Orbital Mechanics

The researchers chose to sacrifice some detail in their simulation in order to model planetary systems from start to finish. They hope to extend their hybrid model approach so that they can eventually model planets spiraling all the way into the central star. Currently, the simulation cannot track such planets beyond a certain point.

Edward Thommes, a physicist at the University of Guelph in Ontario, Canada

Dude, what’s your sign? I am a Leo. That means that the Sun is seen in the constellation of Leo on my birthday. If you are an Aries then the Sun is seen in the constellation of Aries on your birthday. The sun moves, as seen from the Earth, against the background of fixed stars. You are assuming that you can ignore the stars and look only at the Sun from the Earth in such a way that it seems to stand still in the sky, and that the apparent lack of motion can be assumed to be the same as a real lack of motion. This does not work for the following reason.
To make the Sun stand still in Earth's sky you need to turn the direction you are looking at just the same speed as Earth orbits the Sun, but in the opposite direction. You have in fact placed yourself in a rotating coordinate system. I am looking at the Sun-Earth system from an inertial coordinate system that is not rotating or accelerating. In my inertial coordinate system, the Sun is at the origin and the Earth moves in X and Y. (Or if you prefer polar coordinates R and Theta.) The problem with a rotating coordinate system is the virtual forces (centrifugal and coriolis) that go along with it. These forces can only be calculated from the rotation rate, so the rate that the Earth revolves around the Sun is extremely important. In addition, you are making another common but invalid simplifying assumption, that the Earth’s elliptical orbit is close enough to a perfect circle to be considered as such. From the real Earth, the Sun never stands still in the sky. It gets farther and closer as Earth moves from aphelion to perihelion. It also speeds up and slows down so that your coordinate system is rotating at a variable speed. And the only way to know that speed is to do a calculation in an inertial coordinate system.
So if you throw a rock at the Sun, it behaves just like the Earth does. It gets a little closer to the Sun, stops, moves back from the Sun, stops again, moves toward the Sun again, and so on for ever. It is a complex interplay of centrifugal force, coriolis force, and the variable rotation rate of your rotating coordinate system.
To see what will really happen if you throw a rock at the Sun, use my orbit simulator:
http://home.austin.rr.com/campbelp/orbit.html
First, click the drop button to release a “dot” from the rocket. At first the dot will be hidden behind the rocket. Consider this dot as the Earth, since the rocket starts in an orbit identical to the Earth’s orbit around the sun. Then enter 1.0g and 100.0 Sec. in the thrust controls. Turn the rocket to point at the Sun and click the Fire button. At first nothing seems to change because the rocket is still right on top of and hiding the dot. After about a quarter of an orbit the two will separate enough to see them separately. Remember, the dot represents the Earth and the rocket is the object being thrown at the Sun.

Hence the circular motion equation - mv^2/r = GM1M2/r^2
Without the velocity, they would not balance. And if there is a difference in velocity, they have to be moving.

So G changes directly with r^2. But why? What is accelerating the Earth? Where is the energy to change the speed coming from?

But what about an object that is in a perfectly circular orbit around the sun? Does it accelerate? Is a changing direction vector acceleration?