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W9 L17 CENTRAL FORCE MOTION

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W9 L17 CENTRAL FORCE MOTION Powered By Docstoc
					                        CENTRAL – FORCE MOTION

When a particle moves under the influence of a force directed toward a fixed
center of attraction, the motion is called central-force motion.

Motion of a Single Body

Consider a particle of mass m moving under the action of the central gravitational
attraction, as shown in Fig, (9.1).




                                       Figure (9.1)

         mm 0
  FG
          r2

Where m0 is the mass of the attracting body, and the particle of mass m could be
the moon moving about the earth, or a satellite in its orbital motion about the
earth above the atmosphere.

The most convenient coordinate system to use is polar coordinates in the plane
of motion.
The force F is always in the negative r-direction and there is no force in θ-
direction.

    mm 0             2        
G            m r  r    ma r 
                          
       r2                        
                                           .......... .......(1)
                             
0  m r   2 r    ma         
                                 
                r
multiply by
               m
              
                      d  
0  r 2   2r r    r 2  
                      dt    
So
     
r 2   h  cons tan t                 .......... .........( 2)

                                                                              
The angular momentum of m about m0 is r x mv and has the magnitude of mr 2  .
This states that the angular momentum of m about m0 remains constant (is
conserved).

During time dt, the radius vector sweeps out a shaded area equal to dA as
shown in Figure (9.1).
     1 
dA   r rd
     2 
The rate at which area is swept is
 
    1 
A  r 2   cons tan t
    2

So, areas swept out in equal times are equal “ Kepler’s second law”.

The shape of the path followed by m may be obtained by solving equation (1)
with time t, as
    1
r
    u
                 
           
r   1 u 2 u  h u    hdu d
                       
                                                               
                                 
and           r  h d 2u d 2   h 2 u 2 d 2 u d 2

Substitute into equation (1)
                    d2u 1
 Gm 0 u 2  h 2u 2 2  h 2u 4
                    d  u
or
d2u      Gm
     u  20
d 2
          h

The solution of the above second order equation is

                            Gm
            C cos     2 0
         1
u
         r                   h

C and δ are the two integration constants.

The phase angle δ can be eliminated by choosing the x- axis so that r is
minimum when θ = 0, so
1            Gm
   C cos   2 0               .......... .....( 3)
r             h

Conic Sections

The conic section is formed by the locus of a point which moves so that the ratio
e of its distance from a point (focus) to a line (directrix) is constant. So from Fig.
(9.1),

          r
e
   d  r cos 
Which may rewritten as

1 1         1
  cos                    .......... .......( 4)
r d        ed

This has the same form of equation (3). The motion of m is along a conic section
         1          h2
with d  and ed 
        C          Gm 0
         h2C
or e             .......... .......... ..5 
         Gm 0

Three cases to be investigated corresponding to the value of e. The trajectory for
each of these cases is shown in Fig. (9.2).




Figure (9.2)
Case 1: ellipse (e < 1).

From equation (4), we deduce that

r is minimum when θ = 0, and
r is maximum when θ =π.

                            ed   ed                              ed
2a  rmin  rmax                             or          a
                           1 e 1 e                            1 e2

1 1  e cos 
  
r         
    a 1 e2          

rmin  a1  e            rmax  a1  e          Also         b  a 1 e2

When e = 0, r = a, and the ellipse becomes a circle.

For the period

      A           ab                      2ab
    
                               or    
                  1 2                      h
      A             r 
                  2

For

   h2C                         1            ed
e                        d          a                    b  a 1 e2 ,      and Gm 0  gR 2
   Gm 0
        ,
                               C
                                 ,
                                           
                                           1 e2
                                                 ,
                                                    
              a3 2
  2
          R g

Where R is the mean radius of central attracting body and g is the absolute value
of acceleration due to gravity at the surface of the attracting body.


Case 2: parabola (e = 1).

Equations (4) and (5) become

  1  cos 
1 1
                                and     h 2 C  Gm0
r d
The radius vector becomes infinite as θ approaches π, so the dimension a is
infinite.

Case 3: hyperbola e > 1).

From equation (4), the radial distance r becomes infinite for the two polar angles
θ1 and – θ1 defined by cos θ1 = - 1/e.
Branch I represent physical possible motion which correspond to – θ1 < θ < θ1,
Fig. (9.3).
Branch II corresponds to angles in the remaining sector (with r negative)




                                  Figure (9.3)


                                            1 cos 
   cos   
1 1               1                   1
                             or             
r d             ed                   r    ed   d


Energy Analysis

The energy of particle m is constant

E=T+V
     1           2
           2
                        mgR 2
E  m r  r 2   
     2  
                      
                          r
               
                         1       gR 2
at   0,      r  0,        C           from equation (3)
                         r        h
 
      h
r          from equation (2)
      r
2E              g 2R 4
     h2C2  2
 m               h
Substitute value of C from equation (5)

          2Eh 2
e   1
         mg 2R 4

For:

elliptical orbit   e < 1,   E is negative

parabolic orbit    e =1,    E is zero

hyperbolic orbit e > 1,     E is positive

For elliptical orbit

     gR 2m
E
       2a
Substitute in energy equation


            1 1 
 2  2gR 2      
             r 2a 


                g 1 e    g rmax
 Perigee  R          R
                a 1 e    a rmin


                g 1 e    g rmin
 Apogee  R           R
                a 1 e    a rmax

				
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posted:12/1/2011
language:English
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