# Gravitation and Circular Motion

### Why does a orbiting satellite not fall to the Earth?

A common misconception among students is that orbiting satellites do not experience gravity. It is argued that since these satellites are not falling to the Earth, they must experience no gravity. This thought is not true because the very action of orbiting around the Earth shows that the satellite is under the influence of Earth’s gravity. Then why is the satellite not falling to the Earth?

The below simulation explains why satellites are not falling back to the Earth. In fact, these satellites are actually in the process of falling to the Earth. The simulation explains clearly how this works.

Satellite orbiting the Earth

So the answer to the above question is that the satellite is in fact, falling to the Earth. However, the Earth’s surface curves away at the same rate the the satellite is falling towards Earth. Hence, the satellite never falls onto the Earth.

It is important to note that this can happen because the satellite possesses an initial velocity that is parallel to the Earth’s surface. Without sufficient velocity of this component, the satellite would fall towards Earth.

### Gravitational force and circular motion

Planets revolve around the Sun and the Moon revoles around the Earth. These orbits are generally approximately circular, and the centripetal force to pull them around the orbits is the gravitational force.

\begin{aligned} \frac{GMm}{r^2} &= \frac{mv^2}{r} \\ \frac{GM}{r} &= {v^2}\\v&=\sqrt{\frac{GM}{r}}\end{aligned}

We can see that the linear speed, hence the angular speed, of an orbiting body is independent of the mass of the body. It is only dependent on the mass of the body producing the gravitational field and the distance of the orbiting body from the centre of the main mass.

To find the relationship between the radius and the period of orbit,

\begin{aligned} (\frac{2\pi r}{T})^2 &= \frac{GM}{r}\\T^2 &= \frac{4\pi^2}{GM}r^3 \end{aligned}

Hence we can see that the square of the period is proportional to the cube of the radius of orbit. The is the Kepler’s third law.

### Summary

1. When an object orbits around another mass, the centripetal force is gravitational in nature.

2. The square of the period is proportional to the cube of the radius of orbit. This is also known as the Kepler’s third law.

### Review

#### Question 1

Two stars orbit each other in a time of 3.0 years. Calculate the angular speed \begin{aligned} \omega \end{aligned} for each star. Given that the ratio \begin{aligned} \frac{M_1}{M_2}=2.0 \end{aligned} , and the separation of the starts is $3.0 \times 10^{12} \text{ m}$ , calculate the radii \begin{aligned} R_1 \end{aligned} and \begin{aligned} R_2 \end{aligned} .

#### Solution:

Angular speed is the angular displacement per unit time. Hence,

\begin{aligned}\omega &= \frac{2 \pi}{3.0 \times 365 \times 24 \times 3600}\\&=6.6\times 10^{-8}\text{rad s}^{-1} \end{aligned}

The two stars must be orbiting with the same angular speed. Otherwise, there will be a point in time when the inner star catches up and the centre of mass would change position. The gravitational force between the two stars would then change, changing the centripetal force. This is not possible for a stable star system, hence the angular speeds of the two stars must be the same.

Equating the gravitational force to the centripetal force,

\begin{aligned} \frac{GM_1M_2}{(R_1 + R_2)^2 } &= M_2 R_2 \omega^2\\\frac{GM_1}{(R_1 + R_2)^2} &= R_2 \omega^2 \end{aligned}

Similarly,

\begin{aligned} \frac{GM_2}{(R_1 + R_2)^2} &= R_1 \omega^2 \\\frac{M_1}{M_2} &= \frac{R_2}{R_1}\end{aligned}

Since \begin{aligned} \frac{M_1}{M_2} = 2.0 \end{aligned} ,

\begin{aligned} \frac{R_2}{R_1}&=2.0\\R_2&=2.0R_1\\R_1+2R_1 &= 3.0 \times 10^{12}\\R_1&=1.0\times 10^{12}\text{ m}\\R_2 &= 2.0 \times 10^{12} \text{ m} \end{aligned}

# Gravitation – Potential and Potential Energy

## Gravitational Potential and Potential Energy

Gravitational potential at a point is the amount of work needed to bring a unit mass from infinity to that point.

Gravitational potential is defined to be negative since gravity is attractive in nature. A negative potential means that no external work is needed to bring a unit mass from infinity to that point, since the gravity-producing mass would be doing the work to pull the unit mass to that point.

Mathematically,

\begin{aligned} \phi = -\frac{GM}{r} \end{aligned}

Note that potential is a scalar quantity. Hence, the potential at a point is simply the algebraic sum of the potential of the different masses at that point.

\begin{aligned} \phi_\text{sum} = -\frac{GM_1}{r_1} - \frac{GM_2}{r_2} - ... \end{aligned}

Similarly, gravitational potential energy is defined as

Gravitational potential energy of a mass at a point is the amount of work needed to bring the mass from infinity to that point.

\begin{aligned} \text{GPE} = -\frac{GMm}{r^2} \end{aligned}

You may find this concept similar to gravitational force and field strength. Both gravitational force and potential energy invloves the product of two masses \begin{aligned} Mm \end{aligned} while field strength and potential involves just the gravity-producing mass \begin{aligned} M \end{aligned}.

## Gravitational Potential vs Field Strength

It is easy to compare the relative values of potential and field strength because of the similar form of equations.

\begin{aligned} g &= -\frac{GM}{r^2} \\ \phi &= -\frac{GM}{r}\end{aligned}

## GPE = mgh

In many junior Physics text, gravitational potential energy is quoted with the formula
\begin{aligned} \text{GPE} = mgh \end{aligned}
This formula assumes that the change in height in insignificant compare to the radius of the Earth. This formula also calculates the change in the potential energy due to a change in position.

The formula \begin{aligned} F_G = \frac{GMm}{r} \end{aligned} calculates the actual amount of potential energy a mass possess due to its position. This formula does not calculate the change in potential energy. To calculate the change in potential energy,

\begin{aligned}\text{change in GPE} &= \frac{GMm}{r_1} - \frac{GMm}{r_2}\\\end{aligned}

If \begin{aligned} r_1 \approx r_2 \end{aligned} ,

\begin{aligned} \text{change in GPE} &= GMm ( \frac{1}{r_1} - \frac{1}{r_2} ) \\&= GMm (\frac{r_2-r_1}{r_1 r_2})\\&=gm(r_2-r_1)\\&=mgh\end{aligned}

From this, we have our old formula \begin{aligned} \text{GPE} = mgh \end{aligned}

### Summary

1. Gravitational potential at a point is the amount of work needed to bring a unit mass from infinity to that point.
2. Gravitational potential energy of a mass at a point is the amount of work needed to bring the mass from infinity to that point.
3. \begin{aligned} \phi = -\frac{GM}{r} \end{aligned}
4. The gravitational potential has a larger magnitude than the field strength.
5. When the change in height is small, we may use \begin{aligned} \text{change in GPE} = mgh \end{aligned}