To find the velocity of a satellite in a geostationary orbit, we can use the formula for orbital velocity derived from Newton's law of gravitation and the circular motion concept. The orbital velocity \( v \) can be expressed as:
\[ v = \sqrt{\frac{G m_p}{r}} \]
where:
- \( G \) is the gravitational constant,
- \( m_p \) 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 orbits at a height where its orbital period matches the rotation period of the Earth (24 hours). The approximate distance \( r \) for a geostationary orbit around Earth is about \( 42,164 \) km from the center of the Earth. This distance includes the Earth's radius (approximately \( 6,371 \) km) plus the altitude of the geostationary orbit (approximately \( 35,786 \) km).
Thus, the correct expression for the velocity of a satellite in a geostationary orbit would be:
\[ v = \sqrt{\frac{G m_p}{42,164 \text{ km}}} \]
Therefore, the correct choice from the provided responses is:
v = Gmp(42,164 km)−−−−−−−−√
5 answers