Gravity on Different Planets

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.

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.

Let's immediately check this result with something we know!

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.

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:

\[g_\text{Saturn}=\frac{GM_\text{Saturn}}{r_\text{Saturn}^2}=11.2\,\mathrm{\frac{m}{s^2}}.\]

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.