Suppose you could dig a tunnel from the Earth's surface straight to the center of the Earth. When you crawl down the tunnel that you dug, do you think your weight would increase, decrease, or stay the same? On the one hand, you have less and less mass below you pulling you towards the Earth's center, but on the other hand, you get closer and closer to the Earth's center of mass. One easy and intuitive way to answer this question is to think about what would happen to your weight once you reach the center of the Earth: it would be zero because all parts of the Earth's pull will cancel each other out. Thus, we suspect that our weight will gradually decrease when we crawl down our own tunnel toward the center of the Earth. Let's take a closer look at gravitational acceleration on Earth and on other planets.
Gravity on Earth
On and near Earth's surface, the value for the gravitational acceleration is approximately \(g_\text{Earth}=9.81\,\frac{\mathrm{m}}{\mathrm{s}^2}\). This means that all objects that fall freely near Earth's surface experience the same acceleration, \(g_\text{Earth}\). You experience this as well: if you jump upwards, gravity slows you down, and as you fall back to Earth after reaching your peak elevation, gravity speeds you up downwards. This slowing down and speeding up both happen with the exact acceleration \(g_\text{Earth}\).
The gravitational acceleration at a point in space is the acceleration experienced by an object situated at that point when gravity is the net force acting on the object.
The gravitational acceleration (which is an acceleration) should not be confused with the force of gravity (which is a force).
Acceleration Due to Gravity on Different Planets
You may now be curious about whether or not other planets have a constant gravitational acceleration at their surface. To find out, we need to understand where such gravitational acceleration comes from in the first place.
Newton's Law of Universal Gravitation
Isaac Newton realized that our weight is the consequence of Earth's mass pulling on us. According to his third law of motion, this means that we are also pulling on the Earth with our mass. He concluded that there is always a force acting between two objects with mass. Using the data for planetary orbits presented by Johannes Kepler, and using Kepler’s three laws of planetary motion, Newton also concluded that the gravitational force,\( F, \) must be proportional to both masses, \( M \) and \( m, \) and inversely proportional to the square of the distance between them, \( r. \)
To express this relationship mathematically he needed a proportionality constant:
\[G=6.67\times 10^{-11}\,\frac{\mathrm{N}\,\mathrm{m}^2}{\mathrm{kg}^2},\]
which is now called Newton's gravitational constant.
Newton's Law of Universal Gravitation is given by
\[F=\frac{GMm}{r^2}.\]
Gravitational Acceleration
Fig. 1 - Loose rocks are no problem at all, except for the fact that those rocks experience a gravitational acceleration towards the road.
Here comes the ingenious part. We know from Newton's second law of motion that
\[F=ma,\]
where \(F\) is the net force acting on an object of mass \(m,\) causing an acceleration \(a\) as a result. Thus, if an object with mass \(m\) accelerates as a result of the gravitational force \(F\) exerted by another object with mass \(M\) at a distance \(r\), then we can write down
\begin{aligned} \textcolor{#00b695}{F}&=ma,\\ \textcolor{#00b695}{\frac{GMm}{r^2}}&=ma. \end{aligned}
Then, can divide by the mass of our object, \( m, \) on both sides and this gets us
\[a=\frac{GM}{r^2}.\]
We see that the acceleration of our object does indeed not depend on its own mass, but only on the mass of the other object and its distance to our object! This means that we are dealing with a constant gravitational acceleration \(g\) at that point in space, given by
\[\boxed{g=\frac{GM}{r^2}.}\]
It is important to realize that this gravitational acceleration is a property of that particular point in space, and not of the object itself: all objects will experience the same gravitational acceleration at that point.
In principle, we have only shown that this is true for point masses at a certain distance from each other. However, it turns out (you can check this using calculus) that spheres and balls will have the same effect on objects as if they were a point mass at the location of their center of mass! So for calculations with spheres and balls (planets, for example), we can just take their center of mass to determine the effective distance to other objects.
