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A satellite is in a circular orbit 600 miles above the surface of the Earth. What is the period of the orbit?

You may use the following constants:
- Radius of the Earth: 4000 miles
- Gravitational Constant: [tex]6.67 \times 10^{-11} \, \text{m}^3/(\text{kg} \cdot \text{s}^2)[/tex]
- Mass of the Earth: [tex]5.98 \times 10^{24} \, \text{kg}[/tex]
- Number of meters in a mile: 1609

Period = ______ seconds

Answer :

Final answer:

To calculate the orbital period of a satellite, convert all measurements to the same unit (in this case, meters), then use Kepler's third law. Add the satellite's altitude to the earth's radius to find the total distance from the earth's center. Then, use the formula T = 2 * pi * sqrt(r³ /(G*M)) to find the period.

Explanation:

To compute the orbital period of a satellite, we can make use of Kepler's third law. First, we have to first convert the altitude and the radius of the earth from miles to meters: (600 miles * 1609 m/mile) = 965400 m and (4000 miles * 1609 m/mile) = 6437392 m. Then, the total distance from the center of the Earth to the satellite (its orbital radius) becomes r = altitude + radius of earth = 965400 m + 6437392 m = 7402792 m.

Then, Kepler's third law, T = 2 * pi * sqrt(r³ /(G*M)), allows us to calculate the orbital period, T, where G is the gravitational constant, and M is the mass of the Earth.

So, T = 2 * pi * sqrt((7402792 m)³ / (6.67 * 10⁻¹¹ m³/(kg * s²) * 5.98 x 10²⁴ kg)) = __ seconds.

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