Physics for Beginners page 10

Table of Contents

Physics For Beginners page 10

Table of Contents for Page 10

Gravity (continued)
- Return of Gravity in Detail: Armed with the Scientific Notation
Next Page

6. Gravity (continued)

Return of Gravity in Detail: Armed with the Scientific Notation

In this section, we will make use of the scientific notation to examine the general gravity formula in more detail. If you are not familiar with the scientific notation, please refer to this section. The scientific notation is just a fancy way to keep track of very large and very small numbers. If you don't want to learn scientific notation, you can still read through most of this section, however you might not be able to do some of the examples. Finally, there are certain rules that every sequel must obey, and hopefully, I have remembered them all.

A "Little" Something about Big G, the Mysterious Gravitational Constant
First, let's recall the general gravity formula. Before proceeding, make sure you understand the Gravity in Detail section. In particular, you should have a good understanding of the force of gravity and what it depends on before proceeding.

Okay, we have already discussed the above formula at length in a previous section. The only thing we neglected was the gravitational constant, G. At that time, we put off talking about G by saying it was a constant and to think of it as a number. We did this because we did not have all the necessary tools (namely, the scientific notation) to talk about it. Now, armed with the scientific notation, we are better prepared to give a "little" more detail about G. (As an aside, see how we have worked the name of the title into the actual text, like any good sequel should.)
First, let's figure out the units of G. Even though it is just a constant number, it has units. It isn't too difficult to figure out its units. Just remember that all the units on the right side of the above equation must equal the units on the left side of the equation. Please refer to this section if you need a reminder about units. Also, recall that we can multiply and divide units much like numbers. In fact, this is what we did in the Converting Units section.
As we stated above, the units on the left side of the general gravity formula above must equal the units on the right side of the formula. This is because the equal sign in the above formula requires everything on the right side of the equation to be equal to the left side of the equation, including the units.

In other words, the units of must equal the units of

As we know, the unit of force is Newtons, or N for short. But, as we saw in units section, N (or Newtons) is short for (kg*m)/(s*s) where kg means kilograms, m means meters, and s means seconds. Be careful not to confuse the m which stands for meters with the m which stands for the mass of an object. In this section, the m which stands for the mass of an object will have a subscript. I apologize for the confusing symbology here.
Since the unit on the left side of the formula above is Newtons, the units on the right side of the formula must also equal Newtons.

In other words, the units of must equal Newtons or (kg*m)/(s*s)

Remember, this is because Newtons is just shorthand for (kg*m)/(s*s).
Therefore, after the appropriate multiplication and division, the units of G, the two masses and the distance should conspire to equal Newtons or (kg*m)/(s*s). Specifically, if you take the unit of G and then multiply it twice by kg and then divide it twice by m, the result you obtain should equal Newtons, or (kg*m)/(s*s).
This can be a little confusing, so I have repeated the above argument below with all the units written out explicitly. In addition, since we don't know the unit of G yet, I have written it as the [unit of G] in the discussion below. The [unit of G] represents the units of the gravitational constant, G. In essence, what we are going to do is put units into the general gravity formula and make sure that the units on both sides of the equation are equal to one another. This will eventually help us figure out the [unit of G].

The equation in the last line above tells us that the units on the left side of the general gravity formula must equal the units on the right side of the formula. In fact, we arrived at this equation by putting units into the general gravity formula. That's all we did above.
As you can see from the last line above, the [unit of G] must be such that the units on both sides of the equation are equal to one another. Once again, the [unit of G] in the above discussion represents the units that G is made of because we don't know what they are yet.
Specifically, we can see how to figure out the [unit of G] from the equation in the last line above. By looking at this equation, we see that, after we multiply the [unit of G] twice by kg and divide it twice by m, the units on the right side of the equation must equal the units on the left side of the equation.
Below, I have rewritten the equation (in the last line above) in a slightly different form. All I did was pull the [unit of G] out front to isolate it from the other units and to make it a little easier to work with.

