You must have heard the phrase ‘what goes up must come down’. In physics terms, as Sir Isaac Newton discovered, what goes away from the earth will be pushed back towards it. So, when you throw a ball in the air, whether it goes straight up, to the left, or to the right, it eventually falls back to the ground. And this is true regardless of the height from which the ball is released.
In fact, it takes a significant amount of force, such as the engines of an aeroplane and the lift force produced by the wings, to keep pushing an object away from the earth. Without these forces, it would fall out of the sky.
What is gravity?
Gravity is a force that attracts all objects that have a mass to each other. As the earth has a mass, it attracts other objects towards it. The same is true for other objects, which similarly attract each other towards themselves, including the earth. Even we attract the earth towards ourselves with the force of gravity.
But why isn’t this obvious? Why don’t we see other objects attract each other, given that they all have mass? We will consider this in what follows.
Gravity is not just a force but a force field
A force field is a region in which an object experiences a non-contact force.
Force fields cause an interaction between objects and particles without the objects touching each other. In the case of gravity, that interaction happens between masses. Any object will experience an attractive force if you put it in the gravitational field of another object.
Gravitational Force Facts
Gravitational force is a non-contact force that is always attractive. This means that it can act at a distance and doesn't need the interacting objects to touch each other. In general, when we think of gravity acting on a system, we can consider that the gravitational force is exerted directly on the system's centre of mass.
The centre of mass is the weighted average position of the mass distribution of a system.
The centre of mass of a sphere with uniform mass distribution is in its geometrical centre.
Gravitational force is always exerted along the line connecting the centre of mass of the interacting objects. Moreover, the force of gravity acting between them is an action-reaction pair, which are forces of equal magnitude and opposite directions.
The gravitational force that Earth exerts on the Sun has the same magnitude as the force that the Sun exerts on Earth, just opposite direction.
Fig. 2 - The gravitational force that Earth exerts on the Sun, \(F_{g,\space Earth,\space Sun}\), has the same magnitude and opposite direction as the force that the Sun exerts on Earth\(F_{g,\space Earth,\space Sun}\). However, both of them act along the line that joins their centres of mass.
The gravitational force between two objects with mass can be quite similar to the electrostatic force between two charged objects. After all, both are non-contact forces that can pull objects together at a distance. However, there are many important differences, so these two forces should not be confused. Let's discuss a little bit more how they compare.
The vector representation of force fields
Force fields can be represented as a system of vectors, as in this diagram, in which the arrows represent the gravitational field on the earth.
Figure 1. Gravitational field lines. Source: Sjlegg, Wikimedia Commons (Public domain).
The earth’s gravitational field is radial, which means that the lines of force intersect at the centre of the earth.
As the diagram shows, the field lines are closer together at the surface of the earth. This indicates that the gravitational force is stronger here. Where the lines move further apart from each other, the force decreases.
How do we calculate the force of gravity?
Have a look at the equation below, which represents Newton’s law of gravitation:
\[F = G \frac{m_1m_2}{r^2}\]
- F = magnitude of the gravitational force.
- G = gravitational constant.
- r = distance between the centres of two masses.
- m1 = mass of one of the objects.
- m2 = mass of the other object.
Newton’s gravitational field: when two bodies are placed in a gravitational field, they experience a force that is the product of the two masses and the inverse square of the distance from the centre of both masses.
The constant G is a gravitational constant, which has a very small value:
\[G = 6.67430 \cdot 10^{-11} \frac{Nm^2}{kg^2}\]
Calculate the gravitational force between two 3kg spheres that are 2m apart.
The mass of both objects is 3kg. So m1 and m2 are 3kg, while r is 2m, with G being \(6.67 \cdot 10 ^ {-11} \frac{Nm ^ 2}{kg ^ 2}\). Putting in all the values gives us:
\[F = 6.67 \cdot 10^{-11} \cdot \frac{3 \cdot 3}{2^2}\]
\[F = 1.5 \cdot 10^{-10}\]
The gravitational constant G, which, as we said, has a very small value, is the reason why objects don’t fly and collide with each other. It is also the reason why the earth is not attracted to us but we to it. After all, our mass is negligible compared to that of the earth.
