Question

3 multiple choice 1 point a satellite with a mass of “m” (little m for the satellite’s mass) orbits a planet with a mass of m (big m for the planet’s mass) at a distance of r from the center of the planet. assume that the satellite orbits the planet in a circular path. which of the following equations will tell you the satellite’s velocity at this location? the gravitational constant will be “g” in the equation. \\(\sqrt{gmm}\\) \\(g^2 r^2 m\\) \\(\frac{gmm}{r}\\) \\(\sqrt{\frac{gm}{r}}\\) \\(\sqrt{\frac{r}{gm}}\\)

Explanation:

Step1: Recall Centripetal and Gravitational Force

For a satellite in circular orbit, gravitational force provides centripetal force. So, \( F_{gravity} = F_{centripetal} \). The gravitational force is \( F_g = \frac{GMm}{r^2} \), and centripetal force is \( F_c = \frac{mv^2}{r} \).

Step2: Equate the Two Forces

Set \( \frac{GMm}{r^2} = \frac{mv^2}{r} \).

Step3: Solve for Velocity \( v \)

Cancel \( m \) from both sides: \( \frac{GM}{r^2} = \frac{v^2}{r} \). Multiply both sides by \( r \): \( \frac{GM}{r} = v^2 \). Take square root: \( v = \sqrt{\frac{GM}{r}} \).

Answer:

\(\boldsymbol{\sqrt{\frac{GM}{r}}}\) (the fourth option in the given choices)