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Newton's Law of Gravity: Attracting Force Between Masses

Important Formula

Formula: Newton's Law of Gravity

$$F_{\text g} ~=~ G \, \class{brown}{M} \, \frac{\class{brown}{m}}{r^2}$$ $$F_{\text g} ~=~ G \, \class{brown}{M} \, \frac{\class{brown}{m}}{r^2}$$ $$r ~=~ \sqrt{ G \, \frac{ \class{brown}{M} \, \class{brown}{m} }{ F_{\text g} } }$$ $$\class{brown}{M} ~=~ \frac{1}{G} \, \frac{ F_{\text g} }{\class{brown}{m}} \, r^2$$ $$\class{brown}{m} ~=~ \frac{1}{G} \, \frac{ F_{\text g} }{\class{brown}{M}} \, r^2$$ $$G ~=~ \frac{ F_{\text g} }{\class{brown}{M} \, \class{brown}{m}} \, r^2$$

Rearrange formula

What do the formula symbols mean?

Gravitational force

$$ F_{\text g} $$ Unit $$ \mathrm{N} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{s}^2} $$

It is the force that the mass $ \class{brown}{m_1} $ exerts on the mass $ \class{brown}{m_2} $ and vice versa.

Distance

$$ r $$ Unit $$ \mathrm{m} $$

Distance between the masses $ \class{brown}{m_1} $ and $ \class{brown}{m_2} $. The larger the distance, the smaller the gravitational force $ F_{\text g} $.

1st mass

$$ \class{brown}{M} $$ Unit $$ \mathrm{kg} $$

The mass of the first body.

2nd mass

$$ \class{brown}{m} $$ Unit $$ \mathrm{kg} $$

The mass of the second body.

Gravitational constant

$$ G $$ Unit $$ \frac{\mathrm{N} \, \mathrm{m}^2}{\mathrm{kg}^2} = \frac{\mathrm{m}^3}{\mathrm{kg} \, \mathrm{s}^2} $$

The gravitational constant is a physical constant that occurs in equations describing the interaction between masses. It has the following experimentally determined value: $$ G ~\approx~ 6.674 \, 30 ~\cdot~ 10^{-11} \, \frac{ \mathrm{m}^3 }{ \mathrm{kg} \, \mathrm{s}^2 } $$

Table of contents

Have you ever wondered why a peach 🍑 falls from a tree to the earth and does not fly into space? This everyday observation, which we take for granted, is closely connected to one of the fundamental laws of physics - Newton's Law of Gravitation. This law, formulated by Isaac Newton in the 17th century, describes the attraction force between objects in the universe that have mass. So, all the objects you are familiar with.

This law, is super simple and at the same time so universal! It allows us to understand many observations quantitatively:

The movement of the celestial bodies and why the moon does not fall on the earth.
The movement of satellites to enable something like Google Maps.
The trajectories of asteroids to shoot them down in time before they wipe out humanity.
The structure of whole galaxies to understand how this universe came into being.
The trajectories of balls.
The launch of a SpaceX rocket by Elon Musk.

Let's consider an everyday example. Throw a ball into the air. What happens next? The ball goes up, away from the Earth, slows down as it reaches its highest point, momentarily hangs in the air, and eventually falls back to the ground. This seemingly simple event is actually the result of an invisible force acting on the ball - gravitational force exerted by the Earth on the ball.

The Newtonian Law of Gravitation describes the gravitational force $ F_{\text g} $ between an object with a mass $ \class{brown}{M} $ and another object with a mass $ \class{brown}{m} $, which interact with each other due to their masses. It states that every mass in the universe exerts an attractive force $ F_{\text g} $ on every other mass. This means that not only the earth attracts the ball, but also the ball attracts the earth - a mutual interaction.