Figure 2. Force F acting on m2 due to m1.
The distance between the two objects has more impact than their masses because Newton’s gravitational equation follows an inverse square law. This means that if the distance doubles, the force is one-quarter of the strength of the original force.
What is the gravitational force of a single mass?
The force of a single mass is its gravitational field strength, which is defined as force per unit mass when it is placed in a gravitational field.
\[g = \frac{F}{m}\]
- g is measured in units of newtons per kilogram (\(N \cdot kg^{-1}\)).
- F is the force experienced by mass m when it is placed in a gravitational field.
As the gravitational field on the earth’s surface is almost uniform, we can assume g to be constant. Hence, g is just the acceleration of mass m in a gravitational field.
Point masses
Point masses are objects that behave as if all mass is concentrated at their centre. Uniform shapes have a point mass.
The significance of point masses is that they have a radial gravitational field. In this, the field lines radiate from its centre. For point masses, our earlier equation becomes:
\[g = \frac{Gm}{r^2}\]
- m = mass of the object (kg).
- G = gravitational constant (\(6.67 \cdot 10^{-11} \frac{Nm^2}{kg^2}\)).
- r = distance from the centre (m).
The gravitational field strength of other planets
The gravitational force depends on the mass of the planet. Mars, for instance, has a gravitational field strength of 3.71 N/kg because it is only about half the diameter of the earth. But, and here comes the interesting part, your weight also depends on the gravitational force g.
Mass and weight
Your mass is the same wherever you go in the universe. What differs is your weight, which depends not only on your mass but also on gravity. So, for instance, if you weigh 99.8kg on earth, you would only weigh 37.74kg on Mars.
The moon has a gravitational force of 1.62 N/kg. This is why on the moon, it is easier to fly than to walk. On Mars, walking becomes a bit easier but is still a challenge because of the low gravitational pull.
Mass and distance
The tides that form on the surface of the earth show how both mass and distance affect the gravitational force.
We get tides on the earth’s surface because of the gravitational pull of the moon and the sun. And although the sun has far more mass than the earth, the distance between the two plays a significant role due to the inverse square proportionality. As the moon is much closer to the earth, the earth’s oceans respond to the moon revolving around the earth, which causes the tides. The sun does have an impact, too, but the tides produced by the sun are much smaller.
Gravitational Force Examples
It is time to justify those crazy claims we presented before. Let's start by discussing why a hollow Earth could present a lack of gravity inside. This is a consequence of a result known as Newton's shell theorem which states that the net gravitational force exerted on an object located anywhere inside a thin, spherical shell is zero. This result sounds surprising because outside of the shell the net gravitational force can be considered as the result of all the mass being located in the centre of the sphere.
Rather than a formal demonstration requiring calculus, we will provide a symmetry argument. Let's assume that Earth's mass is uniformly distributed and that it is spherical. But it will help to think of it as if it were a ring of mass fist.
Fig. 3 - Consider that all the mass of Earth was distributed evenly forming a circumference.
Considering any point inside of it, we can always divide the ring into two regions by tracing a diameter through the point and a line perpendicular to it.
Fig. 4 - We divide the ring into two regions.
Let's focus on Region 1 in Fig. 6. It shows three pieces of mass at different points on it: a, b, and c. Point b is in the direction of the diameter traced before, and a and c are at the same distance from b. All these pieces of the ring have the same mass. The forces due to these pieces of mass are represented by \(\vec{F}_a, \space \vec{F}_b\), and \(\vec{F}_c\), respectively. Since the mass at position b, is the closest to our test mass inside the ring, this is greater than the others. Since a and b are at the same distance from the test mass, \(\vec{F}_a\), is equal in magnitude to \(\vec{F}_c\) and their components are perpendicular to the direction of \(\vec{F}_b\) are equal too, cancelling each other out.
Fig. 5 - Region 1, produces a net force in the direction of \(\vec{F}_b\). The same happens when considering the corresponding part of the spherical shell.
