Question 13.5: A Geosynchronous Satellite Consider a satellite of mass m mo...
A Geosynchronous Satellite
Consider a satellite of mass m moving in a circular orbit around the Earth at a constant speed v and at an altitude h above the Earth’s surface as illustrated in Figure 13.9.
(A) Determine the speed of the satellite in terms of G, h, R_E (the radius of the Earth), and M_E (the mass of the Earth).
(B) If the satellite is to be geosynchronous (that is, appearing to remain over a fixed position on the Earth), how fast is it moving through space?
(A) Conceptualize Imagine the satellite moving around the Earth in a circular orbit under the influence of the gravitational force. Figure 13.9 is a polar view of this motion. This motion is similar to that of the International Space Station, the Hubble Space Telescope, and other objects in orbit around the Earth.
Categorize The satellite moves in a circular orbit at a constant speed. Therefore, we categorize the satellite as a particle in uniform circular motion as well as a particle under a net force.
Analyze The only external force acting on the satellite is the gravitational force from the Earth, which acts toward the center of the Earth and keeps the satellite in its circular orbit.
Apply the particle under a net force and particle in uniform circular motion models to the satellite:
F_g=m a \rightarrow G \frac{M_E m}{r^2}=m\left(\frac{v^2}{r}\right)Solve for v, noting that the distance r from the center of the Earth to the satellite is r=R_E+h :
(1) v=\sqrt{\frac{G M_E}{r}}=\sqrt{\frac{G M_E}{R_E+h}}
(B) To appear to remain over a fixed position on the Earth, the period of the satellite must be 24 h = 86 400 s and the satellite must be in orbit directly over the equator.
Solve Kepler’s third law (Equation 13.11, with a = r and M_S \rightarrow M_E ) for r:
T^2=\left(\frac{4 \pi^2}{G M_S}\right) a^3=K_S a^3 (13.11)
r=\left(\frac{G M_E T^2}{4 \pi^2}\right)^{1 / 3}Substitute numerical values:
\begin{aligned}r& =\left[\frac{\left(6.674 \times 10^{-11} N \cdot m^2 /kg^2\right)\left(5.97 \times 10^{24} kg\right)(86400 s)^2}{4 \pi^2}\right]^{1 / 3} \\& =4.22 \times 10^7 m \end{aligned}Use Equation (1) to find the speed of the satellite:
\begin{aligned}v & =\sqrt{\frac{\left(6.674 \times 10^{-11} N \cdot m^2 /kg^2\right)\left(5.97 \times 10^{24} kg\right)}{4.22 \times 10^7 m}} \\& =3.07 \times 10^3 m/s\end{aligned}Finalize The value of r calculated here translates to a height of the satellite above the surface of the Earth of almost 36 000 km. Therefore, geosynchronous satellites have the advantage of allowing an earthbound antenna to be aimed in a fixed direction, but there is a disadvantage in that the signals between the Earth and the satellite must travel a long distance. It is difficult to use geosynchronous satellites for optical observation of the Earth’s surface because of their high altitude.