Решение

$$m \frac{(v + \omega R)^2}{R} = G \frac{M m}{R^2} + N_1$$

$$m \frac{(v - \omega R)^2}{R} = G \frac{M m}{R^2} - N_2$$

$$\Delta N = \frac{m}{R} \left( (v + \omega R)^2 - (v - \omega R)^2 \right)$$

$$\omega = \frac{2\pi}{T}, \quad T \approx 24 \cdot 3600 \ \text{c}$$

$$\Delta N = \frac{m}{R} \left( v^2 + 2v\omega R + (\omega R)^2 - v^2 + 2v\omega R - (\omega R)^2 \right) =$$

$$= \frac{m}{R} \cdot 4 v \omega R = 4 v \omega m = 4 v m \frac{2\pi}{T} = \frac{8 \pi v m}{T}$$

$$\Delta N = \frac{8 \pi v m}{T} \approx 9 \cdot 10^3 \ \text{H}$$