Lesson Video: Acceleration Due To Gravity

In this video we learn about the acceleration caused by Earth’s mass, and how the kinematic equations let us calculate motion due to gravity’s acceleration.

14:27

Video Transcript

In this video, we’re going to learn about the acceleration due to gravity. What it is, the physical effects that it has, and how to include this acceleration as we work through practice problems.

As we start out, imagine that you are competing in an egg drop competition. This is a competition where competitors will wrap an egg in padding then drop their egg with its padding from various heights and see who can drop their egg from the highest height without the egg breaking. After long hours of preparation, you have devised a support system for your egg which you hope will allow you to drop it from the highest height and win the competition.

After careful testing, you found that your egg support system can reach a maximum speed before impact after which the egg remains intact and doesn’t break of 8.0 meters per second. The question is, given this maximum speed, what’s the highest height you can drop your egg from without the egg breaking? To answer that question, we’ll need to know about the acceleration due to gravity.

To better understand the acceleration due to gravity, let’s step back and consider the Earth as a gigantic mass that we’ll call 𝑚. When a massive object exists, it creates a field around it. This field draws other massive objects in toward it. Through careful measurements, we found out that the closer we are to the mass creating a field, the more we feel pulled towards that mass, whereas the farther away we are, the less we feel attracted.

If we were to take a mass, call it capital 𝑀, and put it on the surface of the Earth, then the effect of the Earth’s attractive field on that mass, capital 𝑀, can be quantified by something called the object’s weight, symbolized 𝑤. The weight of an object on the surface of the Earth is equal to the product of the mass of that object with the acceleration, or the attraction, created by the mass of Earth on that object.

Through experimentation, it’s been determined what that attraction due to the Earth is near the Earth’s surface. As a number, the value of that attraction is about 9.8 meters per second squared. That’s how much the Earth will accelerate a mass of object on its surface. Now, we may be thinking that makes sense for an object which is in free fall, that is falling towards the Earth. But what about one like our mass capital 𝑀 that rests on the surface of the Earth? How can we say that’s affected by an acceleration when clearly it’s not moving?

In the case of our mass capital 𝑀 or any mass that rest on the surface of the Earth, another reactive force helps to keep it from falling towards the center of the Earth. But still the mass is within the gravitational field of the Earth and, therefore, experiences this acceleration of about 9.8 meters per second squared. This number by the way is referred to in shorthand as lowercase 𝑔. And it’s called the acceleration due to gravity.

If we insert this symbol into our equation, so it now reads 𝑤 equals 𝑀 times 𝑔, this equation may you look familiar to something we’ve seen elsewhere. Newton’s second law of motion says that the net force on an object is equal to the object’s mass times its acceleration 𝑎. Comparing these two equations, we see that the two masses line up with one another. And weight is a force and can be represented by the symbol 𝑤 sub 𝐹. So, this shows us more clearly that 𝑔, which is a quantity due to the Earth’s gravitational attraction, is an acceleration, that is, a quantity with units of meters per second squared.

So, the Earth has mass and, therefore, creates a field around itself that attracts other masses. And near the surface of the Earth, when we measure this attraction, we find it’s approximately 9.8 meters per second squared. What does that mean practically for an object that we drop from a certain height? Let’s say we walk up to the edge of a steep cliff and drop a rock so that it enters into free fall. We can imagine further that we have a stopwatch where we keep track of the time after the rock’s release. And we want to understand its position and its speed at each of these time increments.

Since we release the rock from rest, we know its initial speed will be zero meters per second. As soon as we release that rock, it will begin to accelerate at a rate of 9.8 meters per second squared, which means it will speed up 9.8 meters per second every second that it falls. This means that at 𝑡 equals one second, the rock’s velocity will be 9.8 meters per second. And at 𝑡 equals two seconds, its velocity will be 9.8 meters per second plus 9.8 meters per second, or 19.6 meters per second. And at 𝑡 equals three seconds, we’ll add 9.8 meters per second to that. And the rock will now be speeding along at 29.4 meters per second. That’s how the acceleration due to gravity affects the speed of a descending object over time.

We can also use 𝑔, the acceleration due to gravity, to understand the distances covered by the rock vertically over these one-second increments. From when we first drop it to a time of one second, the rock will fall 4.9 meters. As the rock gains speed due to the acceleration caused by gravity, it covers more ground per second. And from zero to two seconds, it covers 19.6 meters of vertical distance. And if we wait till three seconds after dropping the rock, it will be 44.1 meters below the point where we released it.

This is how the acceleration due to gravity 𝑔 affects objects in free fall. The numbers we’ve calculated in this example, by the way, come from using the kinematic equations. Let’s get some practice using those equations with acceleration caused by gravity.

A coin is dropped from a hot-air balloon that is 356 meters above the ground and rising at 12.3 meters per second vertically upward. Assume that vertically upward displacement corresponds to positive values. Find the distance between the coin and the ground 3.07 seconds after being released. Find the velocity of the coin 3.07 seconds after being released. Find the time before the coin hits the ground.

In this multipart exercise, first, we wanna solve for the distance between the coin and the ground 3.07 seconds after the coin is released. We’ll call that distance 𝑑. We also wanna solve for the velocity of the coin that same amount of time after it’s released. We’ll call this velocity 𝑣. And finally, we want to solve for the time it takes before the coin hits the ground. We’ll label this time 𝑡.

