Question

orbits of satellites quick check
let g be the universal gravitational constant and ( m_p ) 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)

( circ v = sqrt{\frac{gm_p}{(48,115 \text{km})}} )

( circ v = sqrt{\frac{gm_p}{(7,324 \text{km})}} )

( circ v = sqrt{\frac{gm_p}{(15,522 \text{km})}} )

( circ v = sqrt{\frac{gm_p}{(42,164 \text{km})}} )

Explanation:

Step1: Recall geostationary orbit radius

A geostationary satellite orbits the Earth at an altitude where its orbital period matches Earth's rotation. The radius of a geostationary orbit (distance from Earth's center) is approximately \( 42,164\ \text{km} \).

Step2: Derive orbital velocity formula

The orbital velocity \( v \) of a satellite is given by equating gravitational force (\( F_g=\frac{Gm_pm_s}{r^2} \)) to centripetal force (\( F_c=\frac{m_sv^2}{r} \)), canceling \( m_s \) (mass of satellite) gives \( \frac{Gm_p}{r^2}=\frac{v^2}{r} \), simplifying to \( v = \sqrt{\frac{Gm_p}{r}} \), where \( r \) is the orbital radius.

Step3: Match radius to options

Using \( r = 42,164\ \text{km} \) (geostationary orbit radius) in the formula \( v = \sqrt{\frac{Gm_p}{r}} \), the correct equation is \( v=\sqrt{\frac{Gm_p}{(42,164\ \text{km})}} \).

Answer:

\( v = \sqrt{\frac{Gm_p}{(42,164\ \text{km})}} \) (the last option: \( v=\sqrt{\frac{Gm_p}{(42,164\ \text{km})}} \))