College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 2 - Problems - Page 67: 60


$h = 2643~km$

Work Step by Step

The gravitational field strength is $\frac{GM}{R^2}$, where $M$ is the Earth's mass, and $R$ is the distance from the center of the Earth. Our weight would be half of what it is at the Earth's surface if the gravitational field strength was half of what it is at the Earth's surface. Let $R_E$ be the radius of the Earth (which is 6380 km). Let $h$ be the altitude above the surface where the gravitational field strength is one-half of its value at the surface. We can find $h$: $\frac{GM}{(R_E+h)^2} = \frac{1}{2}~\frac{GM}{R_E^2}$ $(R_E+h)^2 = 2R_E^2$ $R_E+h = \sqrt{2}~R_E$ $h = R_E~(\sqrt{2}-1)$ $h = (6380~km)~(\sqrt{2}-1)$ $h = 2643~km$
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