Flying Over a Rod

The gravitational potential of the bar, with line mass density $\lambda$ , is $\begin{aligned} V(x,y) & = \; -G \int_{-1}^1 \frac{\lambda\,du}{\sqrt{(u-x)^2 + y^2}} \; = \; -G\lambda\Big[\sinh^{-1}\left(\tfrac{u-x}{y}\right)\Big]_{-1}^1 \; = \; -G\lambda\left[ \sinh^{-1}\left(\tfrac{1-x}{y}\right) + \sinh^{-1}\left(\tfrac{x+1}{y}\right)\right] \\ & = \; G\lambda \ln\left(\frac{x-1 + \sqrt{(x-1)^2 + y^2}}{x+1 + \sqrt{(x+1)^2 + y^2}}\right) \end{aligned}$ and the particle will satisfy the equations of motion $\ddot{x} \; = \; -\frac{\partial V}{\partial x} \hspace{2cm} \ddot{y} \; = \; -\frac{\partial V}{\partial y}$ together with the boundary conditions $x(0) = -2$ , $y(0)=0$ , $\dot{x}(0) = \dot{y}(0) = 1.2$ .

Manipulating the derivatives of $V$ so that they are not explicitly singular when $y=0$ , we obtain the equations (with $G=1$ and $\lambda=1$ ): $\begin{aligned} \ddot{x} & = \; \frac{1}{\sqrt{(x+1)^2 + y^2}} - \frac{1}{\sqrt{(x-1)^2 + y^2}} \\[2ex] \ddot{y} & = \; -\frac{4xy}{\sqrt{(x+1)^2 + y^2}\,\sqrt{(x-1)^2 + y^2}\,\big[(x+1)\sqrt{(x-1)^2 + y^2} + (x-1)\sqrt{(x+1)^2 + y^2}\big]} \end{aligned}$ These differential equations can be solved numerically, together with their initial conditions, and we obtain that $y = 0$ when $t = \boxed{2.7651704558}$ .

At a general time $t$ , let the position of the point mass be $\vec{r}_P = (x,y,0)$ and consider a rod element of length $dp$ near the point $\vec{r}_L = (p,0,0)$ . Then, applying Newton's law of gravitation, the force on the point mass is:

$d\vec{F} = \frac{GM \ dp \ (\vec{r}_L - \vec{r}_P)}{\lvert \vec{r}_L - \vec{r}_P \rvert^3}$

Then:

$\vec{F} = \int_{-1}^{1} \frac{GM \ (\vec{r}_L - \vec{r}_P)}{\lvert \vec{r}_L - \vec{r}_P \rvert^3} \ dp$

Applying Newton's second law, the acceleration vector of the point mass is:

$\vec{a} = (a_x,a_y,0) = \frac{\vec{F}}{M}$

Finally the resulting system of differential equations governing the motion of the point mass is:

$\ddot{x} = a_x \ ; \ \dot{x}(0) = 1.2 \ ; \ x(0)=-2$ $\ddot{y} = a_y \ ; \ \dot{y}(0) = 1.2 \ ; \ y(0)=0$

This problem has been handled numerically as follows:

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% Coded using Octave:
clear all
clc

% Spatial discretization of the rod:
dp     = 1e-3;

% Time step:
dt     = 1e-3;

% Initial conditions:
x(1)   = -2;
y(1)   = 0;
dx(1)  = 1.2;
dy(1)  = 1.2;
t(1)   = 0;

% Time index initialisation:
k      = 1;

% Simulating till the particle hits the ground again:
while y(k) >= 0

    % Initialisation of gravitational force:
    F      = [0;0;0];

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

    % Calculating gravitational force:
    for p = -1:dp:1

        % Position vector of rod element:
        rL       = [p;0;0];

        % Gravitational force on point mass due to rod element:
        dF       = (dp*(rL - rP))/(norm(rL - rP)^3);

        % Numerical integration: gravitational force due to entire rod:
        F        = F + dF;
    end

    % Acceleration components (unit point mass):
    ddx     = F(1);
    ddy     = F(2);

    % Numerical integration: Semi implicit Euler:
    dx(k+1) = dx(k) + ddx*dt;
    dy(k+1) = dy(k) + ddy*dt;

    x(k+1)  = x(k) + dx(k+1)*dt;
    y(k+1)  = y(k) + dy(k+1)*dt;

    t(k+1)  = t(k) + dt;
    k       = k + 1;
end

ANSWER   = t(end)
% ANSWER = 2.761