## What is Gravitational Field?

We can easily define the *gravitational field as a region of space in which every object with a mass experiences a pull due to gravitational force.*

For example, when you throw a ball vertically upward, it will go up and then gradually start losing momentum due to deceleration. The ball will reach a point where it will no longer continue to move up again, and then start falling back to the ground. Ever wonder why that happens?

This is because as the ball is moving upward due to the force you applied by throwing it up, another force is also pulling the ball down and it’s called gravitational force. The space where the ball was thrown is called the gravitational field.

Another example is, ever wonder why a mango fruit falls to the ground instead of moving to space? The answer to this question is also very simple. **The gravitational force will always pull the mango fruit back to the ground**.

What happens to you when you jump up? **You will still come back to the ground**.

When you drop two objects, say a pencil and a big stone at the same time from the same height. What did you observe? **Both the pencil and the big stone will fall on the ground at the same time.**

The examples I have mentioned here were thanks to the effect of gravitational force in a gravitational field.

### Important Notes on Gravitational Field

Did you know that the Earth exerts a gravitational force on the moon, even though they are 385,000 kilometres away from each other? Yes, the Earth pulls the moon toward its centre. The space between the earth and the moon is known as the gravitational field.

Did you know that the sun pulls the earth due to gravity even though the distance between them is 1.5 x 108 kilometres? Yes, it does pull the earth due to the force of gravity.

*You may also like to read:*

Newton’s Second Law Practice Problems

## Newton’s Laws of Universal Gravitation

**Law:** *Newton’s law of universal gravitation states that the force of attraction (F) between two masses (M) and (m) is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. This law is also known as inverse law.*

F ∝ Mm / r^{2}

Which will become

F = (GMm) / r^{2}

Where

F = Force of attraction between the two masses M and m

M and m are the masses of the two objects

G = Gravitational constant = 6.7 x 10^{-11} Nm^{2}kg^{-2} [The unit of gravitational constant is Newton meter square per kilogram square (Nm^{2}kg^{-2}) **or** meter cube per kilogram per second square (m^{3}kg^{-1}s^{-2}) ]

r = distance between two masses

According to newton’s third law of motion, actions and reactions are equal and opposite. Therefore, mass M attracts m by an equal but opposite force.

A gravitational field is everywhere. It surrounds bodies with mass, and it fills up space. The gravitational field makes it possible for the moon to orbit around the earth and the earth to orbit around the sun.

After throwing a ball, it will return to the earth. This shows that there is a force of attraction between the mass of the ball and that of the earth.

The force of attraction between the two masses may be too weak to notice unless one of the masses is very large.

Therefore, the mass of an object creates a force field that is attracted toward another object. The gravitational force on earth is equal to the force of attraction between you and the earth.

### More Explanation on Gravitational Force

When a body is at rest, on or near the surface of the earth, the gravitational force is equal to the weight of that body.

The gravitational acceleration of the moon and Venus is different from that of earth. Thus, if you are to measure your weight on these three different planets, you will notice different weights.

When two bodies are gravitationally locked, their gravitational force is centered in an area that is not at the center of either body, but at the barycenter of the system. For example, when two people sit at both ends of a seesaw, the lighter person will be pushed up while the heavier person will go down.

To balance the heavier and the lighter person on the same seesaw. The heavier person must sit closer to the center of the mass of the seesaw (fulcrum) just so that it can balance. Barycenter is the balance point between the earth and the moon. The earth and the moon move around the barycenter in their orbits.

The knowledge of the gravitational field has given us the chance to send people to other planets. It has also helped us to keep satellites in any orbit of our choice.

## Weight of an Object

When an orange slips from your hand, it will drop on the floor, this is due to the weight of the orange pulling it to the floor. The weight of a body is the product of its mass and gravitational acceleration of the body.

Weight (w) = mass (m) x gravitational acceleration (g)

w = mg

In the 17th century, newton published Newton’s law of universal gravitation together with his three laws of motion.

He extensively studied the works of Galilei Galileo and Kepler on the motion of planets and came up with his contributions.

## Relationship Between Gravitational Constant (G) and Gravitational Acceleration (g)

Some students find it difficult to understand the difference between the gravitational constant (G) and the gravitational acceleration (g).

The value of the gravitational constant (G) is 6.7 x 10^{-11} Nm^{2}kg^{-2}. It’s the relationship between two masses and the distance between them.

While gravitational acceleration is due to a pull on a body because of the force of gravity. The value of gravitational acceleration is 9.81 ms^{-2}.

Now, to understand the relationship between G and g when an object is on the surface of the earth. We will equate the force of attraction between the two masses with weight.

w = F

This implies

mg = (GMm) / r^{2}

and m will cancel each other from both sides to give us

g = GM / r^{2}

Where M = mass of the earth

Therefore, we can say that the mass of the earth is

M = gr^{2} / G

Therefore, we can describe acceleration due to gravity (g) as force per unit mass on the surface of the earth.

Additionally, the equation g = GM / r^{2} shows that we can determine the acceleration due to gravity on the surface of the earth from the mass of the earth and its radius.

