Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems: 76


height = 2640 km

Work Step by Step

$\frac{1}{2} \times G\frac{m}{r^2} = G\frac{m}{(\sqrt{2}~r)^2}$ We can see that the acceleration due to gravity will be half when the distance from the center of the Earth is $\sqrt{2}~r$. $\sqrt{2}~r = \sqrt{2}\times (6380 ~km) = 9020 ~km$ The height above the Earth's surface is 9020 km - 6380 km which is 2640 km. At a height of 2640 km above the Earth's surface, the acceleration due to gravity is half the value of the acceleration due to gravity at the Earth's surface.
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