Решение

$$R_c>>R$$

$$t=\frac{T}{2}$$

$$a=\frac{R_c+R}{2}$$

$$\frac{T^2}{a^3}=\frac{4\pi^2}{\gamma M}$$

$$\gamma M=gR^2$$

$$\frac{T^2}{4}=\frac{\pi^2a^3}{gR^2}=t^2$$

$$t^2=\frac{\pi^2(R+R_c)^3}{gR^2}$$

$$\boxed{t=\pi\sqrt{\frac{R}{g}}[\frac{1+\frac{R_c}{R}}{2}]^{\frac{3}{2}}}$$