Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 372: 9

Answer

$2.431\;\rm km$

Work Step by Step

According to table 13.1, the free-fall acceleration at sea level is given by $$g_{\rm ground}=\bf9.83\;\rm m/s^2$$ where $$g_{\rm ground}=\dfrac{GM_E}{R_E^2}=\bf 9.83\;\rm m/s^2\tag 1$$ And our sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is $$g_{\rm mountain}=9.83-0.0075 =\bf 9.8225\;\rm m/s^2$$ where $$g_{\rm mountain}=\dfrac{GM_E}{(R_E+h)^2}=\bf 9.8225\;\rm m/s^2\tag 2$$ Now we need to find $h$ which is the height of the mountain. Solving (1) for $GM_E$ and then plug it into (2); $$\dfrac{9.83R_E^2}{(R_E+h)^2}= 9.8225 $$ $$ 9.83R_E^2 = 9.8225(R_E+h)^2 $$ $$\dfrac{ 9.83}{9.8225}R_E^2 = (R_E+h)^2 $$ Taking the square toot for both sides; $$ \sqrt{\dfrac{9.83}{9.8225}}R_E = R_E+h $$ Thus, $$h=\sqrt{\dfrac{9.83}{9.8225}}R_E - R_E=R_E\left[\sqrt{\dfrac{9.83}{9.8225}}-1\right]$$ Plugging $R_E$ from table 13.2; $$h =(6.37\times 10^6)\left[\sqrt{\dfrac{9.83}{9.8225}}-1\right]$$ $$h=\color{red}{\bf 2.431\times 10^3}\;\rm m$$
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