yellow donut as it loops around. As the sun moves, it drags the planets with it, leaving some of them with lobed orbits. This may look crazy but it is astonishing that we did not see this in elementary school.

The planets move the sun. The planets move with the sun. Their orbital periods and velocity around the sun do not change. Radial movements cause the rosettes. The rosettes are only seen with respect to the fixed reference of the background stars. The center of the sun loops around a variable radius of about two solar diameters or 14E9 meters. The lobes on the planets move the same radial distance. The orbital radius of Pluto is 6E12 meters. 6E12/14E9 is 480 so the lobes on Plutos orbit would be invisible at this scale. When you can see the whole orbit the lobes are too small to see unless they are greatly magnified. There are 20 lobes in Plutos orbit because the sun makes 20 orbits for each orbit of Pluto. Neptune has about 13 lobes. Uranus has about 6 lobes. Saturn has about 3 lobes. Jupiter has no lobes. Its path looks like a ring but it does move back and forth with the sun. The lobes on the planets are synchronized with the orbit of Jupiter and the Sun.

These calculations and graphics were created with a Liberty Basic program. Basic is easy to read and translate into other computer languages. This is the text which can be pasted into the Basic editor. It may also be pasted into the Notepad. If you saveas it with .bas not a .txt extension in Notepad then you can click it to run in Basic. The extensions will need to be turned on in the control panel.

Astronomers calculate the orbital parameters of extra-solar planets from the visual wobble of a star or the doppler frequency changes as it approaches and receeds from our point of view. When tiny distant Pluto moves, the sun also moves, as they are a binary system. When the sun moves, the cosmos moves, as they are a binary system. This is how inertia works.

For every action there is an equal but opposite reaction. When you push something, you accelerate it to get it moving, something pushes back. It is the opposite acceleration of the mass of the universe which pushes back. What else could there be to push back? Everything is connected. Everything is a part of a binary system with the universe. Inertia is the acceleration dependant reaction of the observable universe. The observable universe is that part of the totality of everything which has a knowable radius and mass. I like to call it the cosmos.
Inertia
Binary systems show how inertia works:
There are two each of masses m, tangent and radial velocities vt and vr and distance to the barycenter, the center of mass of the system, dbc. There are several ratios and products of these.
(1) vts = mu*vtu, their orbital momentums are equal. The subscript s is for a star near the edge of the cosmos. The subscipt u is for the universe or cosmos.
(2) ms*dbcs = mu*dbcu, the mass times distance products are equal and balanced.
(3) vts/dbcs = vtu/dbcu, This is (1)/(2). The angular velocities and orbital periods are the same in a binary system.
(4) ms*vts2/dbcs = mu*vtu2/dbcu, This is (1)*(3). Their centrifugal forces are equal. Centrifugal force exerts its force in a direction perpendicular to its tangent velocity vt. This is the same geometry as the Biot-Savart law. See Paul Marmet.
(5) ms*vrs = mu*vru, their radial momentums are equal in an expanding system.
(6) vrs/dbcs = vru/dbcu, This is (5)/(2). The radial velocity/distance are equal. This is the definition of Hubble's constant. The star and the cosmos both expand according to Hubble. If the star has a radial velocity of .9*c and is at .9*radius of the cosmos = .9*c*age then,
vrs/dbcs = .9*c/(.9*c*age) = 1/age = vtu/dbcu or Hubble's constant.
(7) ms*vrs2/dbcs = mu*vru2/dbcu, This is (5)*(6). Their radial velocity centrifugal forces are equal. Centrifugal force exerts its force in a radial direction, in a direction perpendicular to its tangent velocity vt. Radial centrifugal force exerts its force in a tangent direction, in a direction perpendicular to its radial velocity vr. This is because vt = vr.