Therefore, the gravitational acceleration on the top of a mountain is less when we compare it with the gravitational acceleration at sea level.

This is because the radius of the earth from its center is a bit greater at the top of a mountain than at sea level.

## The velocity of a Mass, m Moving Around the Earth’s Surface

When we have an object of mass m moving around the earth with a constant speed (v). We can obtain v by equating centripetal force [ (mv^{2}) / r ] with the gravitational force [ (GMm) / r^{2} ] that keeps it moving in a circular direction.

Thus,

(mv^{2}) / r = (GMm) / r^{2}

and by making v subject of the formula, we will have

v = √(GM / r)

By multiplying √(GM / r) with (r / r). We will get

v= √(GM / r) x (r / r)

And the above equation will become

v = √(GMr / r^{2}) = √(GM / r^{2}) x r

But g = GM / r^{2}

Hence

v = √gr

## Gravitational Field Intensity

The gravitational field intensity (I) is defined as the force experienced per kilogram of mass at any point.

Gravitational field intensity, I = Froce (f) / mass (m) = g

## Gravitational Potential

Definition: * The gravitational potential (V) at a point on the surface of the earth, is the work done in moving a unit mass of a body from infinity to that point*. The unit of gravitational potential is in joule per kilogram (jkg

^{-1}).

The mathematical expression of the above definition is

Gravitational potential of point mass, V = – (Gm) / r

From the above equation:

m is the mass of the body causing the gravitational field.

r is the distance between the point and the mass.

**Note:**

*As the body changes its direction away from the earth. The radius r will increase which will cause the gravitational potential to reduce. It will get to a point where the gravitational potential will become zero when the radius r is infinitely large. *

*It is also important to know that the gravitational force will perform a negative work if the body is moving away from the earth.*

At the same time, if the body returns to the earth, the radius r will reduce.

When you apply force to raise an object of mass m through a height h, the work done in raising the object is

Work done (W) = Force (mg) x distance (h)

which implies that

W = mgh [where g is the acceleration due to gravity]

### Potential Energy of a Mass on the Earth’s Surface

If a mass, m is on the surface of the earth of radius, r. The potential energy possessed by the mass, V = mgh

and since g = GM / r^{2}

and h = r [where r = radius of the earth]

Therefore,

gh = GM / r

Now, multiply both sides by m to obtain

mgh = GMm / r

Which implies that

The potential energy on the earth surface, V = – (GMm) / r

## Kepler’s Laws

Johannes Kepler is a german astronomer and mathematician. He laid out the foundation of Newton’s universal law of gravitation. Kepler came up with three laws to describe the motion of the planets in orbit:

**Each planet travels along an ellipse with the sun at one focus****The line joining the sun and a given planet sweeps equal areas at equal times.****The square of the period of revolution of the planets is proportional to the cubes of their mean distances from the sun ( T**^{2}∝ r^{3}).

## Solar System

The concept of the solar system simply refers to the sun and other objects that circle it due to the force of gravity. These objects that orbit the sun are called planets.

The currently known planets around the sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto.

We have two categories of planets, terrestrial planets, and gas giants.

**Terrestrial planets** are made up of rocks and metals. They also look similar to the earth. They are closer to the sun and their size is smaller. Terrestrial planets include Mercury, Venus, Earth, and Mars.

Gas giants are bigger than terrestrial planets and are made up of gases. Jupiter and Saturn are composed mainly of hydrogen and helium.

The solar system also contains objects like **Asteroids**, **Comets**, **Meteoroids**, and others.

Points to note on the planets:

- They revolve around the sun in their orbits
- When a planet is close to the sun, its revolution is faster.
- The planets revolve in a similar direction
- Apart from pluto, the orbits of the planets lie in the same plane.

## Satellites

An object that moves around the earth or in space is called a satellite. We have two types of satellites, Artificial and Natural satellites.

An artificial satellite is an object created and launched into space by man. Creating such satellites aims to improve communication, supervision, and security, and to observe the distant universe.

Natural Satellite is synonymous with the moon and is a celestial body that orbits a planet or smaller bodies. They are used to identify non-artificial satellites.

### Orbit

According to the National Aeronautics and Space Administration (NASA), ”An orbit is a regular, repeating path that one object in space takes around another one.”

#### Parking Orbit

**A parking orbit is an orbit in which the period of a satellite in its orbit is precisely equal to the period of the earth as it rotates about its axis.**

### How to Find the Period and Velocity of a Satellite in Its Orbit

If a satellite of mass m circles the earth at an orbit of radius R, from the center of the earth and in the same direction as the rotation of the earth. Then, the centripetal force that sustains it in that orbit will equal the gravitational force. Hence

(mv^{2}) / R = (GMm) / R^{2}

Remember that Gm = gr^{2} [where r = radius of the earth]

We can now rewrite mv^{2} / R = (GMm) / R^{2} into

mv^{2} / R = (mgr) / R^{2}

and by making v subject of the formula, we will have

v = √ (gr^{2} / R)

Now, to determine the period T, of a satellite moving around an orbit, we will say

Velocity (v) = (Displacement of the satellite around an orbit) / Time taken

v = 2πR / T

Which implies that

Period, T = 2πR / v

Parking orbit helps in finding the location of a communication satellite so that the transmission of events will continue to flow across the globe.

