Lesson Video: Free Fall and Weightlessness | Nagwa Lesson Video: Free Fall and Weightlessness | Nagwa

Lesson Video: Free Fall and Weightlessness Physics

In this video, we will learn how to describe the motion of objects acted on only by a constant gravitational force, moving only parallel to the force.

14:49

Video Transcript

In this video, we will learn how to describe the motion of objects acted on by only a constant gravitational force. We will look specifically at objects that are near the surface of Earth, as well as the difference between the actual way of an object and the perceived way of an object. Let’s start by analyzing how objects that are dropped from the same distance from the center of the body that is creating the gravitational field accelerate.

If we drop two objects of different masses, let’s say a bowling ball and a feather, at the same height 𝑑 above the surface of the Moon, which would hit the ground first? It turns out that they hit the ground at the same time. This is because in the absence of any forces other than gravity, all objects will have the same acceleration. When we say in the absence of forces other than gravity, we mean that there are no other forces acting on the object, including air resistance. Going back to our example of dropping a bowling ball and a feather on the surface of the Moon, since there is no atmosphere on the Moon, there is no air resistance. This statement holds true as long as we’re dropping our objects from the same distance from the center of the body creating the gravitational fields.

In this case, we are at distance 𝑑 above the surface of the Moon. So our objects, the bowling ball and feather, are the same distance from the center of the Moon. If we were to drop the same objects, bowling ball and feather, from the same height 𝑑 above the surface of Earth this time assuming there’s no air resistance such as in a vacuum, what would happen? They would once again hit the ground at the same time, but on Earth, they would fall faster than they did on the Moon. This is because the acceleration due to gravity near the surface of the Earth is 9.8 meters per second squared, whereas on the surface of the Moon, it’s approximately 1.6 meters per second squared, or roughly one-sixth of the acceleration due to gravity at the surface of the Earth.

Now that we have seen that mass does not play a role in the acceleration of an object only acted on by gravity, let’s take a look at what we mean by weightlessness. To discuss weightlessness, we first need to remember that weight or force due to gravity is equal to the mass of the object times the acceleration due to gravity. A common misconception is that if we are weightless, we have no force of gravity acting on us. This is simply not true. So what does it mean to be weightless? To understand weightlessness, we first have to understand how we measure weight. One of the common ways is to step on a scale. When we step on a scale in our bathroom. We’re assuming that the normal force or the reaction force that scale pushes on us with is equal to the force of gravity that is pulling us down towards the center of the planet.

Let’s analyze why this assumption is correct as long as we’re standing in our bathroom by starting with Newton’s second law. We should remember that Newton’s second law says that the net force acting on an object is equal to the mass of the object times the acceleration of the object. If we are standing on a scale in our bathroom, then our acceleration is zero, which means that mass times acceleration is also equal to zero. Looking at our diagram, we can see that we have two forces, the normal force pushing up and the force of gravity pulling down. The net force will be the addition of these two forces. Since we typically make the upward direction positive, our normal force will be positive and our force of gravity, since it’s going downwards, will be negative.

So we have a force equation that says that the normal force minus the force of gravity is equal to zero. To find the relationship between the normal force and the force of gravity, we can add the force of gravity to both sides of the equation, canceling out the force of gravity on the left side. We end up with the normal force, the force that the scale pushes on us, being equal to the force of gravity. So when we step on our scale in the bathroom, the normal force or the reading on our scale is going to be our weight. When we say that we’re weightless, what we’re actually saying is that scale reading or the normal force acting on us is gonna be equal to zero. So how can that happen?

For someone to experience weightlessness, they must be falling at the same acceleration as the acceleration due to gravity at that position. The left side of the equation is the same as it was in the bathroom example. But the right side of the equation is now negative 𝑚𝑔 as our acceleration has a value of 𝑔 and is pointing downwards, meaning it is negative. When we add force of gravity to both sides of the equation, it cancels out on the left side of the equation, and it also cancels out on the right side of the equation because force of gravity is defined as 𝑚𝑔. So negative 𝑚𝑔 plus 𝑚𝑔 is zero, meaning that the scale reads zero if we’re falling at the same rate as the acceleration due to gravity in that position as the scale is also falling with an acceleration equal to the acceleration due to gravity.

