To determine the correct equation for the velocity of a satellite in a geostationary orbit, we can use the formula for the orbital velocity:
\[ v = \sqrt{\frac{G m}{r}} \]
where:
- \( G \) is the universal gravitational constant,
- \( m \) is the mass of the planet being orbited,
- \( r \) is the distance from the center of the planet to the satellite.
For a geostationary orbit, the satellite must orbit at a height where its orbital period matches the rotation period of the planet, which is typically about 24 hours for Earth.
The geostationary orbit is approximately at an altitude of 35,786 km above the Earth's equator. Given that the Earth's radius is approximately 6,371 km, the total distance \( r \) from the center of the Earth to the satellite in geostationary orbit is:
\[ r = (6,371 \text{ km}) + (35,786 \text{ km}) = 42,157 \text{ km} \]
Now, we can look through the options provided for any that might resemble this value:
- \( \sqrt{\frac{G m_{p}}{42,164 \text{ km}}} \)
- \( \sqrt{\frac{G m_{p}}{7,324 \text{ km}}} \)
- \( \sqrt{\frac{G m_{p}}{48,115 \text{ km}}} \)
- \( \sqrt{\frac{G m_{p}}{15,522 \text{ km}}} \)
The closest value to 42,157 km is 42,164 km (which is slightly off but still the best choice among the options).
Therefore, the correct equation for finding the velocity of the satellite in a geostationary orbit is:
\[ \textbf{v = } \sqrt{\frac{G m_{p}}{42,164 \text{ km}}} \]
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