Before proceeding, let's take a look at the units on the left side of this equation and compare it to the units on the right side of the equation. Also, keep in mind that the units on the right side must equal those on the left side.
- On the left side of the equation, we have one kg on top, one m on top, and two s on the bottom.
- On the right side of the equation, we have two kg on top and two m on the bottom, in addition to the [unit of G].
By looking at the above equation, we can see what the [unit of G] must be, in order to make the units on both sides of the equation equal to one another. On the right side of the above equation, the [unit of G] must "cancel" one kg on top and two m on the bottom, while "adding" one m on top and two s on the bottom. The [unit of G] has to do this to make the units on the right side of the equation equal the units on the left side of the equation.
See if you can figure out the [unit of G] from the clues given in the above paragraph. Let me give you a couple of hints. The [unit of G] will be made up of other units. In fact, it will be made up of kg, m, and s. These units won't necessarily appear only once. In addition, remember that you can multiply and divide units. Okay, now try to figure out the [unit of G] before proceeding. If you run into some trouble, take a look at this section and think about how to "cancel" units.
1. The first thing we have to do is "cancel" one of the kg on top of the right side in the above equation. In order to do this, the [unit of G] must have one kg on the bottom. We need this so it can "cancel" one of the kg on the top of the equation on the right side. If this is confusing, please refer to this section to see how units can be "canceled".
2. Next, we need to "cancel" the two m's on the bottom of the right side of the equation. To do this, the [unit of G] must have two m's on the top. Make sure to keep track of all the units that are in the [unit of G] as we go along. So far, the [unit of G] has one kg on the bottom and two m's on top.
3. Next, we need to "add" one m to the top of the right side of the equation. We are not "canceling" here. We are just "adding". To do this, the [unit of G] must have one more m on top. It is important to point out that this extra m is needed in addition to the two m's we needed in step 2 above. In other words, the [unit of G] must have a total of three m's on top because it needs to "cancel" the two m's on the bottom of the right side of the equation while having one m leftover for the top.
4. Finally, we need to "add" two s's to the bottom of the right side of the equation. To do this, the [unit of G] must have two s's on the bottom.
Putting all this together, the [unit of G] must have three m's on top, one kg on the bottom, and two s's on the bottom. I have written this out explicitly below. If you didn't arrive at this conclusion, go back over the previous line of thinking a couple more times. As an aside, what we are doing here is an example of dimensional analysis.

This is precisely what is needed to make the units on the right side of the above equation equal to the units on the left side of the above equation. Try this yourself by putting the [unit of G] into the above equation. You can look below if you run into troubles.

The above hint should have made it clear why the [unit of G] had to be what it is. It is the only combination of units that makes the units on the right side of the above equation equal to the units on the left side of the equation. In other words, it is the only combination of units that makes the units on the right side of the general gravity formula equal to the units on the left side. I know this was rather boring, but it gave us more practice with units and further shows their importance.
Now that we've figured out the [unit of G], we can finally reveal its value. However, before doing so, I have another short thought question.
- Thought Question:
  Given the nature of the gravitational force, do you expect the gravitational constant, G, to be large or small?
  If you will recall from the previous discussion of gravity, we saw that gravity is a relatively weak force. Its effects are only apparent when objects with very large masses (like the earth) are involved. For instance, the gravitational attraction between two coins on a table is not apparent, while the gravitational attraction between a coin and the earth is apparent.
  So, given that gravity is a relatively weak force, do you expect the gravitational constant, G, to be small or large? Before you answer, think about this fact and look at the general gravity formula above. In particular, notice that the gravitational constant, G, is on top in the general gravity formula. Take a moment now and try to answer this thought question before reading further.
  Well, I won't keep you in suspense any longer. Because gravity is a relatively weak force, the gravitational constant, G, is very small. In a sense, the gravitational constant, G, serves as a mathematical way to describe the strength of the gravitational attraction between two objects.
Now that we know the gravitational constant, G, has to be very small and we know its units, I will finally reveal the value of the gravitational constant, G.

