Development of the Mathematical Ballistics Model (Pt. 3)

The mathematical model more easily solved by our array of existing differential equations - as I noted in the Oct. 22 post- is given by the pair:

 (A)

 r''  -   r q'² =    - cr^-2 

And:

2 r' q'    +  r q''     =  0

This is a system of 2nd order differential equations and the objective will be to solve for r as a function of t, r(t) and q  as a function of t,  q (t).   Elimination of t then results in r as a function of q.

Moving forward, if we multiply the 2nd equation by the integrating factor r, it becomes an exact differential, i.e.

2 rr' q' +  +  r^-2 q''   =  d/ dt (r^-2  q' ) = 0

Then an integration yields:

(B)   r^-2  q'  =  k ₁ = const.

Solving for q' and  substituting into the first equation of (A)   yields:

(C)

r"   -    k ₁ ²  /r^-3   = -  c/ r ²                                                 

Since:  r''  =   dr'/ dt = (dr'/ dt )(dr/ dt) = ( dr'/dt ) r'

Equation (C) can be changed to a variables separable form:

r' dr'  =  ( k ₁ ²  /r^-3  -  c/ r ² ) dr

Which on integration yields:

 ½ r' ² =   - k ₁ ²  r^-2  + cr^{- 1}  + K ₁

_{Where we choose the arbitrary constant:}

_K 1  =  ½ ( k ₂ ²  -  c² ) k ₁ ^-2

To Simplify the final solution:

r' =   +Ö {( k ₂ ²  -  c² )r^-2  +  2k ₁ ²  cr -  k ₁ ⁴ }/ k ₁ r

Note that since:

  q' =  k ₁ r ^-2   

 the quotient:

r'/q' =  dr/dq  =

 +  r/k ₁ ²  Ö {( k ₂ ²  -  c² )r^-2  +  2k ₁ ² cr -  k ₁ ⁴ }

yields a differential equation in which the variables are separable:

(D)

  k ₁ ²  dr/ r Ö {( k ₂ ²  -  c² )r^-2  +  2k ₁ ² cr -  k ₁ ⁴ } = dq

The integral for the left hand side can then be found in a table of integrals for the form:

ò  dx / x Ö( A + Bx + Cx ² ) =

 1/Ö -A arcsin (Bx + 2A)/ x Ö (B ² - 4AC)   + K₂

And for the specific problem under consideration:

A = -  k ₁ ⁴,   B =  2k ₁ ² c ,    C = k ₂ ²  -  c² 

So that:

 Ö -A  = Ö - (-k ₁ ⁴ )  =  k ₁ ² 

Ö (B ² - 4AC)    =  Ö{4k₁ ⁴ c²  +  4k ₁ ⁴(k ₂ ²  -  c² ) = 2k ₁ ² k ₂

And the solution for (D) is:

k ₁ ²/ k ₂ ² arcsin (2k ₁ ² cr - 2k ₁ ⁴ )/ r 2k ₁ ² k _{2   =}   q  + K ₃

Where  K ₃ is another arbitrary constant of integration.

Simplifying:

cr -  k ₁ ²  =   k ₂ r sin ( q  + K ₃)

or:

r  [c  -  k ₂ sin ( q  + K ₃)]  = k ₁ ²

Solving for r:  

r =  k ₁ ² /c -  k _{2 cos (}q  + k ₃)

Where k ₃ is a new constant based on:

(E)

sin ( q  + K ₃)  =   cos ( q  + K ₃ + p/2) = _{cos (}q  + k ₃)

Equation (E) is the geometric solution of the system of differential equations (A) and gives r as a function of q. This equation is the polar form of a conic with one focus at the origin.  There are three arbitrary constants that have been identified in this equation: k ₁ ,  k ₂ and k ₃.  These will need to be evaluated in terms of the designated conditions of the model, and this will be done in the next instalment.