One Each Day Till They Go Away (Part 5)

First, we'll shift the coordinate frame to have its origin at the fixed particle which is positioned at $(0, 1)$ . Thus the fixed particle is now at $(0, 0)$ and the moving particle starts at $(-2, -1)$ , and the destination point is $(4, 0)$ .

Using polar coordinates, we can write

$\vec{r}(t) = (x, y) = r ( \cos \theta , \sin \theta )$

Differentiating with respect to time $t$ ,

$\dot{\vec{r}} = \dot{r} (\cos \theta, \sin \theta) + r ( - \sin \theta, \cos \theta ) \dot{\theta}$

Differentiating again,

$\ddot{\vec{r}} = \ddot{r} (\cos \theta, \sin \theta )+ 2 \dot{r} (- \sin \theta , \cos \theta) \dot{\theta} + r ( - \cos \theta, - \sin \theta) (\dot{\theta})^2 + r (-\sin \theta , \cos \theta ) \ddot{\theta}$

The force acting on the particle is $\vec{F} = \dfrac{-G M_1 M_2}{r^2} (\cos \theta , \sin \theta) = M_2 \ddot{\vec{r}}$

Substituting $G = 1 , M_1 = M_2 = 1$ , this becomes, $\vec{F} = \dfrac{-1}{r^2} (\cos \theta , \sin \theta) = \ddot{\vec{r}}$

Equating radial and traverse components of the vectors, on both sides of the equation, we deduce that,

$\ddot{r} - r (\dot{\theta})^2 = \dfrac{-1}{r^2} \hspace{20pt}(1)$

and

$2 \dot{r} \dot{\theta} + r \ddot{\theta} = 0 \hspace{20pt} (2)$

Multiplying the second equation by $r$ , we get,

$2 r \dot{r} \dot{\theta} + r^2 \ddot{\theta} = 0 \hspace{20pt} (3)$

The left hand side is just the derivative of $r^2 \dot{\theta}$ , therefore,

$r^2 \dot{\theta} = K \hspace{20pt}(4)$

To determine $K$ , we use the initial conditions, we're given that $r(0) = r_0 = \sqrt{5} , \theta(0) = \theta_0 = \tan^{-1}(\dfrac{1}{2}) - \pi , \dot{\vec{r} } = (v_0, 0)$ , Using the expression for velocity developed above, we have,

$\dot{\vec{r}} = \dot{r} (\cos \theta, \sin \theta) + r ( - \sin \theta, \cos \theta ) \dot{\theta}$

so that, at the initial postion,

$\dot{r}_0 = (v0, 0) \cdot (\cos \theta_0 , \sin \theta_0) = \dfrac{-2 v_0}{\sqrt{5}}$ and $\dot{\theta}_0 = \dfrac{1}{r_0} (v_0, 0) \cdot (-\sin \theta_0, \cos \theta_0 ) = \dfrac{v_0}{5}$

Hence, $K = r_0^2 \dot{\theta}_0 = v_0$

Plugging in equation (4) into (1),

$\ddot{r} - \dfrac{K^2}{r^3} = -\dfrac{1}{r^2} \hspace{20pt}(5)$

By relating the time-derivative of $r$ to the $\theta$ -derivative, equation (5) becomes,

$K^2 ( \dfrac{ r''}{ r^4} - \dfrac{2 r'^2}{ r^5} - \dfrac{1 }{ r^3} ) = -\dfrac{1}{r^2}\hspace{20pt}(6)$

where $r' = \dfrac{dr}{d\theta}$ and $r'' = \dfrac{d^2 r}{d\theta^2}$ .