Let's immediately check this result with something we know!
Let's calculate the acceleration due to gravity on the surface of Earth. Earth's mass is \(M=5.97\times 10^{24}\,\mathrm{kg}\) and its radius is \(r=6.37\times 10^6\,\mathrm{m}\). Thus, we can calculate the gravitational acceleration by substituting these values directly in our formula.
\begin{align*}g_\text{Earth}&=\frac{GM_\text{Earth}}{r_{Earth}^2}=\frac{ 6.67\times 10^{-11}\,\frac{\mathrm{N}\,\mathrm{m}^2}{\mathrm{kg}^2} \times 5.97\times 10^{24}\,\mathrm{kg} }{ \left(6.37\times 10^6\,\mathrm{m}\right)^2 }\\&=9.81\,\mathrm{\frac{m}{s^2}}.\end{align*}
Amazingly, we get exactly the value we know and love!
So what about our experience in the tunnel to the center of the Earth that we dug ourselves earlier?
Well, if every planet had the same constant mass density \(\rho\), we could have a simpler rule. The mass \(M\) of the planet could then be expressed in terms of this constant density and its volume \(V.\) Remember that we define density as the ratio of the mass to the volume
\[\rho = \frac{M}{V} ,\]
then
\[M=\rho \textcolor{#00b695}{V}=\rho \textcolor{#00b695}{\frac{4\pi R^3}{3}}.\]
So the gravitational acceleration on the surface of a planet with radius \(R\) would then be
\[g=\frac{G\textcolor{#00b695}{M}}{R^2}=\frac{G}{R^2}\textcolor{#00b695}{\frac{4\pi \rho R^3}{3}} =\frac{4\pi G\rho}{3}R.\]
We see that the gravitational acceleration would be linear in the planet's radius! With a bit of calculus, we can determine that the "shell" of the part of the Earth that would be around us has no net gravitational influence on us, so if we would be halfway through our tunnel to the center of the Earth, we would experience a gravitational acceleration that is half of that on the Earth's surface! We would gradually experience less and less gravity until we are at the very center of the Earth.
It is also important to note that the gravitational acceleration points toward the center of mass of the object creating the gravitational field. Thus, although in daily life we only experience a pretty much constant gravitational acceleration downwards, in reality, it points toward the center of the Earth.
We have discussed how we can use a single value, \( g, \) to account for the product of relevant constants from the gravitational force formula, finding the acceleration of an object due to gravity. But we can follow the same idea to talk about the force that the object feels —its weight. Let's see how we can do this.
Gravitational Field Strength and Weight
A test object feels a gravitational force near any astronomical object, and this force is referred to as the test object's weight.
Weight is the force an object feels due to the gravity of an astronomical object.
According to Newton's Law of Universal Gravitation, \(F=\frac{GMm}{r^2}\), the weight is different at different distances from the astronomical object. This means that the gravitational force, \( \vec{F}, \) has a value that is different at every point in space. We say that the astronomical object generates a gravitational field, and we define the gravitational field strength \( \vec{g} \) as the vector:
$$\vec{g}=\frac{\vec{F}}{m} .$$
Its magnitude, \( |\vec{g}| = g, \) is given as
$$g=\frac{GM}{r^2},$$
and its direction points toward the center of mass of the astronomical object. With these definitions, we can calculate the weight \(\vec{w}\) of an object simply as
$$\vec{w} = m\vec{g}.$$
For Earth, the gravitational field strength is considered constant, and its approximate value is \( |\vec{g}|=9.81\,\mathrm{\frac{N}{kg}}. \)
But why? Why can we consider it constant if it changes according to the distance? After all, doesn't gravitational force depend on \( r^2\)? The thing is that astronomical objects have such a big size that the values for this force near their surface do not change much at usual height values. Consider we wanted to calculate the gravitational force at \( 10\,\mathrm{m} \) above the surface of Earth.
