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+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#3.2">$\leftarrow$Back</a></h3>
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+ <h3> Statement </h3>
+ <p>
+ $3.2.9.$ By how much will a pendulum clock raise to the height of Everest ($8.9$ km) lag behind in a day? Ostankino Tower ($0.5$ km)?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ The period of oscillation of a mathematical pendulum
+$$ T_0=2\pi\sqrt{\frac{l}{g}}; \quad T_1=2\pi\sqrt{\frac{l}{g^*}} $$
+Acceleration of gravity depending on the distance to the center of the Earth of mass $M$
+$$ g=\frac{GM}{R^2} $$
+
+$$ g^*=\frac{GM}{(R+H)^2} $$
+The ratio of accelerations of gravity for different distances to the center
+$$ \frac{g}{g^*}=\frac{(R+H)^2}{R^2}\Rightarrow g^*=\frac{gR^2}{(R+H)^2} $$
+Using the approximation for a small value of $x =\frac{h}{R} \ll 1$; $(1+x)^\alpha\approx 1+\alpha x$:
+$$ T_1=2\pi\frac{R+H}{R}\sqrt{\frac{l}{g}} $$
+From where we find the required lag as
+$$ \Delta T_1=T_1-T_0=T_0(\frac{R+H}{R}-1)=2\text{ min} $$
+
+$$ \Delta T_2=T_2-T_0=T_0(\frac{R+h}{R}-1)=6.75\text{ s} $$
+
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+ Dzikan Mikita<br>
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+ <h4>Answer</h4>
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+ $$\Delta T_1=2\text{ min};\quad\Delta T_2=6.75\text{ s}$$
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