$3.2.9.$ 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)?
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 $$ \Delta T_1=T_1-T_0=T_0(\frac{R+H}{R}-1)=2\text{ min} $$ $$ \Delta T_2=T_2-T_0=T_0(\frac{R+h}{R}-1)=6.75\text{ s} $$
$$\Delta T_1=2\text{ min};\quad\Delta T_2=6.75\text{ s}$$