A remark on problem about trajectory of a projectile in the atmosphere

[wrobel]

This is just to share some teaching experience.
A problem about trajectory of a projectile in the atmosphere is very well known. The trajectory has a vertical asymptote. This fact is easy to obtain by using the stability theory.

Consider equations of motion. Let $\boldsymbol v=v_x\boldsymbol e_x+v_y\boldsymbol e_y$ be the velocity of the projectile relative the standard Earth fixed frame $xy$ with $y$ directed upwards.
The second Newton
$m\boldsymbol {\dot v}=m\boldsymbol g-\gamma |\boldsymbol v|\boldsymbol v;$
implies
$m\dot v_x=-\gamma v_x\sqrt{v_x^2+v_y^2},\quad \gamma=const>0,$
$m\dot v_y=-mg-\gamma v_y \sqrt{v_x^2+v_y^2}.$
This system has an equilibrium: $v_x=0,\quad v_y=v^*:=-\sqrt{mg/\gamma}.$
This equilibrium is exponentially stable: eigenvalues of linearized system around this equilibrium are both negative so that
$v_x=O(e^{-c_x t}),\quad v_y=v^*+O(e^{-c_yt}),\quad c_x,c_y>0,\quad t\to\infty;$
and
$x(t)=x_0+\int_0^tv_xd\tau\to const,\quad y(t)\sim v^*t.$
Besides the quadratic model of atmosphere drag there are a lot of other models where the stability theory works but an explicit integration is impossible

 

[Greg Bernhardt]

Why does the projectile’s velocity approach a constant value instead of increasing or decreasing indefinitely?

 

[wrobel]

Each solution is bounded for $t\ge 0$. This follows from the estimate
$$\frac{m}{2}\frac{d}{dt}|\boldsymbol v|^2=m(\boldsymbol g,\boldsymbol v)-\gamma|\boldsymbol v|^3\le m|\boldsymbol g||\boldsymbol v|-\gamma|\boldsymbol v|^3.$$
From the first equation we have
$$v_x(t)=v_x(0)e^{-\frac{\gamma}{m}\int_0^t|\boldsymbol v|d\tau}.$$
Thus $v_x\to 0$ as $t\to \infty$ and the
$\omega$-limit set of any solution belongs to the line $\{v_x=0\}$. Only the set $\{v_x=0,\quad v_y=v^*\}$ is an invariant set that this line contains.