Students will learn to apply the universal law of gravity equation in order to determine the parameters of orbits.

### Key Equations

\begin{align*} v = \frac{2 \pi r}{T} \end{align*}

for a particle travels a distance \begin{align*}2\pi r\end{align*} in an amount of time *\begin{align*}T\end{align*}*

\begin{align*} \vec{F_G} = \frac{Gm_1 m_2}{r^2} \end{align*}

\begin{align*} \vec{a_c} = \frac{v^2}{r} \end{align*}

### Guidance

#### When can Gravity Act as a Centripetal Force?

We saw in the Centripetal Forces chapter that the force of Gravity causes an attraction between two objects of mass \begin{align*} m_1 \end{align*} and \begin{align*} m_2 \end{align*} at a distance \begin{align*} r \end{align*} of \begin{align*}
\vec{F_G} = \frac{Gm_1 m_2}{r^2}. \text{ [4]}
\end{align*} By Newton's Third Law, both objects experience the force: equal in magnitude and opposite in direction, and both will move as an effect of it. If one of the objects is much lighter than the other (like the earth is to the sun, or a satellite is to earth) we can approximate the situation by saying that the heavier mass (the sun) does not move, since its acceleration will be far smaller due to its large mass. Then, if the lighter mass remains at a relatively constant absolute distance from the heavier one (remember, centripetal force needs to be constant in magnitude), we can say that the lighter mass experiences an *effectively* centripetal force and thus has a centripetal acceleration.

#### Math of Centripetal Gravity

Gravity is not always a centripetal force. This is a really important point. It only acts as a centripetal force when conditions approximate those listed above --- very much like it isn't constant near the surface of the earth, but very close to it.

If gravity provides centripetal force and acceleration, we can set [2] equal to [4]. It's important to remember that in [2] \begin{align*} m \end{align*} refers to the lighter mass, since that is the one traveling. Then, \begin{align*} \frac{G\cancel{m_1}m_2}{\cancelto{r}{r^2}} = \frac{\cancel{m_1}v^2}{\cancel{r}} \intertext{So, the relationship between velocity and radius for a circular orbit of a light object around an heavy mass (note the mass of the lighter object cancels) is:} Gm_2 = {v_{\text{orb}}}^2 r_{\text{orb}} \end{align*}

#### Geosynchronous orbit

This is the orbit where a satellite completes one orbit of the Earth every 24 hours, staying above the same spot (longitude) on Earth. This is a very important orbit for spy satellites and TV satellites among others. You force the speed of the satellite to be a value such that the satellite makes one rotation every 24 hours.

#### Example 1

### Watch this Explanation

### Explore More

- A digital TV satellite is placed in geosynchronous orbit around Earth, so it is always in the same spot in the sky.
- Using the fact that the satellite will have the same period of revolution as Earth, calculate the radius of its orbit.
- What is the ratio of the radius of this orbit to the radius of the Earth?
- Draw a sketch, to scale, of the Earth and the orbit of this digital TV satellite.
- If the mass of the satellite were to double, would the radius of the satellite’s orbit be larger, smaller, or the same? Why?

- A top secret spy satellite is designed to orbit the Earth twice each day (
*i.e.,*twice as fast as the Earth’s rotation). What is the height of this orbit above the Earth’s surface? - Two stars with masses \begin{align*}3.00\times10^{31} \;\mathrm{kg}\end{align*} and \begin{align*}7.00\times10^{30} \;\mathrm{kg}\end{align*} are orbiting each other under the influence of each other’s gravity. We want to send a satellite in between them to study their behavior. However, the satellite needs to be at a point where the gravitational forces from the two stars are equal. The distance between the two stars is \begin{align*}2.0\times10^{10} \;\mathrm{m}\end{align*}. Find the distance from the more massive star to where the satellite should be placed. (
*Hint:*Distance from the satellite to one of the stars is the variable.) - Calculate the mass of the Earth using
*only: (i)*Newton’s Universal Law of Gravity;*(ii)*the Moon-Earth distance (3.85 x 10^{8 }m); and*(iii)*the fact that it takes the Moon 27 days to orbit the Earth.

- A space station was established far from the gravitational field of Earth. Extended stays in zero gravity are not healthy for human beings. Thus, for the comfort of the astronauts, the station is rotated so that the astronauts
*feel*there is an internal gravity. The rotation speed is such that the*apparent*acceleration of gravity is \begin{align*}9.8 \;\mathrm{m/s}^2\end{align*}. The direction of rotation is counter-clockwise.- If the radius of the station is \begin{align*}80 \;\mathrm{m}\end{align*}, what is its rotational speed, \begin{align*}v\end{align*}?
- Draw vectors representing the astronaut’s velocity and acceleration.
- Draw a free body diagram for the astronaut.
- Is the astronaut exerting a force on the space station? If so, calculate its magnitude. Her mass \begin{align*}m=65\;\mathrm{kg}\end{align*}.
- The astronaut drops a ball, which
*appears*to accelerate to the ‘floor,’ (see picture) at \begin{align*}9.8 \;\mathrm{m/s}^2.\end{align*}

#### Answers to Selected Problems

- a. \begin{align*}4.23 \times 10^7\mathrm{m}\end{align*} b. \begin{align*}6.6 \ R_e\end{align*} d. The same, the radius is independent of mass
- \begin{align*}1.9 \times 10^7\mathrm{m}\end{align*}
- You get two answers for \begin{align*}r\end{align*}, one is outside of the two stars one is between them, that’s the one you want, \begin{align*}1.32 \times 10^{10}\mathrm{m}\end{align*} from the larger star.
- .
- a. \begin{align*}v = 28\;\mathrm{m/s}\end{align*} b. \begin{align*}v-\end{align*}down, \begin{align*}a-\end{align*}right c. \begin{align*}f-\end{align*}right d. Yes, \begin{align*}640\mathrm{N}\end{align*}