We’re told that at the moment the coin is released, the hot-air balloon it’s released from is 356 meters above ground and rising at 12.3 meters per second. Let’s begin our solution by drawing a sketch of this scenario. We’re told that our hot-air balloon is ascending at 12.3 meters per second in a direction upward, which we call the positive direction. And during this ascent, a coin is dropped from the balloon at a moment when the coin is 356 meters above ground level.

Knowing all this, we want to solve for the distance the coin falls. The distance 𝑑 is the distance from ground level to the elevation of the coin after it’s been falling for 3.07 seconds. We also want to solve for the velocity of the coin after it’s been falling for that same amount of time. And we want to find out the total time it takes for the coin to reach the ground.

We know that when the coin is released, it speeds up towards the Earth due to the acceleration caused by gravity 𝑔. Since we’ve established motion vertically up as motion in the positive direction, we can write 𝑔 is equal to negative 9.8 meters per second squared. That acceleration 𝑔 is the only acceleration experienced by the coin. And it’s constant, which means that the kinematic equations apply to this scenario.

These four equations, based on the assumption that acceleration 𝑎 is constant, describe the motion of objects in free fall, for example. Given that we want to solve for distance 𝑑, that we know the coin’s initial velocity, we know the acceleration to which the coin is subjected, and we also know the time after its release at which we want to calculate 𝑑, this all points to using the third kinematic equation listed.

We have to be a bit careful when using this equation because it will tell us how to calculate a distance equal to the distance that the coin descends over 3.07 seconds. But we, rather, want to solve for the coin’s elevation after that amount of time. When we solve for lowercase 𝑑, that will equal the total original elevation of the coin minus the distance it descends over 3.07 seconds. We can write this as 356 meters plus 𝑣 zero 𝑡 plus one-half 𝑎𝑡 squared because of the sign convention we’ve chosen. We’ll see how this works out.

Looking at the three variables involved in our expression, we know 𝑣 sub zero. That’s 12.3 meters per second, and it’s positive. We also know 𝑡. We’re given that as 3.07 seconds as well as 𝑎. That’s equal to negative 9.8 meters per second squared. When we plug these values into our expression and then enter this whole expression on our calculator, we find that lowercase 𝑑 is equal to the original distance from the coin to the ground plus negative 8.42 meters. In other words, over 3.07 seconds, the coin descends 8.42 meters. That means that, rounded to three significant figures, it’s 348 meters above the ground after 3.07 seconds of free fall.

Next, we want to solve for the coin’s velocity 3.07 seconds after it’s released. And again, we’ll look through our list of kinematic equations to search for a match. The first equation involves acceleration which is known, initial velocity which is known, final velocity which we want to solve for, and time which is known.

When we write out this relationship using our variables, we see that 𝑣 sub zero, the initial velocity of the coin, is known because that’s the initial velocity of the hot-air balloon. We also know the acceleration and 𝑡. When we plug these three values into our equation and calculate 𝑣, we find the result of negative 17.8 meters per second. That’s the velocity of the coin 3.07 seconds after it began to fall.

Lastly, we want to solve for 𝑡 the time it takes for the coin to reach the ground. Once again, we’ll use a kinematic equation to help us solve for this value. The third kinematic equation written includes the variable we want to solve for 𝑡 as well as acceleration which we know and the coin’s initial velocity 𝑣 sub zero. The distance that the coin falls in total, based on our diagram, is capital 𝐷 plus lowercase 𝑑.

Because this total displacement is in the negative direction by our sign convention, we write that it’s negative 356 meters, a magnitude of distance given in our problem statement. If we take a next step by adding 356 meters to both sides of our equation, we see that we now have an equation in the form of a quadratic equation, where the variable we want to solve for is the time 𝑡.

In our equation, the constant factor in front of the squared term, which we can call capital 𝐴, is equal to one-half times negative 9.8 meters per second squared. Our constant factor in front of the linear term, which we can call 𝐵, is equal to positive 12.3 meters per second. And 𝐶 is equal to 356 meters. We can recall that the quadratic equation tells us, if we have an equation of this form, then the roots of that equation, represented here by 𝑋, are equal to negative 𝐵 plus or minus the square root of 𝐵 squared minus four 𝐴𝐶 all divided by two 𝐴. Where we’ve identified capital 𝐴, capital 𝐵, and capital 𝐶 for our equation.

Using our 𝐴, 𝐵, and 𝐶 values and, in our case, solving for the roots of the time 𝑡, when we plug in our 𝐴, 𝐵, and 𝐶 values into the quadratic equation to solve for 𝑡, we find that the two roots are 9.87 seconds and negative 7.36 seconds. Since we know the time value for the coin to drop to the ground can’t be negative, we eliminate that root and choose the first one. So, 9.87 seconds is the amount of time it takes for the coin to go from being released to hitting the ground.

Now, let’s take a moment to summarize what we’ve learnt about acceleration due to gravity. Gravity has an effect on every object in its field. And it accelerates objects which are in free fall or released. The acceleration caused by gravity is represented by the symbol lowercase 𝑔. Its magnitude is 9.8 meters per second squared. And it points downward towards the Earth. And finally, the kinematic equations explain object motion when acceleration is equal to 𝑔, the acceleration due to gravity.

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