Question

Let G be the universal gravitational constant and mp be the mass of the planet a satellite is orbiting. Which equation could be used to find the velocity of the satellite if it is placed in a geostationary orbit?(1 point) Responses v=\sqrt{\frac{Gm_p}{\left(15,522\ \text{km}\right)}}v=\sqrt{\frac{Gm_p}{\left(15,522\ \text{km}\right)}} v=\sqrt{\frac{Gm_p}{\left(48,115\ \text{km}\right)}}v=\sqrt{\frac{Gm_p}{\left(48,115\ \text{km}\right)}} v=\sqrt{\frac{Gm_p}{\left(7,324\ \text{km}\right)}}v=\sqrt{\frac{Gm_p}{\left(7,324\ \text{km}\right)}} v=\sqrt{\frac{Gm_p}{\left(42,164\ \text{km}\right)}}

Answer 1

To find the velocity of a satellite in a geostationary orbit, we can use the equation:

\[ v = \sqrt{\frac{G m_p}{r}} \]

where:
- \( v \) is the orbital velocity,
- \( G \) is the universal 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.

In a geostationary orbit, the satellite must be at a specific altitude where it matches the rotational period of the planet (24 hours). The radius \( r \) is the distance from the center of the planet to the satellite, which is the sum of the planet's radius and the altitude of the orbit.

The correct value for \( r \) to use in the equation for a geostationary orbit around Earth is approximately 42,164 km from the center of the Earth. This often includes Earth's average radius (approximately 6,371 km) plus the altitude of about 35,786 km (which gives the total of around 42,164 km).

So, from the provided choices, the correct equation to find the velocity \( v \) for a geostationary orbit would be:

\[ v = \sqrt{\frac{G m_p}{\left(42,164 \text{ km}\right)}} \]

Therefore, the correct response is:

**v=\sqrt{\frac{Gm_p}{\left(42,164\ \text{km}\right)}}**