If you know the scientific notation, you will immediately know that this number is very small. If you will notice, I have also included units for the value of G. It is very important to include units whenever you are dealing with G because, without units, it would just be a number and would not be the gravitational constant. Plus, we worked so hard to figure out its units, and it would just be a shame not to include them.
That's all I have to say about G, the gravitational constant. The main idea to remember is that G is a very small number. This is because the force of gravity is a relatively weak force. Second, it is important to realize that G has units associated with it. It is not just a number. If you had trouble with the above section on how we figured out G's units, don't worry about it too much. Just remember that G has units. However, if you have some extra time, go back over how we figured out G's units and try to understand what we did above. It will be useful later.
Some Examples
Now that we are armed with the knowledge of G, we can actually do some examples with the general gravity formula. If you recall, we did not do any examples involving numbers in the Gravity in Detail section. In that section, we were only able to discuss the physics behind the formula because we did not know the value of G. That was fine because understanding the physics of gravity is more important than just blindly using the general gravity formula. However, now we know the value of G, we can do some concrete examples. In particular, these examples will illustrate some of the ideas we discussed earlier and will show that gravity is a relatively weak force whose effects only become apparent when objects of large masses are involved.
- Example 1:
  What is the gravitational attraction between an object with a mass of 10 kg and another object with a mass of 20 kg if they are separated by 0.01 meters? (Assume both objects are on some surface and not suspended in midair.)
  Notice that I said the gravitational attraction between the two objects. They are attracted toward each other with the same amount (or magnitude) of force, just in opposite directions. In other words, the 10 kg object is attracted to the 20 kg object with the same amount (or magnitude) of force that the 20 kg object is attracted to the 10 kg object. The only difference is the directions.
  Solving this problem is a relatively simple affair. All we have to do is put the above values for mass and distance into the general gravity formula below.
  
  I have done so explicitly below.
  
  Therefore, the two objects are attracted toward one another with a force of 0.0001334 N, just in opposite directions. As you can see, this force is relatively small. It is like the example of the two coins in a previous thought question on page 08. And like the example of the two coins, the force of gravity pulling these two objects toward each other is insufficient to overcome the friction holding these objects in place. Therefore, the results we obtain from the general gravity formula are consistent with what we observe in the real world, specifically, that gravity is a relatively weak force whose effects between everyday objects (with small masses) are not readily apparent.
- Example 2:
  What is the gravitational attraction between an object with a mass of 10 kg and the earth if the object is on the earth's surface? (Assume the 10 kg object is a very tiny but dense ball. In other words, assume it is a point.)
  For this problem, we will need two more pieces of information: the mass of the earth is 5.98E+24 kg and the radius of the earth is about 6.378E+6 m.
  We need the mass of the earth because we want to calculate the gravitational attraction between the 10 kg object and the earth. However, can you figure out why we need to know the radius of the earth? Think about that before reading further. If you get stuck, go back to this section to see how to use the general gravity formula. In particular, pay attention to how the distance, r, is measured in the general gravity formula.
  As you will recall, the distance, r, (in the general gravity formula) is measured from the center of one object to the center of the other object. In this example, the two objects that are attracted toward each other are the 10 kg object and the earth. Therefore, the distance, r, should be measured from the center of the earth to the center of the 10 kg object (sitting at the surface of the earth). Since we are assuming the 10 kg object is basically a point, the distance from the center of the earth to the center of the 10 kg object is basically the distance from the center of the earth to the surface of the earth (because the 10 kg object is sitting at the earth's surface). Well, by definition, this is the radius of the earth. That is why we needed to know the radius of the earth. Please refer to the picture below.
  
  Now that we have everything need to solve this problem, all we have to do is put the appropriate values for mass and distance into the general gravity formula above. The procedure is very similar to what we just did in the example above. After doing so, we find that the gravitational attraction between the 10 kg object and the earth is 98.1 Newtons.
  As you can see, when an object with a very large mass (like the earth in this example) is involved, the gravitational force is more apparent. This example is like the example of the coin and the earth in the thought question on page 08. Compared to example 1 above, the gravitational attraction between the earth and the 10 kg object (in this example) is noticeably greater than the gravitational attraction between the 20 kg object and the 10 kg object (in example 1). Once again, we see that the general gravity formula gives us results which are consistent with what we observe in the world, namely that gravitational attraction is only apparent when objects (like the earth) with large masses are involved.