\begin{aligned} r &= r_\text{Earth}+h\\ &= 6.37\times10^6\,\mathrm{m}+10\,\mathrm{m}\\ &= 6\,370\,000\,\mathrm{m}+10\,\mathrm{m}\\& = 6\,370\,010\,\mathrm{m} \approx 6.37\times10^6\,\mathrm{m}\\\implies r&\approx r_\text{Earth}. \end{aligned}
We can see that such a height difference, in reality, would barely have any effect on our calculations. Therefore, we can think of the gravitational field near the surface of an astronomical object as constant and use its radius in the formula for gravitational field strength, just as we did for acceleration.
As you can see, gravitational field strength and gravitational acceleration are closely related but they are different concepts: gravitational field strength lets us know the gravitational force on an object— its weight— if we know its mass, while gravitational acceleration lets us know the magnitude of its acceleration if gravity is the only force acting on it. And, once more, when the weight is the net force acting on an object, we know that its acceleration will be the same numerical value as the gravitational field strength, but in \( \mathrm{ \frac{m}{s^2}} \) instead of \( \mathrm{\frac{N}{kg}} .\) Therefore, we can just talk about the gravitational acceleration without missing vital information.
Gravity on Jupiter
Fig. 3 - Jupiter and one of its moons, called Europa. On Jupiter, you would weigh more, and on Europa, you would weigh less than on Earth.
Jupiter is the largest planet in our Solar System: its diameter is more than ten times Earth's diameter. However, it is a gas giant, which means it is mostly made up of gas. Therefore, its mass is not as large as you would think based on its radius and on Earth's mass density. We can calculate the gravitational acceleration on the surface of Jupiter because we know its mass is \(M_\text{Jupiter}=1.90\times 10^{27}\,\mathrm{kg}\) and its radius is \(r_\text{Jupiter}=6.99\times 10^7\,\mathrm{m}\):
\begin{align*}g_\text{Jupiter}&=\frac{GM_\text{Jupiter}}{r_\text{Jupiter}^2}=\frac{6.67\times 10^{-11}\,\frac{\mathrm{N}\,\mathrm{m}^2}{\mathrm{kg}^2} \times 1.90\times 10^{27}\,\mathrm{kg} }{\left(6.99\times 10^7\,\mathrm{m} \right)^2}\\&=25.9\,\mathrm{\frac{m}{s^2}}.\end{align*}
We see that the gravitational acceleration on Jupiter is about 2.5 times as large as that on Earth. That means that your scale would read 2.5 times as much on Jupiter as it usually reads here on Earth! You'd have to be pretty strong to be able to walk in such conditions: imagine having to carry 1.5 times your body weight on your back and walking around!
Gravity on Neptune
Neptune is the planet in our Solar System that is farthest from the Sun. Let's calculate the gravitational acceleration on Neptune's surface slightly differently this time. Its radius compared to Earth's radius is given by \(r_\text{Neptune}=3.86r_\text{Earth}\), its mass compared to Earth's mass is given by \(M_\text{Neptune}=17.1M_\text{Earth}\). We can then calculate the following:
\begin{align*}g_\text{Neptune}&=\frac{GM_\text{Neptune}}{r_\text{Neptune}^2}=\frac{\frac{GM_\text{Neptune}}{r_\text{Neptune}^2}}{\frac{GM_\text{Earth}}{r_\text{Earth}^2} }\frac{GM_\text{Earth}}{r_\text{Earth}^2}\\&=\frac{M_\text{Neptune}}{M_\text{Earth}}\left(\frac{r_\text{Earth}}{r_\text{Neptune}}\right)^2g_\text{Earth} .\end{align*}
This way, we can nicely see how the gravitational acceleration depends on mass and radius in practice. Let's plug in the numbers!
\[g_\text{Neptune}=17.1\left(\frac{1}{3.86}\right)^2\times 9.81\,\mathrm{\frac{m}{s^2}}=11.3\,\mathrm{\frac{m}{s^2}}.\]
We conclude that the gravitational acceleration on the surface of Neptune is only slightly larger than that on the surface of the Earth. You would weigh only 13% more on Neptune than on Earth, which would feel equivalent to carrying your high school backpack.