Gravitational Force Between Two Masses

The formula you can use to gain an understanding of gravitational attraction is as follows:

$$ \begin{align} F_{\text g} ~=~ G \, \class{brown}{M} \, \frac{\class{brown}{m}}{r^2} \end{align} $$

Here is $ G = 6.674 \cdot 10^{-11} \, \frac{\mathrm{m}^3}{ \mathrm{kg} \ \mathrm{s}^2 } $ the Gravitational constant. It is a physical constant and determines how strong two masses of one killogram attract each other at a distance of one meter. It therefore determines the strength of gravity in our universe:

$$ \begin{align} F_{\text g} ~&=~ G \, \frac{ \class{brown}{1 \, \mathrm{kg}} \cdot \class{brown}{1 \, \mathrm{kg}} }{ 1 \, \mathrm{m}^2 } \\\\

~&=~ G \, \frac{ \class{brown}{1 \, \mathrm{kg}^2} }{ 1 \, \mathrm{m}^2 } \\\\

~&=~ 6.674 \cdot 10^{-11} \, \frac{\mathrm{m}^3}{ \class{brown}{\mathrm{kg}} \ \mathrm{s}^2 } \cdot \frac{ \class{brown}{1 \, \mathrm{kg}^2} }{ 1 \, \mathrm{m}^2 } \\\\

~&=~ 6.674 \cdot 10^{-11} \, \frac{ \class{brown}{\mathrm{kg}} \cdot \mathrm{m} }{ \mathrm{s}^2 } \\\\

~&=~ 6.674 \cdot 10^{-11} \, \mathrm{N} \end{align} $$

The two masses, as you can see, attract each other with a very small gravitational force of $ 6.674 \cdot 10^{-11} \, \mathrm{N} $.

Example: Gravitational force between the earth and the sun

The Earth does not move in a perfect circular orbit around the Sun but rather in an elliptical path. Therefore, the distance between the Earth and the Sun varies over time. The point on the Earth's orbit where it is closest to the Sun is called perihelion, while the point where it is farthest from the Sun is called aphelion.

What is the gravitational force exerted by the Sun on the Earth when the Earth is at perihelion?
What is the gravitational force exerted by the Sun on the Earth when the Earth is at aphelion?

The mass of the Earth is $ \class{brown}{5.972 \cdot 10^{24} \, \mathrm{kg}} $. The mass of the Sun is $ \class{brown}{1.989 \cdot 10^{30} \, \mathrm{kg}} $. The distance of the Earth from the Sun at perihelion is $ 1.47 \cdot 10^{8} \, \mathrm{m} $. The gravitational force exerted by the Sun on the Earth at the smallest distance is:

$$ \begin{align} F_{\text g} ~&=~ 6.674 \cdot 10^{-11} \, \frac{\mathrm{m}^3}{ \mathrm{kg} \ \mathrm{s}^2 } ~\cdot~ \class{brown}{1.989 \cdot 10^{30} \, \mathrm{kg}} ~\cdot~ \frac{ \class{brown}{5.972 \cdot 10^{24}\,\mathrm{kg}} }{ 1.47 \cdot 10^{8} \, \mathrm{m} } \\\\
~&=~ 3.67 \cdot 10^{28} \, \mathrm{N} \end{align} $$

The distance of the earth to the sun in the aphelion (greatest distance) is $ 1.52 \cdot 10^{8} \, \mathrm{m} $. The gravitational force acting on the Earth at aphelion is thus:

$$ \begin{align} F_{\text g} ~&=~ 6.674 \cdot 10^{-11} \, \frac{\mathrm{m}^3}{ \mathrm{kg} \ \mathrm{s}^2 } ~\cdot~ \class{brown}{1.989 \cdot 10^{30} \, \mathrm{kg}} ~\cdot~ \frac{ \class{brown}{5.972 \cdot 10^{24}\,\mathrm{kg}} }{ 1.52 \cdot 10^{8} \, \mathrm{m} } \\\\
~&=~ 3.43 \cdot 10^{28} \, \mathrm{N} \end{align} $$

Thus, the Earth is attracted about 7% more when it is at perihelion than when the Earth is at aphelion.

Gravitational force between more than two objects

Of course, our universe does not consist of only two objects. If we look at the Earth, for example, it is attracted not only by its nearest neighbor, the Moon, but also by the Sun and all the other planets in the solar system. Strictly speaking, it is attracted by all the masses in the universe. However, some masses are so far away ($r$ is large) that their attraction is practically zero and can be neglected.

