In the vicinity of the ore deposit the pendulum oscillation period changed by 0.1 \%. The density of the ore in the deposit is 8 \cdot 10^3 \frac{kg}{m^3} Estimate the size of the deposit if the average density of the Earth is 5.6 \cdot 10^3 \frac{kg}{m^3} and its radius is 6400 km.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $3.2.8.$ In the vicinity of the ore deposit the pendulum oscillation period changed by $0.1 \%$. The density of the ore in the deposit is $8 \cdot 10^3$ $\frac{kg}{m^3}$ Estimate the size of the deposit if the average density of the Earth is $5.6 \cdot 10^3$ $\frac{kg}{m^3}$ and its radius is $6400$ km.
    </p>

    <h3>Solution</h3>
    <p>
        Let us write down, using the law of gravity, the equations for determining the acceleration of free fall, taking into account that the acceleration above the deposit will be greater than far from it.

$$g_{0}=\frac{4}{3}\pi RG\rho_{0}\approx 4\pi RG\rho_{0}$$

$$g=g_{0}+g_{k}\approx g_{0} + 4\pi RG(\rho - \rho_{0})$$
Let us write down the ratio of the periods of oscillation of the pendulum given by the conditions of the problem
$$\alpha =\frac{T-T_{0}}{T_{0}} = 0,1;\quad\xi = \frac{T}{T_{0}}=1+10^{-3}$$
The ratio of the periods is expressed through the values ​​of the acceleration of gravity
$$\xi^{2} = \frac{g_{0}+g}{g_{0}}=1+\frac{r(\rho -\rho_{0})}{R\rho_{0}}$$

$$\boxed{r=\frac{(\xi^{2}-1)R\rho_{0}}{\rho -\rho_{0}}\approx 30\text{ km}}$$

Dzikan Mikita
Aliaksandr Kanashenka

    <h4>Answer</h4>
    <p>
        $$r\approx30\text{ km}$$
    </p>


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