The two centrifugal forces and one gravitational force are equal. The two coriolis forces are equal.
forces = forceu, the centrifugal forces or coriolis forces are equal but opposite. forces is a local force. forceu is the cosmic force inertia, the reaction of the cosmos. Force equals mass * acceleration, so
ms*as = mu*au, the mass of a star is ms = 1.9E30_kg and as is the acceleration of the solar mass star and mu = c3*age/G = 1.912E53_kg is the mass and au the acceleration of the cosmos.
au = as *ms/mu = as *(1.9E30_kg/1.9E53_kg) = as *(1E-23), the acceleration of the cosmos in reaction to the star's acceleration is microscopic. When a force is applied to the star, the star is accelerated. The cosmos mirrors this force with an equal but opposite force. The cosmos has its own tiny acceleration for the forces, since accelerations are proportional to mass. These tiny accelerations over time produce a velocity. The cosmos mirrors these movements in the same way the sun mirrors the movement of the planets. When a planet or a star or anything moves, the cosmos moves.

mu *dbcu = ms *dbcs, this is the balance equation used in scales and levers.
dbcs = .9*c*age = 1.278E26_m. The star is at .9 the radius of the cosmos.
dbcu = ms*dbcs/mu. You can see the derivation of the radius and mass of the cosmos in the black hole paper,
dbcu = ms* .9*c*age* G/(c3*age), substituted values for dbcs and mu.
dbcu = ms*.9*G/c2 = 1328_m, is the distance from the center of the cosmos, to the center of mass of the star-cosmos system. The center of mass of the system is close to the center of the cosmos because of the huge mass ratio even though the star is near the perimeter of the cosmos. When a force moves the star one way the mass of the cosmos moves the other way a very small amount. They are on opposite sides of the center of mass, of a binary system, which can be thought of as a lever and fulcrum.
distu*dbcs = dists*dbcu, dists = 1_m, distu and dists are the distances that the cosmos and sun move. dbcu and dbcs are the length of the levers, across the fulcrum from each other or across the center of mass from each other. The angular movements are equal but the radial lengths of the levers are different.
distu = dists*dbcu/dbcs = 1_m*1328_m/1.278E26_m = 1.04E-23_m, or
If the star moves 1_meter, then the cosmos moves in the opposite direction, across the center of mass, 1.04E-23_meter.

Coriolis forces:
mu *dbcu = ms *dbcs, this is the balance equation used in scales, levers and binary systems.
mu *dbcu/age = ms* dbcs/age , multiplied by 1/age.
     vru = dbcu/age = 1328_m/age = 1328_m/4.73E17_s = 2.8E-15_m/s,
     The radial velocity of the cosmos with respect to center of mass of the system.
     vru = dbcu/age so vru/dbcu = 1/age = Hubble's constant
     vrs = dbcs/age = .9*c*age/age = .9*c,
     The radial velocity of the star with respect to center of mass of the system.
     vrs = dbcs/age so vrs/dbcs = 1/age = Hubble's constant.
     The star is at .9 the radius of the cosmos.
     It has a radial and tangent velocity of .9 the speed of light. Since it has a radial and
     tangent velocity, it is spiraling out, like everything else in the cosmos.
mu*vru = ms*vrs, collected terms. Their radial momentums are equal.
2*mu*angular velocity*vru = 2*ms*angular velocity*vrs, multiplied by 2*angular velocity. These are two equal but opposite coriolis forces.

Newton had the idea that the mass of a body may be considered to be concentrated at a point at the center of the body and that gravitational forces can be considered to act between these points. Concentric shells of mass may be considered to be concentrated at a point, in the center of the shells, for a body outside the shells. Concentric shells of mass cause no resultant force on a body within static shells. For a body within the shells, this is true only for static shells. This is not true for expanding shells or for an expanding cosmos.

As the objects in the cosmos move, like the planets in the solar system, the center of the expanding concentric shells of mass of the cosmos traces the sum of their movements around the barycenter of the cosmos. The movement of the cosmos around the barycenter of the system uses the same mechanics as the movement of the sun on its wobbly path around the barycenter of the solar system.