A geostationary satellite is orbiting around an arbitrary planet $$P$$ at a height of $$11R$$ above the surface of $$P$$, $$R$$ being the radius of $$P$$. The time period of another satellite in hours at a height of $$2R$$ from the surface of $$P$$ is ________. has the time period of 24 hours.

For a geostationary satellite orbiting at a height of $$11R$$ above the surface, the orbital radius is $$r_1 = R + 11R = 12R$$ and its time period is $$T_1 = 24$$ hours.

For the second satellite at a height of $$2R$$, the orbital radius is $$r_2 = R + 2R = 3R$$.

By Kepler's third law, $$\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$$, which gives $$T_2 = T_1 \left(\frac{r_2}{r_1}\right)^{3/2} = 24 \left(\frac{3R}{12R}\right)^{3/2} = 24 \left(\frac{1}{4}\right)^{3/2}$$.

Now $$\left(\frac{1}{4}\right)^{3/2} = \frac{1}{4\sqrt{4}} = \frac{1}{8}$$, so $$T_2 = \frac{24}{8} = 3$$ hours.