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The Downward Motion of Particles in a Uniform Gravitational Field

1. Introduction.

In this article the downward motion of a particle with mass, m, in a uniform gravitational field, g, is discussed.  

2.2. Solution.

Let us begin by considering the downward velocity and acceleration of a particle in a uniform gravitational field, g. The acceleration can of course not be constant since v must always be less than c. But it is given that g, the gravitational field, is constant. And the particle’s gravitational mass at any time t is a function of its speed:

m(t)=g(t)m0.               (2.1)

The gravitational force law thus indicates that the gravitational force points “downward”:

Fgrav(t)=g(t)m0g.                  (2.2)

By Newton 3, however, the mechanical force must be

Fmech(t)=g(t)3m0a(t).          (2.3)

Assuming there can be only one force acting on the particle at any given time, it must be that

g(t)m0g= g(t)3m0a(t).         (2.4)

This being the case,

a(t)=g(t)-2g,                 (2.4)

and

v(t)=(1-a(t)/g)1/2c.    (2.5)

It is noteworthy that v and a are independent of m0; all particles with the stated initial conditions have the same velocity and acceleration after a time t has elapsed.

The analytical solutions for v(t) and a(t) are mathematically challenging and suggest little about the particle’s motion. The values can, however, be numerically computed. Starting with initial conditions of v(0)=0 and a(0)=-c/sec, Fig. 1 plots v(t) vs. t and Fig. 2 plots a(t) vs. t. Note how v asymptotically approaches c, and how a asymptotically approaches zero. A program for the numerical solution is provided in Appendix A.

Figure 1

 

 

 

 

 

 

 

v(t) vs. t

 

Figure 2

 

 

 

 

 

 

 

 a(t) vs. t

 

Appendix A

#COMPILE EXE

#DIM ALL

FUNCTION PBMAIN () AS LONG

DIM c AS DOUBLE

c=3e8

DIM g AS DOUBLE

g=-3e8

DIM n AS LONG

n=2500000

DIM t(n) AS DOUBLE

DIM dt AS DOUBLE

dt=1e-6

DIM v(n) AS DOUBLE

DIM a(n) AS DOUBLE

v(0)=0

a(0)=g

DIM index AS LONG

FOR index=1 TO n-1

    t(index)=index*dt

    v(index)=v(index-1)+a(index-1)*dt

    a(index)=(1-v(index)^2/c^2)*g

NEXT

OPEN "c:\\users\Marjorie Dixon\Documents\speeds.dat" FOR OUTPUT AS #1

OPEN "c:\\users\Marjorie Dixon\Documents\accelerations.dat" FOR OUTPUT AS #2

FOR index=0 TO n-1

    IF index MOD 1000=0 THEN

         WRITE #1, t(index),v(index)

    END IF

NEXT

FOR index=0 TO n-1

    IF index MOD 1000=0 THEN

        WRITE #2, t(index), a(index)

    END IF

NEXT

CLOSE #1

CLOSE #2

MSGBOX("ready for plotting")

 

END FUNCTION