But in the case of the earth not only the attraction of the moon is not negligible, but also that of the sun. Let's assume that the earth is attracted by two celestial bodies, the sun and the moon. (We neglect the other planets for the time being).

What is the total gravitational force $ F_{\text g} $ acting on the Earth and where is the Earth being pulled towards? Towards the sun? Towards the moon? Somewhere in between? To answer these questions, you need the superposition principle. It states that the total attractive force $ F_{\text g} $ on the Earth, is the vectorial sum of the individual attractive forces:

$$ \begin{align} \boldsymbol{F}_{\text g} ~&=~ \boldsymbol{F}_{\text m} ~+~ \boldsymbol{F}_{\text s} \\\\
~&=~ \begin{bmatrix} F_{\text{m1}} \\ F_{\text{m2}} \\ F_{\text{m3}} \end{bmatrix} ~+~ \begin{bmatrix} F_{\text{s1}} \\ F_{\text{s2}} \\ F_{\text{s3}} \end{bmatrix} \end{align} $$

Here the vector $ \boldsymbol{F}_{\text m} $ is the attraction of the moon to the earth and the vector $ \boldsymbol{F}_{\text s} $ is the attraction of the sun at the earth. Note that it is NOT the sum of the magnitudes, but the sum of the vectors! To determine the gravitational force acting on the earth, you must first be familiar with vector addition. Briefly summarized, how you do this graphically:

Attraction of Three Masses - Addition of the Gravitational Forces to the Total Force

Draw the force vectors $ \boldsymbol{F}_{\text m} $ and $ \boldsymbol{F}_{\text s} $. Their length is equal to the magnitude and the direction is away from the Earth to the respective object, that is, toward the Moon and toward the Earth, as shown in illustration 2.
Move the force $ \boldsymbol{F}_{\text m} $ parallel so the beginning of vector $ \boldsymbol{F}_{\text m} $ is at the arrowhead of vector $ \boldsymbol{F}_{\text s} $.
Move the force $ \boldsymbol{F}_{\text s} $parallel so the beginning of vector $ \boldsymbol{F}_{\text s} $ is at the arrowhead of vector $ \boldsymbol{F}_{\text m} $.
A parallelogram of forces is created. The total force $ \boldsymbol{F}_{\text g} $ points along the diagonal as shown in illustration 2.

How the exact vector addition works, you will learn in another lesson. This chapter should only clarify that the law of gravity is not limited to the attraction between TWO masses.

Gravitational field and the gravitational acceleration

The gravitational field $ g $ caused by a body of mass $ \class{brown}{M} $ is the gravitational force $ F_{\text g} $ per mass. In other words, the gravitational field indicates what gravitational force a small sample mass $ \class{brown}{m} $ (e.g., a ball) would experience if placed at a distance $ r $ from the large mass $ \class{brown}{M} $.

We get the gravitational field by dividing the gravitational force 1 by the test mass $ \class{brown}{m} $:

$$ \begin{align} g ~=~ G \, \class{brown}{M} \, \frac{1}{r^2} \end{align} $$

The gravitational field has the unit of force per mass, i.e. N/kg (Newton per kilogram). In SI units, this is equal to $ \mathrm{m}/\mathrm{s}^2 $ (meters per square second). The gravitational field is thus a acceleration field and indicates what gravitational acceleration a test mass at distance $r$ from mass $ \class{brown}{M} $ would experience.

