$2.6.37.$ The Molniya-1 communications satellite has a perigee over the southern hemisphere of the Earth at an altitude of about $500$ km, and an apogee at an altitude of about $40,000$ km over the northern hemisphere. What is the ratio of the angular velocities of rotation of this satellite at perigee and apogee?
Law of energy preservation $$\frac{mv^2_a}{2}-G\frac{mM}{R_a}=\frac{mv^2_p}{2}-G\frac{mM}{R_p}\quad (1)$$ Conservation of angular momentum $$mv_aR_a=mv_pR_p$$ From where velocity at closest position to the planet, at perigee $$v_p=v_a\frac{R_a}{R_p}\quad (2)$$ Let's substitute value of $v_p$ to $(1)$: $$\frac{v_a^2}{2} = \frac{GMR_p}{R_a(R_a+R_p)}$$ From where, we could value of the velocity at farthest position to the planet, at apogee $$v_a=\sqrt{\frac{2GMR_p}{(R_a+R_p)R_a}}$$ Considering dependence $v_p(v_a)$ from $(2)$, we obtain $$v_p=\sqrt{\frac{2GMR_a}{(R_a+R_p)R_p}}$$ By definition of angular velocity $$\omega_a = \frac{v_a}{R_a}=\sqrt{\frac{2GMR_p}{(R_a+R_p)R_a^3}}$$ $$\omega_p = \frac{v_p}{R_p}=\sqrt{\frac{2GMR_a}{(R_a+R_p)R_p^3}}$$ The desired relationship could be found by dividing one equation on another $$\xi = \frac{\omega_p}{\omega_a} = \frac{R_a^2}{R_p^2}\quad (3)$$ Considering the radius of the Earth $$R_a=R_a'+R_e$$ $$R_p=R_p'+R_e$$ After substituting into $(3)$, we are getting $$\boxed{\xi = \frac{(49400\text{ km})^2}{(6900\text{ km})^2} \approx 45}$$
$$\frac{\omega_p}{\omega_a}\approx 45$$
Almaskhan Arsen