In other words, the scale is no longer on the floor but is actually falling through the air the same as we do, which brings up an important distinction between perceived weight and actual weight. The perceived weight is the value of the force of the scale on us, whereas the actual weight is the force due to gravity, which is based on our mass and the acceleration due to gravity at the position we’re at. Assuming the value of 𝑔 is the same in both cases, our perceived weight would be different, but our actual weight would stay the same. In other words, perceived weight is the weight it feels like we have, whereas our actual weight is always the same, the force of gravity exerted on our body.

Therefore, when we talk about weightlessness, what we really mean is that our perceived weight is zero. We feel like we have no weight as we freely fall through a gravitational field. Let’s dig a little deeper into the perceived weight versus actual weight by looking at some examples of a person on an elevator. Let’s look at the impact of having three different accelerations. First, we’ll look at when our elevator has an acceleration of zero or is traveling at a constant velocity. In each situation, the person will have two forces acting on them, the force of gravity pulling them towards the center of the Earth and the normal force or reaction force of the scale pushing upwards on the person. In each case, the force of gravity, the actual weight of the person, will be unchanging as the elevator will be the same distance from the center of the Earth.

We will see how the acceleration of the elevator impacts the perceived weight or the reaction force of the scale on the person. In our first scenario, elevator is traveling at a constant velocity, meaning that it has zero acceleration. The left side of the equation will be the normal force minus the force of gravity, as our net force is the addition of the normal force in a positive upward direction plus the force of gravity, which is in a downward or negative direction. Just as we saw with our bathroom example, our acceleration is zero. Therefore, 𝑚𝑎 becomes zero.

When we add the force of gravity to both sides of our equation, the force of gravity cancels out on the left side of the equation. We end up with the same relationship as we did in the bathroom. The normal force is equal to the force of gravity. The perceived weight in this case is the same as the actual weight. Let’s see what happens when our elevator has an acceleration which is upward. The left side of the equation stays the same as our forces have not changed direction. The right side of the equation is 𝑚𝑎 as our acceleration is upward, which we chose to be positive. When we add the force of gravity to both sides of the equation, it cancels out on the left side, leaving us with our normal force being equal to 𝑚𝑎 plus 𝐹𝑔.

This tells us that when our elevator is accelerating upward, our normal force, the perceived weight, is greater than the force of gravity, the actual weight. You may have experienced this yourself when we first start moving upward in an elevator. When it accelerates upward, it feels like we’re pushed into the floor of the elevator, essentially the same as feeling heavier. This is because our perceived weight is higher like it feels like we weigh more. Therefore, the scale in this situation is not an accurate representation of how much we weigh.

But what happens if our elevator is accelerating downwards? The left side of the equation stays the same as we saw in those two forces in the same direction as they’ve been acting for the last two examples. But this time, the right side of the equation becomes negative 𝑚𝑎 as our acceleration has the value of 𝑎 but the direction of downward or negative. When we add the force of gravity to both sides of the equation, it cancels out on the left side. And we have the normal force being equal to the force of gravity minus 𝑚𝑎. This tells us that the apparent weight, the normal force, is less than the actual weight, force due to gravity.

Just as we saw when our elevator was accelerating upward, when our elevator is accelerating downward, the scale does not tell us our actual weight. Let’s apply what we’ve just learned about objects and free fall and weightlessness to a couple of example problems.

Two objects, object I and object II, are both 120 meters above the surface of the Moon, where the only force that acts on either of them is the gravitational force of the Moon. Neither object has any motion other than that due to its gravitational acceleration. Object I and object II each have masses of 1.5 kilograms. What is the ratio of the mass of object I to the mass of object II? What is the ratio of the gravitational force on object I to the gravitational force on object II? What is the ratio of the acceleration toward the lunar surface of object I to the acceleration toward the lunar surface of object II?

Looking at our first question, we are asked to find the ratio of the mass of object I to object II. This is the same thing as saying we’re gonna take mass one and divide it by mass two to come up with our ratio. In the problem, we’re told that both of our objects, object I and object II, each have masses of 1.5 kilograms. Therefore, we can plug in 1.5 kilograms for mass one and 1.5 kilograms for mass two. When we divide 1.5 kilograms by 1.5 kilograms, we get 1.0. The ratio of mass of object I to the mass of object II is 1.0.