Example: gravitational acceleration on earth

The gravitational acceleration $ g $ experienced by a ball or other test mass near the surface of the Earth can be determined using the Earth's mass $ \class{brown}{M = 5.972 \cdot 10^{24} \, \mathrm{kg}} $ and the Earth's radius $ r = 6.371 \cdot 10^{6} \, \mathrm{m} $. Put the values into Eq. 6:

$$ \begin{align} g ~&=~ 6.674 \cdot 10^{-11} \, \frac{\mathrm{m}^3}{ \class{brown}{\mathrm{kg}} \ \mathrm{s}^2 } \cdot \frac{ \class{brown}{5.972 \cdot 10^{24} \, \mathrm{kg}} }{ (6.371 \cdot 10^{6} \, \mathrm{m})^2 } \\\\

~&=~ 9.8 \, \frac{\mathrm{m}}{ \mathrm{s}^2 }
\end{align} $$

Table : Gravitational acceleration not far above the surface of the celestial body. For the distance of the celestial body to the test mass, the radius of the respective celestial body was used.
Celestial body	Gravitational acceleration $ \class{red}{g} $
Mars	3.7
Venus	8.9
Earth	9.8
Jupiter	24.8
Sun	274

Gravitational potential and potential energy in the gravitational field

You get the potential energy (gravitational energy) $ W_{\text{pot}} $ of a test mass $ \class{brown}{m} $, which is at a distance $ r $ from the large mass $ \class{brown}{M} $, if you divide the gravitational force 1 by $ r $:

$$ \begin{align} W_{\text{pot}} ~=~ -G \, \class{brown}{M} \, \frac{\class{brown}{m}}{r} \end{align} $$

The potential energy is negative (has a minus sign) so that the test mass $ \class{brown}{m} $ has a smaller ("more negative") potential energy when it is closer to the mass $ \class{brown}{M} $. It's just a convention that physicists have established that way. If you are only interested in the magnitude of the potential energy, then you can omit the minus sign.

The potential energy of a test mass (e.g. ball) in the gravitational field of the massive body (e.g. planet) can be expressed with the help of the gravitational acceleration $ g $. Multiply both sides with the radius $ r $ in Eq. 6, then you get the following equation:

$$ \begin{align} g \, r ~=~ G \, \class{brown}{M} \, \frac{1}{r} \end{align} $$

Now you can replace the righthand side of Eq. 8 with $ \class{brown}{m} \, g \, r $:

$$ \begin{align} W_{\text{pot}} ~=~ -\class{brown}{m} \, g \, r \end{align} $$

If we use an approximation for the gravitational acceleration $ g $ for the earth, then $ r $ corresponds to the height (let's call it $ h $) above the ground:

$$ \begin{align} W_{\text{pot}} ~=~ \class{brown}{m} \, g \, h \end{align} $$

We have omitted the minus sign because only the magnitude of the potential energy is given here.

The gravitational potential $ V $ is potential energy $ W_{\text{pot}} $ per mass $ \class{brown}{m} $. So divide equation 7 to get the gravitational potential:

$$ \begin{align} V ~=~ \frac{W_{\text{pot}}}{\class{brown}{m}} ~=~ -G \, \class{brown}{M} \, \frac{1}{r} \end{align} $$

The advantage of gravitational potential over potential energy is that it is independent of the test mass $ \class{brown}{m} $.

What is the difference between mass and weight?

Mass $ m $ is the property of a body. Weight, on the other hand, is an informal abbreviation for the gravitational force $F_\text{g}$. This depends on the mass $ m $ as follows: $ F_\text{g} ~=~ m \, g $.

What is the difference between gravitational and inertial mass?

Inertial mass $ \class{brown}{m_\text{t}} $ is in the Newton axiom: $F~=~\class{brown}{m_\text{t}} \, a$. Inertial Mass $ \class{brown}{m_\text{t}} $ is a component of the Newtons Second Law: $F~=~\class{brown}{m_\text{t}} \, a$. Inertial mass resists a change in the motion state of a body. This state changes during change in direction or during acceleration of the body. Gravitational Mass $ \class{brown}{m_\text{s}} $, on the other hand, relates to gravity and influences how strong the gravitational force $F_\text{g} ~=~ \class{brown}{m_\text{s}} \, g$ is. Here, $g$ is the gravitational acceleration.

Experiments have shown that inertial mass $ \class{brown}{m_\text{t}} $ and heavy mass $ \class{brown}{m_\text{s}} $ agree with an accuracy of $10^{-13}$ and are therefore assumed to be equal: $\class{brown}{m_\text{t}} = \class{brown}{m_\text{s}} $.