Question 54: Gravitation

Question

A geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance $r_2$ from the center of the earth, such that $r_1=1.21 r_2$. The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is _________.

Accepted Answer

For satellites, the time period $ T $ is proportional to the distance to the power of $ \frac{3}{2} $:
$ T \propto r^{\frac{3}{2}} $
This gives us the ratio of the time periods for the second and first satellites:
$ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{\frac{3}{2}} $
Converting this to angular velocity, $ \omega $, which is inversely proportional to the time period, we have:
$ \frac{\omega_2}{\omega_1} = \left(\frac{r_1}{r_2}ight)^{\frac{3}{2}} = (1.21)^{\frac{3}{2}} $
Calculating $ (1.21)^{\frac{3}{2}} $, we find:
$ \omega_2 = \omega_1 	imes 1.331 $
The combined angular displacement of both satellites is $ 2\pi $ radians in the same time period $ t_0 $:
$ (\omega_2 + \omega_1) t_0 = 2\pi $
Solving for $ t_0 $:
$ t_0 = \frac{2\pi}{\omega_2 + \omega_1} = \frac{2\pi}{\left(\frac{4}{3} + 1ight) \omega_1} = \frac{6\pi}{7\omega_1} $
Since $ T_{GSS} $, the period of the geostationary satellite, is 24 hours, we substitute:
$ t_0 = \frac{6\pi}{2\pi} \frac{T_{GSS}}{7} = \frac{3 	imes 24 	ext{ hours}}{7} = \frac{24}{p} 	ext{ hours} $
Therefore, solving for $ p $:
$ p = \frac{7}{3} = 2.33 $

Gravitation

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