For our second question, we’re asked to find the ratio of the gravitational force on object I to the gravitational force on object II. To set up the ratio, we divide the gravitational force on object I by the gravitational force on object II. We need to recall that the gravitational force on an object is equal to the mass of the object times the acceleration due to gravity of the object at that position. Plugging in our values, 𝑚 one and 𝑚 two both have values of 1.5 kilograms. And the acceleration due to gravity in both is the acceleration due to gravity from the Moon. Since there’s an acceleration due to gravity of the Moon in the numerator and the denominator, they cancel each other out. When we divide 1.5 kilograms by 1.5 kilograms, we get 1.0 again. The ratio of the gravitational force on object I to the gravitational force on object II is 1.0.

In our third question, we are asked to find the ratio of the acceleration toward the lunar surface of object I to the acceleration toward the lunar surface of object II. We can find the ratio by dividing the acceleration of object I by the acceleration of object II. From the problem, we are told that the only force that acts on either of them is the gravitational force of the Moon and that neither object has any motion other than that due to its gravitational acceleration. This means that both object I and object II have acceleration due to gravity that are the same, 𝑔 of the Moon. When we divide 𝑔 of Moon by 𝑔 of Moon, we get 1.0. The ratio of the acceleration toward the lunar surface of object I to the acceleration toward the lunar surface of object II is 1.0.

A woman with a mass of 55 kilograms stands on a weighing scale that is on the floor of a descending elevator as shown in the diagram. The elevator is descending with an acceleration of 9.8 meters per second squared. What is the reading on the weighing scale?

The diagram shows a woman standing on an elevator with a scale at the bottom on the floor and an arrow representing the acceleration downwards. Let’s label the information from our problem on our diagram. The problem told us that the woman on the elevator has a mass of 55 kilograms and that the value of the acceleration is 9.8 meters per second squared in a downward direction. We’re being asked to find the reading on the scale, which is labeled as our normal force in the diagram. This is the reaction force of the scale on the woman. And it’s pointing in an upward direction. We have also included the weight of the woman or the force due to gravity pointing downwards as it points towards the center of the Earth.

To determine what the reading is on the weighing scale, we need to begin with Newton’s second law. Newton’s second law says that the net force on an object is equal to the mass of the object times the acceleration of the object. To find the net force, we add our two forces together as represented by the left side of our equation. The normal force or the reading on the scale is positive as it’s in an upward direction. And we typically choose upwards to be positive. The force of gravity is negative as it is in the opposite direction pointing downwards.

On the right side of the equation, our mass was given to us as 55 kilograms and our acceleration was negative 9.8 meters per second squared, where the negative is because the acceleration is in a downward direction. Multiplying out the right-hand side of the equation, we get negative 539 newtons. To continue solving our problem, we’ll need to replace the weight of the woman with an expression based on the values. We need to remember the weight or the force due to gravity is equal to the mass of the object times the acceleration due to gravity. For this problem, the mass of the woman is 55 kilograms and the acceleration due to gravity is 9.8 meters per second squared.

Multiplying out our two values, we once again get 539 newtons. To solve for the reaction force of the scale on the woman, we need to add 539 newtons to both the left side of the equation as well as the right side of the equation. This cancels out the 539 newtons on the left side equation as well as on the right side of the equation, leaving us with zero newtons. The reading on the weighing scale on the floor of a descending elevator at a rate of 9.8 meters per second squared is zero newtons. As the elevator is falling at the same rate as 𝑔, it is in free fall, and the woman inside is weightless, which is why the contact force 𝐹 𝑁 is measured to be zero.

Summary

In the absence of any forces other than the force of gravity, objects of the same distance from the center of a planet will have the same acceleration regardless of mass. Objects are accelerated at a value of 9.8 meters per second squared near Earth’s surface. If an object is accelerating toward the center of a planet with a value greater than or equal to 𝑔, this object will not exert a reaction force on a second object that is accelerated only gravitationally.

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