Now, we will use the change of variables, $u = \dfrac{1}{r}$ . This will change equation $(6)$ into,

$K^2 ( u'' + u ) - 1 = 0 \hspace{20pt}(7)$

where $u'' = \dfrac{d^2 u}{d\theta^2}$ . Equation $(7)$ is a linear differential equation with constant coefficients. Dividing by $K^2$ ,

$u'' + u - \dfrac{1}{K^2} = 0 \hspace{20 pt} (8)$

The solution of this differential equation is:

$u(\theta) = \dfrac{1}{K^2} + A \cos (\theta - \theta_0) + B \sin (\theta-\theta_0)$

Since $r(0) = r_0 = \sqrt{5}$ , then , $u(0) = \dfrac{1}{\sqrt{5}}$ , and therefore, $A = \dfrac{1}{\sqrt{5}}- \dfrac{1}{v_0^2}$ .

We also have $u'(0) = \dfrac{-1}{r_0^2} \cdot \dfrac{\dot{r}_0}{\dot{\theta}_0} = \dfrac{-1}{5} \cdot \dfrac{-2 v_0}{\sqrt{5}} \cdot \dfrac{5}{v_0} = \dfrac{2}{\sqrt{5}}$

Hence, $B = u'(0) = \dfrac{2}{\sqrt{5}}$ . And therefore,

$u(\theta) = \dfrac{1}{v_0^2} + (\dfrac{1}{\sqrt{5}}- \dfrac{1}{v_0^2}) \cos(\theta - \theta_0) + \dfrac{2}{\sqrt{5}} \sin(\theta-\theta_0) \hspace{20pt} (9)$

At the point $(4, 0)$ ( originally $(4, 1)$ ), $\theta = 0$ and $r = 4$ , hence,

$\dfrac{1}{4} = \dfrac{1}{v_0^2} + (\dfrac{1}{\sqrt{5}}- \dfrac{1}{v_0^2}) \cos( - \theta_0) + \dfrac{2}{\sqrt{5}} \sin(-\theta_0) \hspace{20pt} (10)$

Upon subtituting $\theta_0$ , equation $(10)$ becomes,

$\dfrac{1}{4} = \dfrac{1}{v_0^2} ( 1 + \dfrac{2}{\sqrt{5}} ) \hspace{20pt} (11)$

which solves to,

$v_0 = \displaystyle 2 \sqrt{ 1 + \dfrac{2}{\sqrt{5}}} \approx \boxed{2.75276384}$

Finally, we note that the trajectory taken by the moving particle is actually a branch of a hyperbola, and not an ellipse.

Used a completely numerical approach for this one. I treated $v_o$ as a tuning parameter and varied it till the final coordinates of the moving mass are $(4,1)$ . I did this in the following way:

• Set $v_o$ to some arbitrary positive real number.
• Ran a simulation until the Y coordinate becomes +1.
• Checked how close the corresponding X coordinate is to the value of +4.
• This process was repeated till the final X coordinate was sufficiently close to +4.

Simulation code attached below:

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clear all
clc

% vo used as a tuning parameter:
vo      = 2.7527596;

% Initial conditions:
x(1)    = -2;
y(1)    = 0;
dx(1)   = vo;
dy(1)   = 0;

% Time step and initialisation:
dt      = 1e-6;
t(1)    = 0;
k       = 1;

% Position vector of fixed mass:
rp      = [0;1;0];

% Run loop till y coordinate of moving mass becomes +1:
while y(k) <= 1

    % Position vector of moving mass:
    r          = [x(k);y(k);0];

    % Newton's law of gravitaion:
    Fg         = (rp - r)/(norm(rp - r)^3);

    % Newton's second law of motion:
    ddx        = Fg(1);
    ddy        = Fg(2);

    % Numerical integration for velocity components: Explicit Euler
    dx(k+1)    = dx(k) + dt*ddx;
    dy(k+1)    = dy(k) + dt*ddy;

    % Numerical integration for position components: Implicit Euler
    x(k+1)     = x(k) + dx(k+1)*dt;
    y(k+1)     = y(k) + dy(k+1)*dt;

    % Time advancement:
    t(k+1)     = t(k) + dt;

    % Index advancement:
    k          = k + 1;

end

% Tune vo till the final x-coordinate of the mass is +4:
X_END      = x(end);

% X_END     = 3.999988696701238
% vo        = 2.7527596