10 March 2006

How Do We Model A Solar System?

George E. Hrabovsky, President, MAST, george@madscitech.org

The Column So Far

Here is a concise list of the columns in this series that have been presented so far:

Column 1: "A New Year for The Mind of a Theorist," where I introduced the format change and presented the basic ideas of dimensional analysis.

Theory Project #2: Solar System Models

As stated last time, all theory involves either derivation, solution., or understanding of equations.  In the case of the desire to model a solar system, we already know the equations to use. They have already been derived. Newton's second law of motion is the first, and Newton's Universal Law of Gravitation is the second.

We must understand these equations before we can start using them. Let us examine the first of these equations,

F = m a .

This is almost always quoted as, "Force is mass times acceleration," where force is F, mass is m, and acceleration is a. This is not true. The equation tells us, instead, that the quantity of force is the same as the product of the quantities of mass and acceleration. Put another way, it tells us that a force applied to a mass accelerates that mass. Equation (1) tells us nothing about the force being applied to the mass.

This brings us to the second equation. This defines the force due to gravitational attraction between two masses, called m_1 and m_2;

F = (G m_1 m_2)/r^2 .

In this equation r is the distance between the two masses and G is the gravitational constant

FormBox[RowBox[{(, RowBox[{G,  , =,  , RowBox[{6.673, ×, 10^(-11),  , m^3,  , kg^(-1),  , s^(-2)}]}], )}], TraditionalForm].

We can now put these equations together to form,

(G m_1 m_2)/r^2 = m a .

Let us agree that the first mass will be that of the central star of the solar system. We want to predict the orbit of a planet m_2 around that star.

(G m_1 m_2)/r^2 = m_2 a .

Recall from basic mechanics that acceleration is the second time-derivative of position, a = ^2x/ t^2, so

(G m_1 m_2)/r^2 = m_2  ^2x/ t^2 .

We can simplify this by canceling terms; in this case m_2 goes away.

(G m_1)/r^2 = ?^2x/? t^2 .

Of course, in the real world, the terms for distance and position are actually vectors. There is a vector rule that states

| b | = (b · b)^(1/2) .

If we have the vector b = b_xe_x +  b_ye_y, then the dot product b · b = b_x^2 + b_y^2.

Since the force works in the direction of the position vector r, we need to multiply the force of gravity by the unit vector in the direction of that force, so we have

F = (G m_1 m_2)/(| r |^2) r/(| r |) = (G m_1m_2)/(| r |^3) r .

So the final form of the equation becomes,

(G m_1)/(| r |^3) r = ^2r/ t^2 .

We have now successfully derived the equation of motion for a solar system with a central star and one planet. Next time we will solve this equation.


Created by Mathematica  (March 7, 2006)

   
Copyright 2005 by Society for Amateur Scientists