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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 {mgR^2}{(x + R)^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 to gravity. Also, by Newton's Second Law, $ F = ma = m (dv/dt) $ and so

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

(a) Suppose a rocket is fired vertically upward with an initial velocity $ v_0. $ Let $ h $ be the maximum height above

a) $V_{o}=\sqrt{\frac{2 g R h}{h+R}}$

b) $\sqrt{2 g R}$

c) $36580 f t / s, 6.93 \mathrm{mi} / \mathrm{s}$

Differential Equations

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Missouri State University

Harvey Mudd College

Idaho State University

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