You are likely to have already heard some variation of how Isaac Newton discovered gravity, the most common version of the story being the one where an apple landed on his head. Although this story is probably not true, the fact that an apple fell in the first place was of great interest to Newton. It prompted him to ask the question, "why does the apple fall to the ground and not upwards or sideways? What force is acting on the apple to cause that?". It was through trains of thought like these that Newton came up with an expression to represent the force (that he named "gravitational force").
This expression was named Newton's Law of Gravitation, is a universal law that describes the gravitational force between two objects, and it has been a staple in both mathematics and science since its discovery. In this article, we will delve deeper into the topic of Newton's Law of Gravitation, covering its definition, formula, and examples of its application.
Newton's law of gravitation: definition
Newton's law of gravitation is defined as follows:
The gravitational force of attraction that exists between any two particles in the universe is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their centres.
From this definition, you can see two things:
An increase in the mass of either of the particles will result in an increase in the magnitude of the gravitational force between them because the gravitational force between two particles is directly proportional to the product of their masses. Similarly, a decrease in the mass of either particle will result in a decrease in the magnitude of the gravitational force between them.
An increase in the distance between the two particles will result in a decrease in the magnitude of the gravitational force between them. This is because the gravitational force is inversely proportional to the distance between them. Similarly, a decrease in the distance between the two particles will result in an increase in the magnitude of the gravitational force between them.
It is important to note that "particle" is not used to refer to an object of atomic size, but is instead used to refer to any object with mass. A particle could be a planet, it does not necessarily have to mean something small.
Newton's Law of Gravitation is also often called Newton's law of Universal Gravitation. This is because the law can be applied anywhere in the universe, and it will hold true.
Newton's law of gravitation: equation
From the definition of Newton's law of gravitation, you find that the equation is:
\[ F_g \propto m_1m_2 \qquad \text{ and } \qquad F \propto \frac{1}{r^2}.\]
When you combine these, you get the following:
\[ F_g \propto \frac{m_1m_2}{r^2}.\]
This result is then multiplied by the universal gravitational constant, \(G\), to get the equation we use to represent Newton's law of Gravitation:
\[ F_{g} = G\frac{m_1m_2}{r^{2}},\]
where:
\(F_g\) is the gravitational force of attraction between two particles.
\(G\) is the universal gravitational constant with \(G = 6.674 \times 10^{-11} \text{N} \cdot \text{m}^{2} /\text{kg}^{2}\).
\(m_1\) and \(m_2\) are the masses of the particles.
\(r\) is the distance between the centres of the two particles.
This equation is also known as the inverse square law. It is called this because the gravitational force of attraction is inversely proportional to the square of the distance between the two particle centres.
Another well-known inverse square law equation is:
\[g = G\frac{M_E}{r^2},\]
where:
- \(g\) is the value of the gravitational free fall acceleration (which is approximately \(9.81\, \text{m/s}^2\) for Earth).
- \(G\) is the universal gravitational constant
- \(M_E\) is the mass of the Earth.
- \(r\) is the distance from the Earth's centre to a point above its surface.
This equation is used to calculate the value of \(g\) at any point above the Earth's surface. It is derived using Newton's law of gravitation.
First, set the following two equations equal to each other:
\[F_g = mg \qquad \text{and} \qquad F_g = G\frac{mM_E}{r^2}.\]
Therefore
\[ mg = G\frac{mM_E}{r^2}.\]
Cancel the like terms
\[ \cancel{m}g = G\frac{\cancel{m}M_E}{r^2},\]
to get the final equation
\[ g = G \frac{M_E}{r^2}.\]
When working with this equation, it is important to note that \(r\) is the Earth's radius plus the height above the Earth's surface.
\[ r = R_E + h.\]
This equation can also be used to calculate the value of the gravitational free acceleration of other planets. Simply use the other planet's mass and radius in place of the Earth's.
Please note that gravitational force always exists between particles and that it is independent of the medium separating the two particles. This means that the particles could be submerged in water or suspended in a vacuum and the gravitational force of attraction between them would not change as long as the distance between them remains constant.
