Gravitational Acceleration

Standing \(24\) miles above the Earth, Austrian daredevil Felix Baumgartner was about to try something people had scarcely even imagined: a space jump. The gravitational pull of the Earth causes objects to accelerate continuously at an approximately constant rate as they fall. Knowing this, on October 14th, 2012, Felix leaned forward and let gravity pull him off the safety of the space shuttle he was in.

Fig. 1 - Felix Baumgartner is about to start his space dive. Once he leans forward, there is no going back!

Normally, air resistance would slow him down. But, Felix was so high above the Earth that air resistance had too small of an effect, and so he was in total free fall. Before he opened his parachute, Felix had broken the sound barrier as well as numerous world records. This article will discuss what made Felix reach the speed he did—gravitational acceleration: its value, formula, units, and calculation—and also go over some gravitational acceleration examples.

Gravitational Acceleration Value

An object that only experiences gravitational acceleration is said to be in free-fall.

Regardless of the masses or compositions, all bodies accelerate at the same rate in a vacuum. This means that if there were no air friction, any two objects falling from the same height would always reach the floor simultaneously. But how large is this acceleration? Well, this depends on the magnitude of the force that Earth pulls us with.

The magnitude of the force that Earth exerts on us at a fixed place on the surface is determined by the combined effect of gravity and the centrifugal force caused by Earth's rotation. But at usual heights, we can ignore the contributions from the latter, as they are negligible in comparison to the gravitational force. Therefore, we will just focus on gravitational force.

The force of gravity near Earth's surface can be considered to be approximately constant. This is because it changes too little for normal heights which are too small in comparison to Earth's radius. This is the reason why we often say that objects on Earth fall with a constant acceleration.

This free-fall acceleration varies over the Earth's surface, ranging from \(9.764\) to \(9.834\,\mathrm{m/s^2}\) depending on altitude, latitude, and longitude. However, \(9.80665\,\mathrm{m/s^2}\) is the conventional standard value. The areas where this value differs significantly are known as gravity anomalies.

Gravitational Acceleration Formula

According to Newton's Law of Gravitation, there is a gravitational attraction between any two masses and it is oriented to drive the two masses toward one another. Each mass feels the same force magnitude. We can calculate it by using

the following equation:

$$F_g = G\frac{m_1 m_2}{r^2}\\$$

where \(m_1 \) and \(m_2 \) are the masses of the bodies, \(G\) is the gravitational constant equal to \(6.67\times 10^{-11}\,\mathrm{\frac{m^2}{s^2\,kg}}\) , and \(r\) is the distance between the bodies' centers of mass. As we can see, the force of gravity is directly proportional to the product of the masses and inversely proportional to the squared distance between their center of mass. When we talk about a planet like Earth, attracting a regular object, we often refer to the gravitational force as the weight of this object.

You might have seen that we often calculate the magnitude of the weight, \( W, \) of an object on Earth using the formula:

$$W= mg,$$

where \( m \) is the mass of the object and \(g\) is typically referred to as the acceleration due to gravity on Earth. But where does this value comes from?

We know that a body's weight is nothing else than the gravitational force that Earth exerts on it. So let's compare these forces:

\begin{aligned} W&=m\textcolor{#00b695}{g} \\[6pt] F_g &= \frac{GM_\text{E} m}{r_\text{E}^2}= m \textcolor{#00b695}{\frac{GM_\text{E}}{r_\text{E}^2}} \\ \end{aligned}

If we identify \( g\) as \( \frac{GM_\text{E}}{r_\text{E}} \) we get obtain a shortcut for calculating the gravitational force on the object —its weight— simple as \(w=mg\). This is so useful that we define a physical quantity to refer specifically to it: the gravitational field strength.

And now you might be wondering, then, why do we call it the "acceleration due to Earth"? If the weight is the only force acting on our object, Newtown Second's Law tells us that

\begin{aligned} ma &= F\\ma &= w\\ ma &= mg\\ a &= g.\end{aligned}

the acceleration of the object is equal to the magnitude of the gravitational field strength, regardless of the object's mass! This is why we calculate the free-fall acceleration or gravitational acceleration of Earth as

$$ g = \frac{GM_\text{E}}{r_\text{E}^2},$$

since the numerical value is the same, it is just a conceptual difference.

Note that the gravitational acceleration of Earth depends only on the Earth's mass and radius (since we are considering the object to be on Earth's surface). However, there is a caveat here. Earth is not perfectly spherical! Its radius changes depending on where we are located. Due to Earth's shape, the value of gravitational acceleration is different on the poles than on the equator. While the gravity at the equator is around \(9.798\,\mathrm{m/s^2}\), it is close to \(9.863\,\mathrm{m/s^2}\) at the poles.

Gravitational Acceleration Units

From the previous section's formula, we can find the unit of gravitational acceleration. Remember that unit of the gravitational constant \(G\) is \(\mathrm{m^3/s^2\,kg}\), the unit of mass is \(\mathrm{kg}\), and the unit of distance is \(\mathrm{m}\, \mathrm{meters}\). We can insert these units into our equation to determine the units of gravitational acceleration:

$$\begin{align*} [g] &=\left[ \frac{Gm_\text{E} }{r_\text{E}^2}\right] \\ [g] &=\left[ \frac{\frac{\mathrm{m}^3 \,\mathrm{kg}}{\mathrm{s^2 \,kg}}}{\mathrm{m^2}} \right] \end{align*}$$

Then, we can cross off the \(\mathrm{kg}\)'s and squared meters on the top and the bottom:

$$[g]=\left[\mathrm{m/s^2}\right]\\\mathrm{.}$$

So, the unit of gravitational acceleration is \(\mathrm{\frac{m}{s^2}}\) which makes sense! After all, it is an acceleration!

Gravitational Acceleration Calculation

We discussed how to calculate the acceleration due to gravity on Earth. But the same idea applies to any other planet or astronomical body. We can calculate its gravitational acceleration using the general formula:

$$ g=\frac{GM}{R^2}.$$

In this formula, \( M \) and \( R \) are the mass and radius of the astronomical object, respectively. And we can know the direction of this acceleration will always be towards the center of mass of the astronomical object.

Now, it's time to apply some of what we know to real-world examples.

Gravitational Acceleration Examples

In the example above, we saw that as the altitude increases, the value of gravity decreases. When we look at the graph below, we see how it changes exactly. Note that this is not a linear relation. This is expected from our equation since gravity is inversely proportional to the square of the distance.

Fig. 3 - This is a graphic of gravitational acceleration vs. altitude. As the altitude increases, the value of gravity decreases.

Gravitational acceleration has different values for different planets because of their different masses and sizes. In the next table, we can see the gravitational acceleration on surfaces of different astronomical bodies.