Answer
$v=11.2$ km/s
Work Step by Step
$vdv=ady$
$\int_v^0vdv=-g_0R^2\int_0^{\infty}\frac{dy}{(R+y)^2}$
$\frac{v^2}{2}|_v^0=\frac{g_0R^2}{R+y}|_0^{\infty}$
$v=\sqrt{2g_0R}$
$v=\sqrt{2(9.81)(6356)(10)^3}$
$v=11167$ m/s $= 11.2$ km/s
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
Published by
Pearson
ISBN 10:
0133915425
ISBN 13:
978-0-13391-542-6
Chapter 12 - Kinematics of a Particle - Section 12.2 - Rectilinear Kinematics: Continuous Motion - Problems - Page 19: 33
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