To find the velocity of a satellite in a geostationary orbit, we can use the formula for orbital velocity for a circular orbit:
\[ v = \sqrt{\frac{G \cdot m_p}{r}} \]
where:
- \( v \) is the orbital velocity,
- \( G \) is the universal gravitational constant,
- \( m_p \) is the mass of the planet (or the central body, in this case),
- \( r \) is the distance from the center of the planet to the satellite.
In a geostationary orbit, the satellite needs to have an orbital radius \( r \) equal to the radius of the planet plus the altitude of the geostationary orbit above the planet's surface. For Earth, the altitude of a geostationary orbit is approximately 35,786 km above the Earth's surface. The average radius of the Earth is about 6,371 km, making the total distance from the center of the Earth to the satellite in geostationary orbit approximately:
\[ 6,371 \text{ km} + 35,786 \text{ km} \approx 42,157 \text{ km} \]
As such, the closest option provided that represents this distance is:
v = \sqrt{\frac{G \cdot m_p}{42,164 \text{ km}}}
Thus, the correct response is:
v=Gmp(42,164 km)−−−−−−−−√