# 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}

# Gravitation – Force and Field Strength

## Concept of a Gravitational Field and Force

A gravitational field is a region in which a mass experiences a force.

A mass, \begin{aligned} m \end{aligned} that is present inside a gravitational field experiences a gravitational force. This gravitational field is produced by another mass \begin{aligned} M \end{aligned} .

The amount of force experienced by the mass \begin{aligned} m \end{aligned} is directly proportional to the product of the two masses and indirectly proportional the the square of the separation.

\begin{aligned} F_G=\frac{GM m}{r^2} \end{aligned}

The separation \begin{aligned} r \end{aligned} is the distance between the centre of mass of the two masses. We often assume that the two masses are point masses if the separation is large relative to the radius of the masses. If the separation is not large, then it is important to use the distance between the centre of mass of the two masses. We should not use the separation between the two surfaces of the masses. Furthermore, one may safely assume that for a uniform sphere, the centre of mass is the centre of the sphere.

There are two regions about the field produced by a mass: the region outside the mass and the region inside. The region outside the mass follows the inverse exponential relationship of 1/r^2.

Inside the mass, the relation is direct proportional to the distance from the centre of the mass. This is because as you proceed nearer to the centre of the mass, there is less mass “below” you. The part of the mass “above” you pulling you “up” is offset by the mass “below” you pulling you down.

### Gravitational Field Strength

Gravitational field strength is often misunderstood. Its definition is

Gravitational field strength at a point is the gravitational force acting on a unit mass at that point.

From the definition,

\begin{aligned} \text{force} &= m \times g\\mg &= \frac{GMm}{r^2}\\g &= \frac{GM}{r^2}\end{aligned}

We can observe that gravitational field strength only depends on the gravity-producing mass and the distance from it. It is not dependent on the test mass.

There is only one value of gravitational field strength at any particular point since we are always comparing the gravitational force on one unit mass. If there are multiple masses creating gravitational fields, the gravitational field strength at any particular point would be the vector sum of all the field strengths due to the the different masses.

### Field Strength on the surface of the Earth

When we calculate weight of an object, we always use the formula \begin{aligned} w = mg \end{aligned}

\begin{aligned} g \end{aligned} is referred to as the gravitational field strength (although it is commonly stated as the gravitational acceleration). Near the Earth’s surface, the field strength of the Earth is

\begin{aligned} g &=G\frac{M}{r_\text{Earth radius}^2}\\&=G\frac{M}{6400000^2}\end{aligned}

The field strength at 10 km above the surface would be \begin{aligned} G\frac{M}{6410000^2} \end{aligned}.

The difference among them is negligible. Hence we assume that the gravitational field strength on Earth’s surface is constant.

### Activity 1

Access the online PHET simulation.

Objective:
What is the relationship between the force acting on m2 by m1 and the force of m1 on m2?

1. Using the default values, observe the force on m2 by m1 and the force on m1 by m2.
What do you observed?
2. Change the mass of m1 to another value.
What do you observed?
3. Explain.

### Activity 2

Access the same simulation as activity 1. Also open this activity sheet to download a copy of the activity on iCloud.

Objective:
What is the relationship between the field strength with distance?

1. Change m2 to 1 kg, while m1 remains as 50 kg.
2. Drag m2 to the 10 m mark, and m1 to the 0 m mark. You may not be able to do this exactly, but an approximate position would be fine.
3. Record the force on m2 by m1 in the table.
4. Repeat this for distances 9, 8, 7, 6, 5, 4, 3, 2 and 1 m.
5. Observe the shape of the graph.

Question:

1. Will the gravitational field strength ever reaches zero?
2. At which point(from 0 m to infinity) is the field strength the strongest?

### Summary

1. understand the concept of a gravitational field as an example of a field of force and define gravitational field strength as force per unit mass.
2. understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre.
3. recall and use Newton’s law of gravitation in the form \begin{aligned} F=G\frac{Mm}{r^2} \end{aligned}

### Review

##### Question 1

Geostationary satellites are satellites that orbit around the Earth with a period of 24 hours. The satellite would appear as a stationary point relative to an observer on Earth. Calculate the distance above the Earth’s surface of a geostationary satellite. Properties of Earth

##### Solution

Since the centripetal force of the satellite is due to gravity,

\begin{aligned} \frac{GMm}{r^2} &= mr\omega ^2\\\frac{GM}{r^2} &= r \omega ^2 \\r &= \sqrt[3]{\frac{GM}{w}} \end{aligned}

$\omega$ is known, since the geostationary satellite must make one orbit in one day,

\begin{aligned} \omega &= \frac{2 \pi}{24 \times 3600}\\&= 7.3\times 10^{-5} \text{ rad s}^{-1}\\r&=42.1\times 10^6\text{ m}\end{aligned}

Hence, the distance above Earth’s surface is

\begin{aligned} d &= (42.1 - 6.38) \times 10^6 \text{ m}\\ &= 35.7 \times 10^3 \text{ km} \end{aligned}

It is interesting to note that the distance of Moon from Earth is about $370 \times 10^3 \text{ km}$. Hence, a geostationary satellite is about 10% of the distance from Earth to the Moon.