Canonical Projectile Problem: Finding the Escape Velocity of the Earth

Abstract

The escape velocity of the Earth is calculated using an idealized projectile model that allows for the determination of projectile velocity as a function of its altitude. In order to obtain an approximate solution for projectile altitude as a function of time which cannot be determined by an exact solution the concept of regular perturbation theory in ordinary differential equations is introduced as a pastoral interlude. Then a regular perturbation expansion is performed on the model to obtain the desired asymptotic solution of altitude as a function of time when the initial projectile velocity is much less than that of the escape velocity. An energy argument making use of the fact that gravity acts as a conservative force for this canonical model is also introduced to examine this phenomenon in more detail. The problems extend these analyses to the rest of the solar system planets and to two other canonical projectile problems that are nonconservative.

Problems

2.1.

Compute g and $V_e$ for all the other planets in both metric and British engineering units using the values in Table 2.1. Note that there are 3,280 ft in a km. Compare these values to that for Earth.

2.2.

Consider a projectile of mass m that is shot vertically upward from the surface of the Earth with an initial velocity $V_0$. Assume that the gravitational force acts downward at a constant acceleration g while the force of air resistance has a magnitude proportional to the square of the velocity with proportional constant $k > 0$ and acts to resist motion. Let $x=x(t)$ denote the altitude of the projectile at time t and $v(t) = dx(t)/dt$ be its velocity.

• (a) Explain why the governing equation of motion for the projectile is given by

  $$\begin{aligned} {\left\{ \begin{array}{ll} m\frac{dv}{dt} = -kv^2-mg, &{} v> 0 \\ m\frac{dv}{dt} = kv^2 - mg, &{}v < 0 \end{array}\right. } \text{ for } t > 0; \end{aligned}$$
  $$\begin{aligned} x(0) = 0, \, v(0) = V_0. \end{aligned}$$

• (b) Introduce the change of variables $V(x)=v[t(x)]$ and define $V_1(x)=V^2(x)$ for $V(x) > 0$ and $V_2(x)=V^2(x)$ for $V(x) < 0$. Show that $V_1(x)$ and $V_2(x)$ satisfy the linear first-order ordinary differential equations

  $$\begin{aligned} \frac{dV_1}{dx} + 2\frac{k}{m}V_1= & {} -2g, \, V_1(0) = V_0^2; \\ \frac{dV_2}{dx} - 2\frac{k}{m}V_2= & {} -2g, \, V_2(x_m) = 0; \end{aligned}$$

  where $x_m$ is defined implicitly by $V_1(x_m ) = 0$.

• (c) Solve these equations explicitly for $V_1(x)$ and $V_2(x)$ and show that

  $$\begin{aligned} V_2(0) = \frac{V_0^2}{1+kV_0^2/(mg)} < V_0^2 = V_1(0). \end{aligned}$$

  Discus this result in the context of (2.6.14).

2.3.

In the projectile problem, when air resistance is taken into account but the variation of gravitational force with altitude is neglected, the governing dimensional ordinary differential equation and initial conditions are

$$\begin{aligned} m\frac{d^2x^*}{d{t^*}^2} + k\frac{dx^*}{dt^*} = -mg; \, x^*(0) = 0, \, \frac{dx^*(0)}{dt^*} = V_0, \end{aligned}$$

where $x^*= x^*(t^*)\equiv$ its altitude and $t^*\equiv$ time should air resistance be assumed to exert a force proportional to the speed of the projectile with proportionality constant $k > 0$ while acting in a direction to oppose motion.

• (a) When the effect of air resistance is small, the change of variables

  $$\begin{aligned} t = \frac{t^*}{V_0/(2g)}, \, x(t) = \frac{x^*(t^*)}{V_0^2/(2g)} \end{aligned}$$

  is still appropriate (why does this follow from the concepts of scaling and introduction of nondimensional variables?). Show that this change of variables transforms the governing equation and initial conditions into the dimensionless form for $x=x(t;\beta )$

  $$\begin{aligned} \ddot{x} + \beta \dot{x} = -\frac{1}{2}; \, 0< \beta = k\frac{V_0}{2mg}<< 1, \, t >0; \, x(0;\beta ) = 0, \, \dot{x}(0;\beta ) = 1; \end{aligned}$$

  where $\dot{(-)}\equiv d(-)/dt$.

• (b) Seek a regular perturbation expansion of the solution to this dimensionless problem of the form

  $$\begin{aligned} x(t;\beta ) = x_0(t) + \beta x_1(t) + \beta ^2 x_2(t) + O(\beta ^3). \end{aligned}$$

  Proceed as in the corresponding problem treated in Section 2.5 to find

  $$\begin{aligned} t_m = 2[1 + b_1\beta + b_2 \beta ^2 + O(\beta ^3)],\, x_m = x(t_m;\beta ) \text{ to } O(\beta ^2). \end{aligned}$$

  Also compute T such that $x(T;\beta ) = 0$ to that order and compare $t_m$ with $T-t_m$. Discuss this result in the context of (2.6.14).

• (c) Inverting $x = x(t;\beta )$ to obtain $t = t(x;\beta )$ and defining $V(x;\beta ) = \dot{x}[t(x;\beta );\beta ]$, demonstrate that this change of variables transforms the nondimensional problem into

  $$\begin{aligned} V\frac{dV}{dx} + \beta V = -\frac{1}{2}; \, V(0;\beta ) = 1. \end{aligned}$$

  Solve this separable first-order ordinary differential equation and obtain an implicit relationship satisfied by V and x.

• (d) Given that a power series can be integrated term-by-term in its interval of convergence, deduce from the geometric series

  $$\begin{aligned} \frac{1}{1+\varepsilon } = \sum _{n=0}^{\infty }{(-1)^n \varepsilon ^n} \text{ for } |\varepsilon | < 1, \end{aligned}$$

  that

  $$\begin{aligned} \ln (1+\varepsilon ) \sim \varepsilon - \frac{\varepsilon ^2}{2} + \frac{\varepsilon ^3}{3} - \frac{\varepsilon ^4}{4} \text{ for } |\varepsilon | < 1. \end{aligned}$$

• (e) Determine an explicit formula for $x_m$ as a function of $\beta$ from the relation of part (c) where $V(x_m;\beta ) = 0$. Check the result of part (b) by using the Maclaurin series of part (d) to obtain a power series for $x_m$ from this formula through terms of $O(\beta ^2)$ assuming $\beta$ “small” in some sense. Discuss this restriction in the context of the result of part (b) with respect to the behavior of T when compared to $2t_m$.