Obtaining a Complete Position function in complex system

[bleist88]

Hi,

{First off I should share my ultimate goal, which is to have a position function of time of several bodies, let's say in a solar system, so that the position of each body and the acceleration due to gravity on all the other bodies in the system are given at any time t. Although Kepler's formulas are useful, I am not wanting only to understand how bodies in orbits work, but also bodies that are randomly thrown through the system such as the Voyager stations or meteors or what have you.

So to take this problem into steps and build up to this:

There is first being able to describe a rock dropped close to Earth. Here, acceleration of gravity is the constant 9.8m/s. This is simple and takes less thought.

Next, Imagine dropping this body from much higher where the gravity is weaker and so the acceleration is smaller, and as the body falls, the acceleration is not constant. This is where I get stuck.

Next, I think, would be to imagine now that body given some x component of velocity and to determine the path it would take in the ( x, y, z, t ) as well as the ( r ) it would have.

This would give the path the moon takes but also the path a meteor close to Earth would take if it fell into Earth.

Next would be an even greater challenge, to Imagine a multi-body system so that the mass of both bodies would be considerable and be constantly changing all the others position. This would involve not only a function based on changin acceleration, but also changing position of the major gravity players.}

I understand basic Calculus, which will obtain a position function of time based on Newton's F=ma and the constant acceleration of gravity close to the Earth's surface: Y(t) = Yo + Vo(t) + [a/2](t)^2

...where Yo is original height, Vo is original dy/dt, and a is the constant acceleration of gravity.

Where I am stuck is the next ste, where you would drop a ball close to Earth, yet far enough away so that the acceleration due to gravity is no longer constant but smaller, and a function on the distance from the center of the Earth the object is...

Fg = GMm/y^2 = ma ... so... a = GM/y^2

But the problem lies where you would apply this to the former equation I used when gravity was constant and unchanging. (Now Y is not just the height but the distance from the two objects centers)

It gives: Y(t) = Yo + Vo(t) + {GM/2[Y(t)]^2}(t)^2...

This is a bit ugly in computer type, but basically the same as before except that the acceleration is not a constant but another function based on Y. But Y(t) is dependent on a(t) which is really a function of Y(t). This means the function is a function of itself, which at my misunderstanding of Calculus makes no sense. What did Newton do from here?



This is the question I have and I would love for it to be answered. I have searched through my Dynamics book, Calculus, and gone to professors and still have yet to get further.

 

[Stephen Tashi]

Hi,

{First off I should share my ultimate goal, which is to have a position function of time of several bodies, let's say in a solar system, so that the position of each body and the acceleration due to gravity on all the other bodies in the system are given at any time t.

Sounds like you want to solve the "n-body problem", which is currently unsolved. Under that title, you can find many methods for solving it approximately.

This means the function is a function of itself, which at my misunderstanding of Calculus makes no sense.

Look into the subject of "differential equations".

What did Newton do from here?

I don't know and it may not be a simple question to answer. Newton's "Principia Mathematica..." is written to imitate the geometric approach of Euclid. It's difficult to read (even in translation from the Latin) and may not be how he actually first made his discoveries.

 

[HallsofIvy]

The acceleration due to gravity, over a distance is given by
a= -\frac{GM}{r^2}
or
\frac{dv}{dt}= -\frac{GM}{r^2}

Since t does not appear explicitely in that, we can use a technique called "quadrature".
By the chain rule, dv/dt= (dv/dr)(dr/dt). But v= dr/dt so dv/dt= v dv/dr.

That is, the equation becomes
v dv/dr= -\frac{GM}{r^2}
v dv= -\frac{GM}{r^2}dr

Integrating both sides, (1/2)v^2= \frac{GM}{r} so that
v= dr/dt= \sqrt{\frac{2GM}{r}}
so that r^{-1/2}dr= \sqrt{2GM}dt