Question 7.14: Goal Apply Kepler’s third law to an Earth satellite Problem ...
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 the Earth. This can occur only if the satellite’s orbital period is the same as the Earth’s period of rotation, 24.0 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.
(a) Find the distance r to geosynchronous orbit.
Apply Kepler’s third law:
T^2 = (\frac{4π^2}{GM_E})r^3
Substitute the period in seconds, T = 86 400 s, the gravity constant G = 6.67 × 10^{-11} kg^{-1} m^3/s^2, and the mass of the Earth, M_E = 5.98 × 10^{24} kg. Solve for r:
r = 4.23 × 10^7 m
(b) Find the orbital speed:
v = \frac{d}{T} = \frac{2πr}{T} = \frac{2π(4.23 × 10^7 m)}{8.64 × 10^4 s} = 3.08 × 10^3 m/s
Remarks Both these results are independent of the mass of the satellite. Notice that Earth’s mass could be found by substituting the Moon’s distance and period into this form of Kepler’s third law.