To find the velocity of a satellite in a low Earth orbit, we can use the formula for the orbital velocity of a satellite. The formula can be derived from the gravitational force and the centripetal force required for circular motion.
The orbital velocity \( v \) is given by the equation:
\[ v = \sqrt{\frac{G m_p}{r}} \]
where:
- \( G \) is the universal gravitational constant,
- \( m_p \) is the mass of the planet,
- \( r \) is the distance from the center of the planet to the satellite.
In the case of a low Earth orbit, the distance \( r \) can be approximated as the radius of the Earth (about 6,371 km) plus the altitude of the satellite above the Earth. For a low Earth orbit at an altitude of approximately 200 km, this gives:
\[ r \approx 6,371 \text{ km} + 200 \text{ km} = 6,571 \text{ km} \]
Thus, we can express the orbital velocity in a low Earth orbit (for a satellite at about 200 km above the Earth's surface) as:
\[ v = \sqrt{\frac{G m_p}{6,571 \text{ km}}} \]
Given the options provided, the one that best resembles this expression is:
\[ v = \sqrt{\frac{G m_p}{(200 \text{ km})}} \]
If interpreted correctly with the understanding that the distance should incorporate the radius of the Earth plus the altitude of the orbit, the corresponding option that indicates usage of the gravitational constant and the mass yields the correct formula.
Therefore, the correct response to find the velocity of the satellite in a low Earth orbit is:
\[ v = \sqrt{\frac{G m_p}{(200 \text{ km})}} \]
Considering the question context, if only a response directly implies the formula structure, you should select the corresponding equivalent among the simplified notations which correctly reflects the orbital condition at mentioned altitudes.
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