For any point to the right of mass b, there is a corresponding point at the same distance to the left, and adding their forces produces the same result. Therefore, for all the forces due to the pieces of mass forming Region 1, only the components in the direction of \(\vec{F}_b\) will remain. Adding all the contributions from Region 1, we obtain a force acting in the direction of \(\vec{F}_b\). We can extrapolate this idea if we consider the corresponding section of the shell. For every point that is not b, we can find another on this part of the spherical shell such that their components perpendicular to \(\vec{F}_b\) cancel out. The resulting gravitational force from it, \(\vec{F}_1\) would be in the same direction.
By the same argument, we can conclude that the part of the shell corresponding to Region 2 exerts a force, \(\vec{F}_2\), in the opposite direction.
Fig. 6 - Region 2 generates a net force equal and opposite to that of Region 1.
Not only that, \(\vec{F}_2\) is equal in magnitude to \(\vec{F}_1\). Perhaps this is a bit hard to believe given that there is more mass in Region 2. However, the positions of the mass pieces forming Region 2 are overall farther away than those in Region 1. The decrease of the force due to this greater distance is such that it compensates, making both forces equal in magnitude. Finally, we can add these forces together:
\[F_{net} = \vec{F}_1 + \vec{F}_2 = 0\]
The net gravitational force inside the thin, spherical shell is indeed zero.
We can take a step further and see that the thickness does not really matter at all. We can consider any thick shell as formed by multiple layers of thin shells, all of them with a zero net gravitational effect. Then, the net gravitational force exerted by the thick shell is the sum of the thin shells— zero!
Fig. 7 - Cross-sectional representation of a thick spherical shell model of Earth. This thick shell can be considered to be comprised of multiple thin shells.
This result can help us to understand what happens when digging a tunnel. As we dig, the net gravitational effect will be the sum of the gravitational effect from the thick spherical shell we leave behind—which is zero—and that of the spherical portion of the Earth still ahead of us. Therefore, we only feel a gravitation force due to the spherical section ahead of us.
Fig. 8 - When digging a tunnel through Earth, only the mass of the sphere ahead of us contributes to the net gravitational force we feel.
We can calculate the gravitational effect from this spherical portion by considering Earth has a uniform mass distribution (in reality Earth is much denser near its centre, so this is just an approximation). We can define a constant mass density, \(\rho\), as
\[\rho = \frac{M_E}{V_E}\]
where ME and VE are the mass and volume of Earth, respectively. Using this density we can express the mass of the spherical portion, M, using the formula for the volume of a sphere:
\[M = \rho V = \rho \frac{4 \pi R^3}{3}\]
In the equation above, R, is the radius of the portion ahead of us (the distance to Earth's centre). Thus, the gravitational force due to this spherical portion of Earth over a test mass m is
\[F_g = \frac{GMm}{R^2} = \frac{G}{R^2} \frac{4 \pi \rho R^3}{3} m = \frac{4 \pi G m \rho}{3}R\]
Surprisingly, gravitational force does not depend on the square of the distance from Earth's centre anymore. Once we are inside Earth, the gravitational force depends linearly on the distance to its centre. This doesn't mean that Newton's Law of Gravitation is not valid there. We are calculating the force the same way. However, as we decrease the distance the effective mass decreases proportionally to a factor of R3.
Thus, effectively, the gravitational force is proportional to R. The next graph shows the gravitational force that a 70kg person would experience as a function of the distance from Earth's centre.
Fig. 9 - Once inside Earth, the gravitational force decreases linearly as the distance from Earth's centre decreases.