Newton's law of gravitation in vector form
Newton's law of gravitation can also be represented in vector form. This form is essentially the same as the equation in the previous section, but is instead given a direction that is defined by a unit vector, \(\overrightarrow{r}_{12}\).
Fig. 1 - Vector diagram showing the force \(\overrightarrow{F}_{12}\) exerted by \(m_1\) on \(m_2\), in the negative direction of the unit vector \(\overrightarrow{r}_{12}\).
The unit vector is calculated according to \(\overrightarrow{r}_{12} = \overrightarrow{r}_1-\overrightarrow{r}_2\) and it is directed from particle \(1\) towards particle \(2\). The gravitational force exerted on particle \(2\) by particle \(1\) is thus:
\[ \overrightarrow {F_{12}} = - G\frac{m_1 m_2}{r^2} \overrightarrow {r}_{12}.\]
Note the negative sign. In this case, the positive direction would be away from particle 1, but, since particle 2 is attracted to particle 1, the gravitational force exerted on particle 2 is directed toward particle 1 and is thus in the negative direction.
When Newton's third law of motion is applied, you will see that \(\overrightarrow{F_{21}}\) (i.e. the force exerted by particle 2 on particle 1) is equal in magnitude but oppositely directed to \(\overrightarrow{F_{12}}\). This is shown below:
\[\overrightarrow{F_{21}} = - G\frac{m_1m_2}{r^2} \overrightarrow {r}_{21} .\]
You know that \(\overrightarrow{r}_{21} = \overrightarrow{r}_2 - \overrightarrow{r}_1\) and that \(\overrightarrow{r}_{12} = \overrightarrow{r}_1-\overrightarrow{r}_2\) so it can be concluded that:
\[ \overrightarrow{r}_{21} = -\overrightarrow{r}_{12}\]
and when you substitute \(-\overrightarrow{r}_{12}\) into the equation for \(\overrightarrow{F}_{21}\) and simplify, you get the following:
\[ \begin{align}\overrightarrow{F_{21}} & = - G\frac{m_1m_2}{r^2} \left( -\overrightarrow {r}_{12}\right) \\&= G\frac{m_1 m_2}{r^2} \overrightarrow {r}_{12}, \end{align}\]
therefore
\[ \overrightarrow{F_{21}} = -\overrightarrow{F_{12}} .\]
Newton's law of gravitational force
The following examples will detail how one can calculate gravitational force using Newton's law of gravitation.
The distance between a spaceship and an asteroid is \(1.8\, \text{km}\). Calculate the gravitational force between them if the spaceship has a mass of \(14000\, \text{kg}\) and the asteroid has a mass of \(8000\, \text{kg} \). Let the value of the universal gravitation constant be \(6.67\times 10^{-11}\, \text{N} \cdot \text{m}^2 \text{/km}^2 \).
Fig. 2 - Diagram of spaceship and asteroid.
Solution:
STEP 1: Convert all values to SI units.
\[ 1.8\, \mathrm{km} = 1800\, \mathrm{m}\]
STEP 2: Substitute the values into the gravitational force equation and simplify to get the value of \(F_g\).
\[ \begin{align}F_g & = G \frac{m_1m_2}{r^2} \\ & = (6.67 \times 10^{-11} )\frac{(14000)(8000)}{(1800)^2} \\& = 2.31 \times 10^{-9} \, \text{N}\end{align}\]
The gravitational force between the spaceship and the asteroid is thus equal to \(2.31 \times 10^{-9} \; \text{N}\).
What happens if you are asked to calculate the net effect of two particles on one? In these scenarios, you must make use of the vector form of Newton's law of gravitation.
Three hockey balls are placed on the corners of a right-angled triangle, as shown below. The triangle has side lengths \(m = 0.05\, \text{m, } n = 0.12\, \text{m,}\) and \(p = 0.13\, \text{m}\). If each ball has a mass of \(450\, \mathrm{g}\), calculate the magnitude of the gravitational force vector acting on the ball \(2\). Let \(G = 6.67 \times 10^{-11}\, \text{N} \cdot \text{m}^2 \text{/km}^2 \).
Fig. 3 - Diagram showing three hockey balls placed on the corners of a right-angled triangle.
