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## Q. 7.13

GEOSYNCHRONOUS ORBIT AND TELECOMMUNICATIONS SATELLITES

GOAL Apply Kepler’s third law to an Earth satellite.

PROBLEM From a telecommunications point of view, it’s advantageous for satellites to remain at the same location relative to a location on Earth. This can occur only if the satellite’s orbital period is the same as the Earth’s period of rotation, approximately $24.0 \mathrm{~h}$. (a) At what distance from the center of the Earth can this geosynchronous orbit be found? (b) What’s the orbital speed of the satellite?

STRATEGY This problem can be solved with the same method that was used to derive a special case of Kepler’s third law, with Earth’s mass replacing the Sun’s mass. There’s no need to repeat the analysis; just replace the Sun’s mass with Earth’s mass in Kepler’s third law, substitute the period $T$ (converted to seconds), and solve for $r$. For part (b), find the circumference of the circular orbit and divide by the elapsed time.

## Verified Solution

(a) Find the distance $r$ to geosynchronous orbit.

Apply Kepler’s third law:

$T^2=\left(\frac{4 \pi^2}{G M_E}\right) r^3$

Substitute the period in seconds, $T=86400 \mathrm{~s}$, the gravity constant $G=6.67 \times 10^{-11} \mathrm{~kg}^{-1} \mathrm{~m}^3 / \mathrm{s}^2$, and the mass of the Earth, $M_E=5.98 \times 10^{24} \mathrm{~kg}$. Solve for $r$ :

$r=4.23 \times 10^7 \mathrm{~m}$

(b) Find the orbital speed.

Divide the distance traveled during one orbit by the period:

$v=\frac{d}{T}=\frac{2 \pi r}{T}=\frac{2 \pi\left(4.23 \times 10^7 \mathrm{~m}\right)}{8.64 \times 10^4 \mathrm{~s}}=3.08 \times 10^3 \mathrm{~m} / \mathrm{s}$

REMARKS Earth’s motion around the Sun was neglected; that requires using Earth’s “sidereal” period (about four minutes shorter). Notice that Earth’s mass could be found by substituting the Moon’s distance and period into this form of Kepler’s third law.