To find the time it takes for the satellite to complete one orbit, we need to calculate the orbital period. Here are the steps to solve this:

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Step 1: Compute the radius of the orbit

The position of the satellite at a given moment is provided as Cartesian coordinates $(x, y, z)$. The distance of the satellite from the center of the Earth (i.e., the radius of the orbit $r$) can be calculated using the 3D distance formula:

$r = \sqrt{x^2 + y^2 + z^2}$

Substitute the given values $(x, y, z) = (5000, 12000, 7000)$:

$r = \sqrt{5000^2 + 12000^2 + 7000^2} \, \text{km}$

$r = \sqrt{25000000 + 144000000 + 49000000} \, \text{km}$

$r = \sqrt{218000000} \, \text{km}$

$r \approx 14758.3 \, \text{km}$

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Step 2: Calculate the circumference of the orbit

The orbit is circular, so its circumference $C$ is given by:

$C = 2\pi r$

Substitute $r \approx 14758.3 \, \text{km}$:

$C \approx 2 \cdot \pi \cdot 14758.3$

$C \approx 92726.1 \, \text{km}$

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Step 3: Compute the orbital period

The satellite’s speed $v$ is constant and equal to $1.5 \, \text{km/s}$. The orbital period $T$ is the time it takes to travel one full circumference of the orbit:

$T = \frac{C}{v}$

Substitute $C \approx 92726.1 \, \text{km}$ and $v = 1.5 \, \text{km/s}$:

$T = \frac{92726.1}{1.5}$

$T \approx 61817.4 \, \text{seconds}$

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Step 4: Convert the time to hours

To express $T$ in hours:

$T \, \text{(hours)} = \frac{T \, \text{(seconds)}}{3600}$

$T \approx \frac{61817.4}{3600}$

$T \approx 17.17 \, \text{hours}$

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Final Answer:

The satellite takes approximately 17.17 hours to complete one orbit around the Earth.

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Would you like me to explain any part of this in more detail?

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Related Questions:

1. How does the speed of the satellite affect its orbital period?
2. How would the orbital period change if the satellite were closer to the Earth?
3. Can this formula be applied to elliptical orbits? Why or why not?
4. What is the relationship between orbital speed and the radius of the orbit?
5. How does the gravitational force keep the satellite in orbit?

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Tip:

Remember that for circular orbits, the orbital period can also be calculated using Kepler’s third law if the mass of the central object (Earth, in this case) is known.