To find the velocity of a satellite in a geostationary orbit, we use the formula for the orbital velocity \( v \) of a satellite orbiting a celestial body, given by:
\[ v = \sqrt{\frac{G \cdot m_p}{r}} \]
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.
For a geostationary orbit, the satellite must be positioned above the equator and have an orbital radius equal to the distance from the center of the planet to the satellite's orbit. The height of a geostationary orbit above the Earth's surface is approximately 35,786 kilometers, and the Earth's radius is approximately 6,371 kilometers. Therefore, the total radius \( r \) for a geostationary satellite is:
\[ r \approx 35,786 , \text{km} + 6,371 , \text{km} \approx 42,157 , \text{km} \]
Thus, this value can be rounded to approximately 42,164 km for clarity.
Comparing this with the provided options, the correct equation to find the velocity of the satellite in a geostationary orbit would be:
v = √(Gmp / (42,164 km))
So the correct response is: v = √Gmp/(42,164 km).
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