Big G, Meet Little g

If you will recall from a earlier section, we learned that the gravitational attraction between the earth and an object near its surface causes that object to experience a constant downward acceleration of g = 9.81 (m/sec)/sec. Furthermore, we learned, in fact, that all objects (near the earth's surface) experience a constant downward acceleration of g = 9.81 (m/sec)/sec, regardless of what their mass was. Finally, we learned that the constant downward acceleration of g wasn't exactly constant. However, since it changed so little near the surface of the earth, we could assume the acceleration was a constant downward 9.81 (m/sec)/sec when dealing with objects near the surface of the earth. In this section, we will explicitly show this to be true. In addition, we will show how g (the acceleration due to gravity near the earth's surface) can be derived from the general gravity formula.

First, I will demonstrate how to calculate g (the acceleration due to gravity near the earth's surface) from the general gravity formula. This should be a fairly simple affair. For instance, let's say I had a ball in my hand. When I drop it, the only force acting on the ball is gravity. Therefore, when we calculate the acceleration that the ball experiences after I release it, we will be calculating the acceleration due to gravity and not to any other force. To calculate the acceleration the ball experiences in response to gravity, we first have to calculate the force of gravity pulling it down. We would then have to divide the force by the mass of the object (the ball in this case) undergoing the acceleration. We have to divide the force that the ball experiences by its mass because this is what Newton's 2nd Law of Motion tells us. It tells us the acceleration an object experiences (in response to the force acting on the object) is obtained by the formula: a = F/m. As an aside, notice that the ball only experiences an acceleration after it is dropped. Before the ball was dropped, it does not experience a net force because the force of gravity pulling it down is canceled by the force of the hand holding the ball up. When the ball was in my hand, it experiences no acceleration because the net force acting on it was zero.

The first thing we need to do is find a formula which tells us how to calculate the force of gravity pulling the ball downward. Well, we are, indeed, in luck because we spent two entire sections talking about this formula. The formula of interest here is the general gravity formula because it tells us how to calculate the gravitational attraction between two objects. In this case, the two objects that are attracted to each other are the ball and the earth. The ball is being pulled down while the earth is being pulled up. They are attracted toward each other with the same amount of force, just in opposite directions. I have written the formula again below.

Next, let's be specific and call object A the earth. Likewise, we will make object B the ball. In other words, is the mass of the ball because object B is the ball. We know the mass of the earth is 5.98E+24 kg. However, instead of putting in 5.98E+24 kg for the mass of object A (the earth in this case), let's put in for the mass of object A, where represents the mass of the earth. The subscript E just means that it is the mass of the earth. We are doing this because it will save us some space and make things a little more clear. After putting in for the mass of object A, we get the following equation.

Next, let's put in the distance between the centers of the two objects which are attracted toward each other. In this case, the two objects that are attracted to each other are the ball and the earth. So, what is the distance between the center of the ball and the center of the earth? Once again, assume the ball is basically a point object that is near the surface of the earth. In other words, assume the ball is right at the surface of the earth. With these assumptions, this is exactly like example 2 above. Therefore, the distance between the center of the ball and the center of the earth is just the radius of the earth. Please refer to the picture above.

This is perfect because the radius of the earth was given in example 2 above. The radius of the earth is 6.378E+6 m. However, once again, instead of putting this number into the equation above, let's just put in for the distance, r, where represents the radius of the earth. After making the substitution of for the distance between the center of the earth and the center of the ball (near the earth's surface), we get the following equation.

Let's take a quick look at what we have so far. This is the force with which the ball is being pulled downward by the earth. As usual, this is also the force with which the ball pulls the earth upward, but we are not concerned with this right now. Looking at the equation above, let's see if we are missing any information. We know because that is just the mass of the earth. Likewise, we know because that is just the radius of the earth. In addition, we know G because that is the gravitational constant, and we know its value from our discussion above.