Definitely interesting. But it is still not clear how can this force cause oscillations as mentioned at the beginning. We can solve this mystery by comparing the form of the previous equation to that of a force on a spring. The magnitude of a spring's force Fs is
\[F_s = kx\]
where k is a constant related to the spring stiffness and x is the distance from the equilibrium position. While the equation for the gravitational force over a test mass, m, inside Earth is
\[F_g = \frac{4 \pi G m \rho}{3} R\]
Note that if the mass is uniformly distributed, \(\frac{4 \pi G m \rho}{3}\) is a constant as well, which we can represent as kg. Then, both equations are identical in form.
\[\begin{align} F_s = kx \\ F_g = k_g R \end{align}\]
Even more, R, is the distance from Earth's centre. Once we dig enough to reach R = 0 the gravitational force is 0. And if we keep digging, the distance starts increasing again. Since all the time this force is pulling us towards Earth's centre, R plays exactly the same role as x in the spring equation: R is the distance from an equilibrium point.
This is why if we built a tunnel across the diameter of the Earth and fell into it, we would oscillate around the Earth's centre just as a mass attached to a spring would. Now that you know that both force equations have the exact same form, hopefully, it is easier to believe that the motion they produce is the same.
Let's do one more example.
Consider a 70 kg person digging a tunnel. If they dig \(1.000 \cdot 10^5 m\) going across Earth's diameter. What is the magnitude of the gravitational force they would feel? (Assume that Earth's mass is distributed uniformly).
Solution
Earth's radius is \(r_E=6.371 \cdot 10^6m\), after digging \(1.000 \cdot 10^5m\) the distance from Earth's center would be:
\[R =6.371 \cdot 10^6m−1.000 \cdot 10^5m=6.271 \cdot 10^6m\].
We defined the density of Earth but we did not calculate that explicitly, so let's do that
\[\begin{align} \rho &= \frac{M_E}{V_E} \\ &= \frac{M_E}{\frac{4}{3} \pi (R_E)^3} \\ &= \frac{5.972 \cdot 10^{24} kg}{\frac{4}{3} \pi (6.371 \cdot 10^6 m)^3} \\ &= 5513 \frac{kg}{m^3} \end{align}\]
Now we are ready to use the formula we found for the gravitational force inside Earth.
\[\begin{align} F_g &= \frac{4 \pi Gm \rho}{3}R \\ &= \frac{4 \pi (6.67 \cdot 10^{-11} \frac{Nm^2}{kg^2})(70.0 kg)(5513 \frac{kg}{m^3})}{3} (6.271 \cdot 10^6 m) \\ &= 676 N \end{align}\]
After digging 100km, the person feels a gravitational force of 676N.
Similarities and Differences Between Electromagnetic and Gravitational Forces
There is symmetry between the equation for gravitational force \(F_g\), and that for electrostatic force \(F_e\).
\[F_g = G \frac{m_1m_2}{r^2}\]
\[F_e = k_e \frac{q_1 q_2}{r^2}\]
Don't worry about the meaning of these expressions. For now, we only want to point out the resemblance between them.
Since both interactions are non-contact forces, we can model them, using vector fields. This means that we can think of the effect of these forces as permeating the whole space, having a different intensity at each point. Hence, we define a vector field by specifying the force an object would exert over a test object at every point.
One of the key differences is that gravity is an interaction between objects with mass, but electromagnetic forces are interactions associated with objects with electric charges, typically in motion. Another difference is that gravitational force is only attractive, but electromagnetic forces can be both attractive and repulsive.
Importantly, the gravitational force is much weaker than the electrostatic force. Think about this. We can use a small magnet to lift another and achieve work against the force of gravity. However, the magnetic force between the small magnets is enough to overcome the gravitational force of the entire planet!
Hopefully, this makes the differences clear. But we still have a lot to discuss, so let's not deviate too much from today's topic and continue by filling in the details behind the formula presented above.
Gravitational Fields - Key takeaways
- Gravity is all about masses attracting one another.
- A gravitational field is a force field in which an object experiences a force.
- Only large masses, such as the sun, the moon, and other planets, have a significant gravitational force.
- Newton’s law of gravitation is:
\[F = G \cdot \frac{m_1m_2}{r^2}\]
The law of gravitation is an inverse square law, which says that the gravitational force decreases when the distance between the objects increases.
Gravitational field strength is force per unit mass acting on an object placed in a gravitational field.
In a radial field, the gravitational field can be represented as:
\[g = \frac{Gm}{r^2}\]
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