Gravity on Saturn
Fig. 4 - Surprisingly, on Saturn, you would weigh roughly as much as on Earth.
Saturn is the second-largest planet in the Solar System and the planet with the largest set of rings. Given that its mass is \(M_\text{Saturn}=5.68\times 10^{26}\,\mathrm{kg}\) and its radius is \(r_\text{Saturn}=5.82\times 10^7\,\mathrm{m}\), we can calculate the gravitational acceleration on the surface of Saturn as follows:
\[g_\text{Saturn}=\frac{GM_\text{Saturn}}{r_\text{Saturn}^2}=11.2\,\mathrm{\frac{m}{s^2}}.\]
Note that, like for the other two gas giants we discussed, you might find different values elsewhere, as these planets are not very round: due to their rotation and non-rigidness, they are ellipsoids, so the effective radius depends a lot on where you are on the surface of these planets.
This value is very similar to that for Neptune: although Neptune is smaller in both mass and radius, the ratios compared to Saturn are such that the gravitational acceleration ends up being about the same.
Gravity on Different Planets - Key takeaways
- Newton's Law of Universal Gravitation gives the gravitational force between two bodies as follows:\[F=\frac{GMm}{r^2}.\]
- From Newton's Law of Universal Gravitation and Newton's second law of motion, we can deduce that the gravitational acceleration at any point in space is given by\[g=\frac{GM}{r^2}.\]
- This gravitational acceleration is the same for all objects, regardless of their speed, mass, or anything else. Only their position in space matters.
- An astronomical object generates a gravitational field, and we define its gravitational field strength \( \vec{g} \) as the vector:
$$\vec{g}=\frac{\vec{F}}{m} ,$$
with a magnitude, \( |\vec{g}| = g, \) is given by
$$g=\frac{GM}{r^2},$$
its direction points toward the center of mass of the astronomical object and it is measured in \(\mathrm{\frac{N}{kg}}.\)
For Earth, the gravitational field strength is considered constant, and its approximate value is \( |\vec{g}|=9.81\,\mathrm{\frac{N}{kg}}. \)
Weight is the force an object feels due to the gravity of an astronomical object.
We can calculate the weight \(\vec{w}\) of an object as
$$\vec{w} = m\vec{g}.$$
- The gravitational acceleration \(g\) is just the magnitude of the gravitational field strength \(\vec{g}\), in \( \mathrm{\frac{m}{s^2}}. \) For example, for Earth, the acceleration due to gravity is \( 9.81\,\mathrm{\frac{m}{s^2}}. \)
References
- Fig. 1 - Mauritius Road Signs - Warning Sign - Falling rocks (https://commons.wikimedia.org/wiki/File:Mauritius_Road_Signs_-_Warning_Sign_-_Falling_rocks.svg) by Government of Mauritius licensed by Public Domain.
- Fig. 2 - Gravitational acceleration in daily life, StudySmarter Originals.
- Fig. 3 - Hubble captures crisp new image of Jupiter and Europa (50354436493) (https://commons.wikimedia.org/wiki/File:Hubble_captures_crisp_new_image_of_Jupiter_and_Europa_(50354436493).jpg) by European Space Agency (https://www.flickr.com/people/37472264@N04) licensed by CC BY 2.0 (https://creativecommons.org/licenses/by/2.0/deed.en).
- Fig. 4 - Saturn in natural colors (captured by the Hubble Space Telescope) (https://commons.wikimedia.org/wiki/File:Saturn_in_natural_colors_(captured_by_the_Hubble_Space_Telescope).jpg) by Hubble Heritage Team (AURA/STScI/NASA/ESA) licensed by Public Domain.
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