But now, if push the ball to the side just a little bit it will roll off. What if I take this bowl and flip it over? Now I again place the ball on the center and it stays there. If you push the ball a little bit to the side, it will roll back to the side. Consider a ball at the bottom of a round bowl. What if the JWST is a little bit closer or a little bit farther away than the calculated orbital distance from above? What then? How do you even look at stability of a system? Typically, a system is stable if when displaced, there is a force pushing it back to the original point. The orbital distance of the moon around the Earth is about 3.68 x 10 8 meters This puts the JWST at about 4 times farther away from the Earth than the moon. This means the distance from the JWST and the center of the Earth is about 1.5 x 10 9 meters. Comparing this to the orbital radius of the Earth at about 1.496 x 10 1 meters. The function crosses the f(r) = 0 at about r = 1.511 x 10 11meters. If I want to look at locations inside the orbit of the Earth, the gravitational force from the Earth would change directions. Here is a plot of the function f(r) for values of r starting at the orbital position of the Earth. Then I can find when f(r) crosses the r=0 axis and that will be my values for the solution. Well, really the easiest way to solve this is to call this some function f(r) and plot it. How can that be? Let me start with a diagram (again, no where near close to scale)Ī 5 th order polynomial. The key is that we want it to have a larger orbital radius, but THE SAME angular velocity of 2π radians per year. What if I wanted to push the Earth into a bigger orbit? In that case, I would have to have a lower angular velocity (it would take longer to orbit the Sun). The key here is that for the Earth to be in a circular orbit with a certain angular velocity (in this case about 2π radians per year), the Earth must orbit at a particular radius. Solving for the orbital radius (since I essentially know ω): Let me deal with this in terms of angular velocity (ω) instead of the linear velocity ( v). So, I write this as a scalar equation by putting in the values for the acceleration and the force. The force and the acceleration are both towards the Sun (the center of the Earth's circular motion). Now, since there is only one force acting on the Earth, Newton's second law says: (I hate the term Newton's second law, but people usually know what it is) Oh, v is the speed the Earth goes around the Sun and ω is the angular velocity (in radians/sec) of the Earth.
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