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Given that the Moon orbits Earth every 27.3 days at an average distance of 3.84 x 10^8 m, use Kepler's third law to find the period of an artificial satellite orbiting at a distance of 7880 km from Earth's center.
A27.3 days
B1.93 hours
C193 hours
D0.193 hours
Answer & Solution
Correct answer: B. 1.93 hours
1. Kepler's third law gives T2 = T1 times the quantity (r2/r1) raised to the three halves power, using the Moon as satellite 1.
2. The radius ratio is 7880 km divided by 3.84 x 10^5 km, which is about 0.0205.
3. Raising 0.0205 to the three halves power gives about 0.00294.
4. Multiplying by the Moon's period, 27.3 days times 24.0 hours per day times 0.00294, gives T2 = 1.93 hours.
5. 27.3 days is simply the Moon's own period left unconverted, and 193 hours and 0.193 hours each misplace the decimal point by one order of magnitude from the correctly scaled result.
_Source: OpenStax College Physics (CC BY 4.0), Ch 6 "Uniform Circular Motion and Gravitation", section 6.6 Satellites and Kepler's Laws: An Argument for Simplicity_
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