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 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. SOLUTION (A) Find the distance r to geosynchronous orbit. Apply Kepler's third law. T2 = 4π2 GME r3 Substitute the period in seconds, T = 86,400 s, the gravity constant G = 6.67 10-11 kg−1 m3/s2, and the mass of the Earth, ME = 5.98 1024kg. Solve for r. r = 4.23 107 m (B) Find the orbital speed. Divide the distance traveled during one orbit by the period. v = d T = 2πr T = 2π(4.23 ✕ 107 m) 8.64 ✕ 104 s = 3.08 103 m/s LEARN MORE 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. QUESTION If the satellite was placed in an orbit three times farther away, about how long would it take to orbit the Earth once? Answer in days, rounding to one significant figure.

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter6: Uniform Circular Motion And Gravitation

Section: Chapter Questions

Problem 50PE: Construct Your Own Problem On February 14, 2000, the NEAR spacecraft was successfully inserted into...

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Question

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 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.

SOLUTION

(A) Find the distance r to geosynchronous orbit.

Apply Kepler's third law.

T² =

4π²

GM_E

r³

Substitute the period in seconds, T = 86,400 s, the gravity constant G = 6.67 10^-11 kg⁻¹ m³/s², and the mass of the Earth, M_E = 5.98 10²⁴kg. Solve for r.

r = 4.23 10⁷ m

(B) Find the orbital speed.

Divide the distance traveled during one orbit by the period.

v =

2πr

2π(4.23 ✕ 10⁷ m)

8.64 ✕ 10⁴ s

= 3.08 10³ m/s

LEARN MORE

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.

QUESTION If the satellite was placed in an orbit three times farther away, about how long would it take to orbit the Earth once? Answer in days, rounding to one significant figure.

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