Chapter 3.4, Problem 25E

Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the distance between them. That is, $F_g=G{M}_1{M}_2/r^2$, where $M_1$ and $M_2$ are the masses of the objects, $r$ is the distance between them(center to center), $F_g$ is the attractive force, and $G$ is the constant of proportionality. Consider a projectile of constant mass $m$ being fired vertically from Earth(see Figure $3.12$). Let $t$ represent time and $v$ the velocity of the projectile.

Figure $3.12$Projectile escaping from Earth

a. Show that the motion of the projectile, under Earth’s gravitational force, is governed by the equation

$\frac{dv}{dt}=-\frac{g{R}^2}{r^2}$,

where $r$ is the distance between the projectile and the center of Earth, $R$ is the radius of Earth, $M$ is the mass of Earth, and $g=GM/R^2$.

b. Use the fact that $dr/dt=v$ to obtain

$v\frac{dv}{dt}=-\frac{g{R}^2}{r^2}$

c. If the projectile leaves Earth’s surface with velocity $v_0$, show that

$v^2=\frac{2g{R}^2}{r}+v_0^2-2gR$.

d. Use the result of part $\left(c\right)$ to show that the velocity of the projectile remains positive if and only if $v_0^2-2gR>0$. The velocity $v_e=\sqrt{2gR}$ is called the escape velocity of Earth.

e. If $g=9.81m/se{c}^2$ and $R=6370km$ for Earth, what is Earth’s escape velocity?

f. If the acceleration due to gravity for the moon is $g_m=g/6$ and the radius of the moon is $R_m=1738km$, what is the escape velocity of the moon?