Question

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 low earth orbit? (1 point) ( \bigcirc v = \frac{1}{(200 \text{km})} sqrt{g m_p} ) ( \bigcirc v = sqrt{\frac{g m_p}{(7,000 \text{km})}} ) ( \bigcirc v = \frac{1}{(7,000 \text{km})} sqrt{g m_p} ) ( \bigcirc v = sqrt{\frac{g m_p}{(200 \text{km})}} )

Explanation:

To find the velocity of a satellite in low - Earth orbit, we use the concept of centripetal force and gravitational force. The gravitational force provides the centripetal force for the satellite's circular motion.

The gravitational force between the planet (mass \(m_p\)) and the satellite (mass \(m\)) is given by \(F_g=\frac{Gm_pm}{r^2}\), and the centripetal force is \(F_c = \frac{mv^2}{r}\).

Setting \(F_g=F_c\) (since gravitational force provides centripetal force for the orbit):

\[

$$\begin{align*} \frac{Gm_pm}{r^2}&=\frac{mv^2}{r}\\ \frac{Gm_p}{r^2}&=\frac{v^2}{r}\\ v^2&=\frac{Gm_p}{r}\\ v&=\sqrt{\frac{Gm_p}{r}} \end{align*}$$

\]

For a low - Earth orbit, the radius \(r\) of the orbit is approximately equal to the radius of the Earth. The radius of the Earth is about \(6371\) km, which is approximately \(7000\) km (a reasonable approximation for this problem).

Now let's analyze each option:

• Option 1: \(v=\frac{1}{(200\space km)}\sqrt{Gm_p}\) does not match the derived formula \(v = \sqrt{\frac{Gm_p}{r}}\). The form is incorrect as it has \(\frac{1}{r}\) outside the square root in a non - matching way.
• Option 2: \(v=\sqrt{\frac{Gm_p}{(7000\space km)}}\) matches the formula \(v=\sqrt{\frac{Gm_p}{r}}\) where \(r = 7000\space km\) (a reasonable approximation for the radius of the Earth in low - Earth orbit).
• Option 3: \(v=\frac{1}{(7000\space km)}\sqrt{Gm_p}\) has an incorrect form. It should be the square root of \(\frac{Gm_p}{r}\) not \(\frac{1}{r}\) times the square root of \(Gm_p\).
• Option 4: \(v=\sqrt{\frac{Gm_p}{(200\space km)}}\) uses \(r = 200\space km\) which is not a reasonable approximation for the radius of the Earth (low - Earth orbit radius is close to Earth's radius, not \(200\) km).

Answer:

\(v=\sqrt{\frac{Gm_p}{(7,000\space km)}}\) (the second option)