It looks like the only missing piece of information we need is the mass of the ball, . Well, let's press on and see if we can calculate the acceleration due to gravity near the earth's surface without knowing the mass of the ball. It will turn out that we won't need this information.

Once we know the force of gravity pulling the ball downward, we can calculate the acceleration the ball experiences by dividing the force by the mass of the ball. Once again, this is because of Newton's 2nd Law of Motion: a = F/m. It is important to be careful here. To find the acceleration an object experiences, we have to divide the force the object feels by its own mass. Therefore, to find the acceleration of the ball due to gravity, we have to divide the force of gravity pulling the ball downward by its own mass. In other words, to find the acceleration the ball experiences due to gravity, we have to divide the above equation by the mass of the ball, . I have done this explicitly below.

In this equation, a is the acceleration the ball experiences due to gravity (near the earth's surface). From this equation, we can calculate the acceleration the ball experiences due to gravity (near the earth's surface). All we have to do is multiply the gravitational constant, G, by the mass of the earth and then divide that twice by the radius of the earth.

In the equation above, you should notice that the cancel. This is an important point. Because the mass of the ball cancels in the equation above, we did not need to know the mass of the ball. More importantly, however, we see that the acceleration the ball experiences due to gravity does not depend on the mass of the ball. In other words, the acceleration the ball experiences due to gravity will be the same regardless of the mass of the ball. Therefore, all objects (near the surface of the earth) will experience the same acceleration (due to gravity), regardless of mass.

Finally, let us put the values for the mass of the earth and the radius of the earth into the above equation to solve for the acceleration the ball experiences due to gravity (near the earth's surface).

a = 9.81 (m/sec)/sec downward (near the earth's surface)

This should look very familiar. In fact, this is exactly the value we had for g (the acceleration due to gravity near the earth's surface) from a previous section. Therefore, we can derive the value of g from the general gravity formula. This should come as a relief and should provide further evidence of the general gravity formula's validity.

To summarize, we can derive the value of g = 9.81 (m/sec)/sec from the general gravity formula. In addition, when we derived g from the general gravity formula, we did not have to know the mass of the object that was experiencing the acceleration due to gravity. In fact, the formula for the acceleration due to gravity (near the earth's surface) that we derived above does not depend on the mass of the object experiencing the acceleration. In other words, the acceleration due to gravity is independent of mass. All objects experience the same acceleration (due to gravity) regardless of their mass. This is also consistent with what we previously learned about gravity.

Next, if you will recall, we had initially said the acceleration due to gravity (near the earth's surface) was a constant 9.81 (m/sec)/sec downward. We later revealed that it was not really constant, however since it changed so little near the surface of the earth, we could assume it was constant. In this section, we will show this explicitly.

The good thing about this is that we have done most of the work. In fact, we can use the above equation with a slight modification. If you will recall, the above equation was the acceleration an object experiences (due to gravity) at the earth's surface. The on the bottom of the equation is the radius of the earth. This was the distance between the center of the earth and the center of the object because the object was at the earth's surface. If we were to change the distance between the center of the object and the center of the earth, the only thing we would have to do is change the distance in the equation above. Everything else remains the same.

Therefore, to get the general formula for the acceleration an object experiences in response to the gravitational attraction between it and the earth (where the object is not necessarily at the earth's surface), we would just replace the in the above equation with r, where r is the distance between the center of the object and the center of the earth. After doing so, we get the following general formula for the acceleration an object experiences in response to the gravitational attraction between the object and the earth. This is the general formula, where the object is not necessarily at the earth's surface. For instance, this formula would be valid if the object was 10 meters above the surface of the earth.

As we stated above, this is the general formula which describes the acceleration an object experiences as a result of being pulled downward by the gravitational attraction between the object and the earth. Once again, r is the distance between the center of the object and the center of the earth. There is one very important thing I would like to point out here. The distance, r, can only be equal to or greater than the radius of the earth. It cannot be less than the radius of the earth. In other words, this formula is only valid if the object is at the surface of the earth or somewhere above the surface of the earth. It is not valid if the object is under the surface of the earth.

