Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: NIF: Kepler problem, part I



On Thu, 2 May 1996, Rauber, Joel Phys wrote:

The problem is as follows:
...

1) I'm to derive the following Kepler law

period^2 = (4 pi (earth-sun distance)^3)/G(M+m)

in a frame of reference where the earth and sun are stationary.

2) My physicists and astronomers have figured out the inverse square law of
gravity, and the values of constants appearing in that law; and they
have figured out the centrifugal force law, namely that there exists an
outward
pointing force on a mass in this frame of reference equal to
mass*const*(distance from coordinate origin) and that the constant has been
verified by experiment to be (4 pi^2)/(period^2). Where period is the time

it takes the stars to orbit the coordinate system. (Note: typo existed in
previous post)

3) The dynamical law I'll use is the 2nd law in this non-inertial frame:

Namely

applied to any mass in the system of analysis

m*a = sum of forces,

where in this situation described above, "sum of forces" is the sum of our
experimentally determined law of gravity and law of centrifugal force.

4) I may legitimately consider the earth and the sun as mass points.

SETUP: the coordinate system

a) Draw a typical looking set of x-y cartesian coordinate axes. This will
be the non-inertial reference frame.

b) draw a vector r_2 from the origin to some point on the postive x-axis.
This will be the position vector of the mass m_2 (the earth for the sake of
arguement). Note underscores denote subscripts.

c) draw a vector r_1 from the origin to some point on the negative axis.
This will be the position vector of the mass m_1 (the sun). If you want it
to look somewhat in scale draw the r_1 vector to be noticably shorter than
the r_2 vector.

d) The origin of my coordinate system is the center of mass of my two body
system.

Observations:

1) If you've drawn the setup correctly, you should have two mass points on
the x-axis; one to the right and one to the left of the coordinate origin.

2) Since this is supposed to be a coordinate system where m_1 and m_2 are
stationary we know that the vectors r_1 and r_2 do not change with time.

3) The absolute value, denoted ABS( ), of the difference in the vectors r_2
and r_1 is the "earth-sun" distance, i.e. ABS(r_2 - r_1) = distance
between m_1 and m_2.


Agreed.

A. R. Marlow E-MAIL: marlow@beta.loyno.edu
Department of Physics PHONE: (504) 865 3647 (Office)
Loyola University 865 2245 (Home)
New Orleans, LA 70118 FAX: (504) 865 2453