### Escape Velocity

**Definition: Escape velocity is the minimum velocity required by an object to permanently escape the influence of the gravitational field.**

To derive the formula for escape velocity, we will consider the work done in carrying the mass of a body away from the center of the earth through a distance r. We will also put into consideration that gravitation is weak.

Hence

Work done (W) = Force (F) x Distance (r)

Which implies that

W = F x r

and F = (GMm) / r^{2}

Therefore, W = ((GMm) / r^{2}) x r = (GMm) / r

Additionally, the work done (W = (GMm) / r) is equal to the kinetic energy (K = (1/2)mv^{2})

Thus, (GMm) / r = (1/2)mv^{2}

And by making v subject of the formula, we will obtain

v = √(2GM / r)

And if the object is launched from the surface of the earth. r = R

Therefore, our equation will become

v = √(2GM / R)

But M = gr^{2} /G

Hence, v will become

v = √(2GM / R) x (gr^{2} /G)

**to obtain v = √2gR which is the formula for calculating escape velocity**

## Gravitational Field Practice Problems

Here are some practice problems to help you understand how to calculate problems involving the gravitational field:

### Problem 1

Two 5 kilograms of spherical balls are placed so that their centers are 50 centimeters apart. What is the magnitude of the gravitational force between the two balls? Take G = 6.67 x 10^{-11} Nm^{2}kg^{-2}.

**Data**

Both masses are the same = M = m = 5 kg

Distance between the masses, r = 50 cm = 50 / 100 = 0.5 m

Gravitational constant, G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

**Formula**

Applying the formula F = (GMm) / r^{2}

#### Solution

We can now insert our data into the formula F = (GMm) / r^{2}

F = (6.67 x 10^{-11} x 5 x 5) / 0.5^{2}

**Therefore, F = 6.67 x 10 ^{-9} N**

### Problem 2

Determine the force of attraction between the sun (m_{s} = 1.99 x 10^{30} kg) and the earth (m_{e} = 5.98 x 10^{24} kg). Assume the sun is 1.5 x 10^{8} kilometers from the earth.

**Data**

Mass of the sun, m_{s} = 1.99 x 10^{30} kg

Mass of the earth, m_{e} = 5.98 x 10^{24} kg

Distance between the sun and the earth, r = 1.5 x 10^{8} km

Gravitational constant, G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

F = ?

**Formula**

We will use F = (Gm_{s}m_{e}) / r^{2} to solve the problem

#### Solution

Insert the data into F = (Gm_{s}m_{e}) / r^{2} to obtain

F = (6.67 x 10^{-11} x 1.99 x 10^{30} x 5.98 x 10^{24}) / (1.5 x 10^{8})^{2}

**Therefore, the force of attraction between the sun and the earth is F= 3.53 x 10 ^{-22}**

### Problem 3

Determine the mass of the earth if the radius of the earth is approximately 6.38 x 10^{6} m, G = 6.67 x 10^{-11} Nm^{2}kg^{-2} and g = 9.8 m/s^{2}

**Data**

Mass of the earth, m_{e} = M = ?

Radius of the earth, r_{e} = 6.38 x 10^{6} m

Gravitational constant, G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

Acceleration due to gravity, g = 9.8 m/s^{2}

**Formula**

M = gr^{2} / G

#### Solution

Let us substitute our formula with our data

M = gr^{2} / G = (9.8 x (6.38 x 10^{6})^{2}) / 6.67 x 10^{-11} = 5.98 x 10^{24} kg

**The mass of the earth is 5.98 x 10 ^{24} kilograms**

### Problem 4

Find the gravitational potential at a point on the earth’s surface. Take a mass of the earth as 5.98 x 10^{24} kilograms, its radius as 6.38 x 10^{6} meters and G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

**Data**

Mass of the earth, M = 5.98 x 10^{24} km

Radius of the earth, r = 6.38 x 10^{6} m

G = 6.67 x 10^{-11} Nm^{2}kg^{-2}

**Unknown**

Gravitational Potential, V = ?

**Formula**

V = GM / r

#### Solution

We will now apply our data to the formula

V = GM / r = (6.67 x 10^{-11} x 5.98 x 10^{24}) / 6.38 x 10^{6} = 6.25 x 10^{7} Jkg^{-1}

**Therefore, the gravitational potential is 6.25 x 10 ^{7} Jkg^{-1}**

### Problem 5

Calculate the escape velocity of a satellite from the earth’s gravitational field.

g = 9.8 m/s^{2}, radius of the earth r = R = 6.4 x 10^{6} m

**Solution **

We will apply the formula

Escape velocity, v = √2gR = √(2 x 9.8 x 6.4 x 10^{6}) = 11 x 10^{3} m/s or 11 km/s

**Therefore, the escape velocity of the satellite from the surface of the earth is 11 km/s**