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A satellite is in a circular orbit 500 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 Earth: [tex]5.98 \times 10^{24} \, \text{kg}[/tex]
- Number of meters in a mile: 1609

Answer :

Final answer:

To find the period of the orbit for a satellite 500 miles above earth, we can use Kepler's third law and plug in the given values (including the proper adjustments for units) into the equation for the period of the orbit.

Explanation:

Given this case, we're dealing with the satellite in a circular orbit, and we want to find the period of the orbit. The period can be calculated using Kepler's third law, which states that the square of the orbital period (T) is proportional to the cube of the distance from the center of the planet (which would be the sum of the Earth's radius and the height of the satellite above Earth).

Using the given values and the formula for the period of the orbit (T = 2π√(r³/μ) where r is the radius, in miles, and μ is the gravitational constant), we can compute the period:

r = (radius of the earth + distance from the earth to the satellite) X number of meters in a mile = (4000+500) X 1609 = 7.2x10⁶ m

μ = gravitational constant X mass of the earth = 6.67x10⁻⁶ m³/(kg⋅s²) X 5.98x10²⁴ kg = 3.986x10¹⁴ m³/s²

These values give us a rough approximation of the orbital period for the given satellite.

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