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)?

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

    <h3 id="back-link"><a href="/#3.2">$\leftarrow$Back</a></h3>

    <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
$$\Delta T_1=T_1-T_0=T_0(\frac{R+H}{R}-1)=2\text{ min}$$

$$\Delta T_2=T_2-T_0=T_0(\frac{R+h}{R}-1)=6.75\text{ s}$$

    </p>
	  <p style="text-align: right; font-style: italic; font-size: 14;">
		  Dzikan Mikita<br>
    </p>

    <h4>Answer</h4>
    <p>
        $$\Delta T_1=2\text{ min};\quad\Delta T_2=6.75\text{ s}$$
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>