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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.
Gravitational acceleration is the acceleration an object experiences when gravity is the only force acting on it.
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.
The weight of an object is the gravitational force that an astronomical object exerts on it.
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.
An astronomical object's gravitational field strength at a point is defined as the vector with magnitude
$$ |\vec{g}| = \frac{|\vec{F}_g|}{m}$$
The direction of this vector points toward the object's center of mass.
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!
Note that the units for gravitational field strength, \( \vec{g}, \) are \( \mathrm{\frac{N}{kg}}. \) Again the difference is just conceptual. And after all, \( 1\,\mathrm{\frac{N}{kg}} =1\,\mathrm{\frac{m}{s^2}} . \)
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.
Calculate the gravitational acceleration due to gravity on the moon which has a mass of \(7.35\times 10^{22} \,\mathrm{kg}\) and a radius of \(1.74\times 10^6 \,\mathrm{m}\).
Solution
Let's insert the given values into our gravitational acceleration formula:
$$\begin{align*} g&= \frac{GM}{R^2}\\[6pt]g&=\frac{\left(6.67\times 10^{-11}\,\mathrm{\frac{m^2}{s^2\,kg}}\right)\left(7.35\times 10^{22}\,\mathrm{kg}\right)}{(1.74\times 10^6 \,\mathrm{m})^2} \\[6pt] g&=1.62\,\mathrm{m/s^2.} \end{align*}$$
Calculate the acceleration due to gravity a) on the surface of the Earth and b) \(r= 3500\,\mathrm{km}\) above the surface of the Earth. Earth's mass is \(5.97\times 10^{24} \,\mathrm{kg}\) and its radius is \(R_\text{E}=6.38\times 10^6 \,\mathrm{m}\).
Solution
a) When we are on the surface of the Earth, we will take the distance as the radius of the Earth. Let's insert the values into our equation:
$$\begin{align*} g&=\frac{GM_\text{E} }{R_\text{E}^2} \\[6pt] g&= \frac{\left(6.67\times 10^{-11} \,\mathrm{\frac{m^3}{s^2\,kg}}\right)(5.97\times 10^24 \,\mathrm{kg})}{(6.38\times 10^6 \,\mathrm{m})^2} \\[6pt] g&= 9.78\,\mathrm{m/s^2.} \\ \end{align*}$$
b) When we are \(3500\,\mathrm{km}\) above the surface of the Earth, we should add this value to the radius of the Earth since the total distance is increased. But first, let's not forget to convert \(\mathrm{km}\) to \(\mathrm{m}\):
$$ r=3.5\times 10^6 \,\mathrm{m} + 6.38\times 10^6 \,\mathrm{m} = 9.88\times 10^6 \,\mathrm{m} $$
Now we are ready to substitute and simplify.
$$\begin{align*}g&=\frac{Gm_\text{E}}{r^2} \\[6pt] g&= \frac{\left(6.67\times 10^{-11} \,\mathrm{\frac{m^3}{s^2\,kg}}\right)(5.97\times 10^24 \,\mathrm{kg})}{(9.88\times 10^6 \mathrm{m})^2} \\[6pt] g&=4.08\,\mathrm{m/s^2.}\end{align*}$$
As we can see, when the distance is so big that it is significant when compared to Earth's radius, the acceleration due to gravity can no longer be considered to be constant as it decreases noticeably.
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.
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.
Body | Gravitational acceleration \(\mathrm{m/s^2}\) |
Sun | \(274.1\) |
Mercury | \(3.703\) |
Venus | \(8.872\) |
Mars | \(3.72\) |
Jupiter | \(25.9\) |
Uranus | \(9.01\) |
Gravitational Acceleration - Key takeaways
- Gravitational acceleration is the acceleration an object experiences when gravity is the only force acting on it.
- 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$$F_g = G\frac{m_1 m_2}{r^2}.$$
- The weight of an object is the gravitational force that an astronomical object exerts on it.
- If the force of gravity between the center of mass of two systems has a negligible change as the relative position between the two systems changes, the gravitational force can be considered constant.
- The conventional standard value of gravitational acceleration on Earth is \(9.80665\,\mathrm{m/s^2}.\)
- As the altitude increases, the gravity decreases. This effect is noticeable for heights that are not negligible when compared to the radius of Earth.
- An object that only experiences gravitational acceleration is said to be in free-fall.
- All objects fall at the same rate when in free fall.
- When the weight is the only force acting on an object, its acceleration is equal to the magnitude of the gravitational field strength, but in \( \mathrm{\frac{m}{s}}.\)
References
- Fig. 1 -Space Jump (https://www.flickr.com/photos/massimotiga/8090904418) by Massimo Tiga Pellicciardi (https://www.flickr.com/photos/massimotiga/) is licensed under CC BY 2.0 (https://creativecommons.org/licenses/by/2.0/)
- Fig. 2 - Gravitational Acceleration for the Earth Example, StudySmarter Originals
- Fig. 3 - Gravitational Acceleration Changes with Altitude, StudySmarter Originals
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Frequently Asked Questions about Gravitational Acceleration
What is the formula for gravitational acceleration?
The gravitational acceleration formula is:
g = GM/R2.
In this equation, G is the gravitational constant with a value of 6.67X10-11 Nm2/s2, M is the mass of the planet, R is the distance of the falling object to the center of mass of the planet, and g is the acceleration due to gravity.
What are examples of gravitational acceleration?
Gravitational acceleration varies depending on where you are. If you are at sea level you will perceive a greater acceleration than up in the mountains. The gravitational force decreases with increasing altitude. As another example, if you were on the Moon, acceleration due to gravity would be 1.625 m/s^2 because the Moon has a much weaker gravitational pull than the Earth. Other examples are the Sun, with a gravitational acceleration of 274.1 m/s^2, Mercury with 3.703 m/s^2, and Jupiter, with 25.9 m/s^2.
What is gravitational acceleration units?
The unit of gravitational acceleration is m/s2.
What do you mean by gravitational acceleration?
An object in free-fall experiences gravitational acceleration. This is the acceleration caused by the gravitational force.
How do you calculate gravitational acceleration?
Gravitational acceleration, g, is calculated by multiplying the gravitational constant, G, by the mass of the body that is attracting the falling object, M. Then dividing by the square of the distance, r2.
g = GM/r2
The gravitational constant has a value of 6.67X10-11 Nm2/ss.
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