Solution:
STEP 1: First convert all the units to SI.
\[450\, \mathrm{g} = 0.45\, \mathrm{kg} \]
STEP 2: First calculate the force of ball \(1\) on ball \(2\).
\[ \begin{align}\overrightarrow {F_{12}} & = G\frac{m_1 \;m_2}{r^2} \hat{r}_{12} \\ & = (6.67 \times 10^{-11}) \frac{(0.450)(0.450)}{(0.05)^2} \\ & = 5.4 \times 10^{-9} \, \text{N} \end{align}\]
STEP 3: Next calculate the force of ball \(3\) on ball \(2\).
\[ \begin{align}\overrightarrow {F_{32}} & = G\frac{m_2 \;m_2}{r^2} \hat{r}_{32} \\ & = (6.67 \times 10^{-11}) \frac{(0.450)(0.450)}{(0.12)^2} \\ & =9.38 \times 10^{-10} \, \text{N} \end{align}\]
STEP 4: Calculate the net force on ball \(2\).
Since the forces \(\overrightarrow{F_{12}}\) and \(\overrightarrow{F_{32}}\) are directed along the edges of a right-angled triangle, the net force exerted by these on ball 2 can be calculated using the Pythagorean theorem.
\[\begin{align}F & = \sqrt{(F_{12})^2 + (F_{32})^2} \\& = \sqrt{(5.4 \times 10^{-9})^2 + (9.38 \times 10^{-10})^2} \\& =5.48 \times 10^{-9} \, \text{N}\end{align}\]
The magnitude of the net force acting on ball 2 due to ball 1 and ball is equal to \(5.48 \times 10^{-9} \, \text{N}\).
Other examples could involve the Earth or the sun and the use of their respective radii and masses. In these cases, you will normally end up with a large answer.
A satellite of mass \(680 \, \mathrm{ kg} \) orbits the earth at a height of \(2700\, \mathrm{ km} \) above the earth's surface. What is the gravitational force between the Earth and the satellite? Take the mass of the earth to be \(5.98\times 10^{24}\,\text{ kg}\) and the radius to be \(6.38\times 10^{6}\, \mathrm{ m}\).
Fig. 4 - Diagram showing the Earth and the satellite.
Solution:
STEP 1: Convert all units to SI units.
\[ 2700\, \text{km} = 2700 \times 10^3 \, \text{m}\]
STEP 2: Substitute the values into the gravitational force equation and calculate the magnitude of the force.
\[ \begin{align}F_g & = G \frac{m_1m_2}{r^2} \\ & = (6.67 \times 10^{-11}) \frac{(680)(5.98 \times 10^{24})}{\left ((2700 \times 10^3) + (6.38 \times 10^{6}) \right)^2} \\& = 3289.76 \, \text{N}\end{align}\]
The gravitational force between the Earth and the satellite is thus equal to \(3289.76 \; \text{N}\).
Limitations of Newton's law of universal gravitation
As is the case with many things in life, Newton's law of universal gravitation is not perfect.
The law is only applicable to rigid bodies in inertial frames of reference, meaning that it only applies when Newton's laws hold true. It can also only be applied to point bodies.
In cases where the distance between particles reaches a magnitude of \(10^{-9} \,\text{m}\) and smaller or when the particles are travelling at high speeds, comparable to the speed of light, the law becomes ineffective and Newton's law of gravitation should not be applied.
Newton's Law of Gravitation - Key takeaways
- Newton's Law of Gravitation is the magnitude of the gravitational force between two particles is directly proportional to the product of both particles' masses.
- The magnitude of the gravitational force between two particles is inversely proportional to the square of the distance between the two particles.
- The formula for Newton's law of gravitation is: \[F_g = G\frac{m_1m_2}{r^2}.\]
- The vector form of Newton's law of gravitation is: \[\overrightarrow {F_{12}} = - G\frac{m_1 \;m_2}{r^2} \hat{r}_{12}.\] where:
- \(\overrightarrow{F_{12}}\) is the force exerted by particle \(1\) on particle \(2\).
- \(\hat{r}_{12}\) is the unit vector in the direction from particle \(1\) to particle \(2\).
- Newton's law of gravitation becomes ineffective when the distance between particles becomes smaller than \(10^{-9}\, \text{m}\).
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