As an aside, you should notice again that the above formula for the acceleration an object experiences (in response to the gravitational attraction between the object and the earth) does not depend on the mass of the object undergoing the acceleration. The formula only depends on the mass of the earth and the distance from the center of the object to the center of the earth. Therefore, as long as the objects are all at the same height above the earth's surface, the acceleration those objects experience (due to the gravitational attraction between the objects and the earth) will be the same for all objects, regardless of their mass. For example, all objects 5 meters above the earth's surface will experience the same acceleration, regardless of their mass. Once again, this is because the general formula for acceleration (due to gravity) is independent of the mass of the object undergoing the acceleration. Please refer to the formula.

Now that we have the general formula for the acceleration due to gravity, let's calculate an object's acceleration if it is 1000 meters above the surface of the earth. (1000 meters is just 1 kilometer.) After doing so, we will compare the acceleration of the object (at 1000 meters above the surface of the earth) to the acceleration the object experiences at the surface of the earth.

Example 1:
Given that the object is 1000 meters above the surface of the earth, calculate the acceleration it experiences due to the gravitational attraction between the object and the earth. (Remember, the radius of the earth is 6.378E+6 meters.)
Try solving this problem on your own before reading further. Pay particular attention to the distance, r.
The only tricky part of this problem is figuring out what number to assign to the distance, r. To repeat, the distance, r, is the distance from the center of the earth to the center of the object.
In this problem, the object is 1000 meters above the surface of the earth. Therefore, the distance from the center of the earth to the center of the object is the radius of the earth plus 1000 meters. In other words, the distance, r, is 6.378E+6 meters (the radius of the earth) plus 1000 meters.
r = radius of the earth + 1000 meters
r = (6.378E+6 meters) + (1000 meters)
r = 6.379E+6 meters
To find the acceleration an object experiences (due to gravity) if it is 1000 meters above the earth's surface, all we have to do is put this value of r into the above equation. After doing so, we get the following result.
a = 9.80 (m/sec)/sec downward
(The acceleration an object experiences due to gravity when it is 1000 meters above the surface of the earth.)

Next, let's compare the acceleration (due to gravity) an object undergoes at the earth's surface versus the acceleration it undergoes 1000 meters above the surface of the earth.

The height of the object (relative to the earth's surface) The acceleration an object experiences (due to the gravitational attraction between the earth and the object) at that height above the earth's surface

0 meters above the earth's surface (i.e, at the earth's surface) 9.81 (m/sec)/sec downward

1000 meters above the earth's surface 9.80 (m/sec)/sec downward

As you can tell from the table above, the acceleration an object experiences (due to the gravitational attraction between the earth and the object) does not change appreciably, even if we were to move the object from the earth's surface to a position 1000 meters above the surface of the earth.

Because of this fact, when dealing with an object near the surface of the earth, we may safely assume that its acceleration (due to gravity) is a constant 9.81 (m/sec)/sec downward.

To summarize, we know the acceleration an object experiences (due to the gravitational attraction between the earth and the object) does depend on its height above the surface of the earth. However, if the object is reasonably close to the earth's surface, the acceleration it experiences does not change by much (even if the height of the object is 1000 meters above the earth's surface). As a result, in most cases, we may assume the object experiences a constant downward acceleration of g = 9.81 (m/sec)/sec. In those cases where we may not assume the acceleration is a constant g, I will explicitly say so.

In conclusion, the acceleration an object experiences (in response to the gravitational attraction between the object and the earth) is independent of the mass of the object. The acceleration does depend on the height of the object above the earth's surface. Therefore, if two objects are at the same height above the earth's surface, the accelerations they undergo will be the same, even if their masses are different. However, if the two objects are at different heights above the surface of the earth, the accelerations will be different. Finally, even though the acceleration of the object depends on its height above the earth's surface, the acceleration changes by such a small amount that we may assume the acceleration is a constant downward 9.81 (m/sec)/sec, in most cases. Therefore, for most cases near the surface of the earth, we may assume that all objects (regardless of their mass) experience a constant downward acceleration of g = 9.81 (m/sec)/sec.

Previous Page

Next Page