Cannonball's orbit

$If\space v_{escape}\space is\space the\space escape\space velocity\space at\space the\space launching\space site$ $\Rightarrow v_{escape}=\sqrt{2\dfrac{GM_⊕}{R_⊕}}\approx 1.1186 \times 10^4 ms^{-1}$ $\Rightarrow v>v_{escape}...........[1]\space\blue{(v=1.2\times 10^4 ms^{-1})}$ $From\space Kepler's\space first\space law\space we\space know\space that$ $the\space shape\space of\space the\space orbit\space can$ $be\space described\space by\space the\space following \space polar\space equation:$ $r=\dfrac{l}{1+e_1\cos\theta}..........[2]$ $if\space l=\dfrac{q^2}{p},e_1=\sqrt{1-\dfrac{q^2}{p^2}}$ $then \space from \space here \space we\space know\space that$ $p=\dfrac{R_{⊕}GM_⊕}{2GM_⊕-R_⊕v^2}\approx -2.1118\times 10^7 m$ $Also\space we\space know\space that\space in\space [2]$ $R_{⊕}=\dfrac{q^2}{p+\sqrt{p^2-q^2}}$ $\Rightarrow q^4-q^2(2pR_⊕-R^2_⊕)=0$ $\because q^2=0\iff e_1=1 \iff v=v_{escape}$ $\therefore [1]\Rightarrow q^2=2pR_⊕-R^2_⊕\approx -3.097×10^{14}m^2$ $\Rightarrow e_1=\sqrt{1-\dfrac{q^2}{p^2}}\approx 1.3016$ $\Rightarrow \lfloor 10e_1\rfloor =\lfloor 13.016\rfloor=13$

Let the center of the earch have the coordinates $(0,0)$ . The X axis points horizontally to the right while the Y axis points vertically up. The cannonball is projected tangentially to the surface of the earth and not perpendicular to the surface of the earth. Please edit the problem statement. At any general time $t$ , let the position vector of the ball with respect to the origin be:

$\vec{r} = x \ \hat{i} + y \ \hat{j} + z \ \hat{k}$

Now, applying Newton's law of gravitation leads to:

$\frac{d^2 \vec{r}}{dt^2}=-\frac{GM \vec{r}}{\lvert \vec{r} \rvert^3}$

$\vec{r}(0) = R \ \hat{j}$ $\frac{d\vec{r}(0)}{dt} = v_o \ \hat{i}$

Let us introduce a change of variables as such:

$\vec{p} = R \vec{r}$

This transforms the equations of motion to:

$\frac{d^2 \vec{p}}{dt^2}=-\frac{GM}{R^3}\left(\frac{ \vec{p}}{\lvert \vec{p} \rvert^3}\right)$

Let us introduce a new time scale $t_1$ such that:

$t = \lambda t_1$

$\implies \frac{d\vec{p}}{dt} = \frac{d\vec{p}}{dt_1} \frac{dt_1}{dt}$ $\frac{d\vec{p}}{dt} = \frac{1}{\lambda} \frac{d\vec{p}}{dt_1}$

Now:

$\frac{d^2\vec{p}}{dt^2} = \frac{d}{dt} \left(\frac{d\vec{p}}{dt} \right) = \frac{d}{dt} \left( \frac{1}{\lambda} \frac{d\vec{p}}{dt_1} \right) =\frac{d}{dt_1} \left( \frac{1}{\lambda} \frac{d\vec{p}}{dt_1} \right) \frac{dt_1}{dt}$ $\implies \frac{d^2\vec{p}}{dt^2} = \frac{1}{\lambda^2} \frac{d^2\vec{p}}{dt_1^2}$

Plugging this into the equation of motion leads to:

$\frac{d^2 \vec{p}}{dt^2}=-\frac{GM}{R^3}\left(\frac{ \vec{p}}{\lvert \vec{p} \rvert^3}\right)$ $\frac{1}{\lambda^2} \frac{d^2\vec{p}}{dt_1^2}=-\frac{GM}{R^3}\left(\frac{ \vec{p}}{\lvert \vec{p} \rvert^3}\right)$

Taking:

$\lambda^2 = \frac{R^3}{GM}$

converts the equation of motion to:

$\frac{d^2\vec{p}}{dt_1^2}=-\left(\frac{ \vec{p}}{\lvert \vec{p} \rvert^3}\right)$

The initial conditions also need to be scaled appropriately, which are originally:

$\vec{r}(0) = R \ \hat{j}$ $\frac{d\vec{r}(0)}{dt} = v_o \ \hat{i}$

$\implies \vec{p}(0) = 1 \ \hat{j}$ $\implies \frac{d\vec{r}(0)}{dt} =R \frac{d\vec{p}(0)}{dt} = v_o \ \hat{i}$ $\implies \frac{d\vec{p}(0)}{dt} = \frac{v_o}{R} \ \hat{i}$ $\implies \frac{d\vec{p}(0)}{dt} = \frac{1}{\lambda} \frac{d\vec{p}(0)}{dt_1} = \frac{v_o}{R} \ \hat{i}$ $\implies \frac{d\vec{p}(0)}{dt_1} = \sqrt{\frac{R^3}{GM}} \left(\frac{v_o}{R}\right) \ \hat{i}$

Finally, the rescaled equation of motion is:

$\frac{d^2\vec{p}}{dt_1^2}=-\left(\frac{ \vec{p}}{\lvert \vec{p} \rvert^3}\right)$ $\vec{p}(0) = 1 \ \hat{j}$ $\frac{d\vec{p}(0)}{dt_1} = \sqrt{\frac{R^3}{GM}} \left(\frac{v_o}{R}\right) \ \hat{i}$

This set of ODEs has been solved numerically using a script of code which will be attached to the end of this solution.

The resulting trajectory is a hyperbola. This is established by the fact that the trajectory becomes a straight line as it moves further away from the earth. This corresponds to the trajectory converging to the asymptote of the hyperbola. The magnitude of the slope of the asymptote is:

$m=\lvert \frac{a}{b} \rvert$ \]

Where $a$ and $b$ are the semi axes of the hyperbola. The eccentricity of the hyperbola is:

$e=\sqrt{1 +\frac{1}{m^2}} \approx 1.3$

The required answer is $\boxed{13}$ . The scrip of code created on OCTAVE is as follows.

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clear all
clc

% Parameters:
G       = 6.6743*10^(-11);
R       = 6.371*10^6;
M       = 5.972*10^(24);

% Projection speed:
vo      = 12000;

% Scaled initial conditions:
dx(1)   = vo*sqrt(R^3/(G*M))/R;
dy(1)   = 0;

x(1)    = 0;
y(1)    = 1;

% Time step and time vector initialisation:
dt      = 1e-5;
tf      = 50;
t       = 0:dt:tf;

% Solving the equation of motion numerically:
for k = 1:length(t)-1

    % Scaled position vector:
    p         = [x(k);y(k);0];

    % Equation of motion:
    ddr       = -p/(norm(p)^3);

    % Extracting acceleration components:
    ddx       = ddr(1);
    ddy       = ddr(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;

end

% PLotting actual trajectory (By rescaling back to original coordinates):
plot(x*R,y*R,'Linewidth',1.5)
grid on
xlabel('X [m]')
ylabel('Y [m]')
title('Cannonball Trajectory')

% Calculating slop of hyperbla asymptote:
m       = (y(end)-y(end-1))/(x(end)-x(end-1));

% Calculating eccentricity:
e       = sqrt(1 + (1/m)^2);
ANSWER  = floor(10*e)

% ANSWER = 13