Chapter 9.3, Problem 54E

According to Newton’s Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from the earth’s surface is

$F=\frac{mg{R}^2}{{\left(x+R\right)}^2}$

where x = x(t) is the object’s distance above the surface at time t, R is the earth’s radius, and g is the acceleration due lo gravity. Also, by Newton’s Second Law, F = ma = m(dv/dt) and so

$m\frac{dv}{dt}=-\frac{mg{R}^2}{{\left(x+R\right)}^2}$

(a) Suppose a rocket is fired vertically upward with an initial velocity v₀. Let h be the maximum height above the surface reached by the object. Show that

$v_o=\sqrt{\frac{2gRh}{R+h}}$

[Hint: By the Chain Rule, m(dv/dt) = mv(dv/dx).]

(b) Calculate v_e = lim_h→∞v₀. This limit is called the escape velocity for the earth.

(c) Use R = 3960 mi and g = 32 ft/s² to calculate v_e in feet per second and in miles per second.