In the previous section, we completed a discussion on the magnitude as well as direction of the gravitational force of attraction. We also saw some solved examples. In all those examples, we were dealing with point masses. In this section we will see how the force acts between extended objects. We already saw the difference between point masses and extended objects here
1. Consider an extended object like a block of wood. We will call it: A
2. A point mass is kept in the vicinity of A. We will call the point mass: B
3. So A is an extended object and B is a point mass
• We want to find the 'gravitational force of attraction' $\mathbf\small{\vec{F}_{G(A,B)}}$ exerted by A on B
4. In this case, each point mass in A will be exerting an attractive force on B
• So we will see a large number of force vectors
♦ All of them will be starting from B
♦ All of them will be pointing towards A
☆ But all of them will not be pointing towards a single point in A
☆ Different vectors will be pointing towards different points in A
5. So we can write:
■ The tail ends of all the force vectors will be the same point. But the head ends will be different
• That means, all the force vectors will be having different directions
6. Now, to obtain the resultant force on B, we must add all the force vectors
• It must be a vector addition
• Such a vector addition of different vectors can be done easily using the principles of calculus. We will see it in higher classes
7. However, for some 'simple extended objects', we can obtain the resultant, without the application of calculus
1. Fig.8.18(a) below shows a spherical shell. We will call it: A
• Note that, it is spherical in shape. And it is hollow inside. So in a 2D drawing, it will appear as a circle
2. A point mass is kept in the vicinity of A. We will call the point mass: B
3. So A is an extended object and B is a point mass
• We want to find the 'gravitational force of attraction' $\mathbf\small{\vec{F}_{G(A,B)}}$ exerted by A on B
4. The point mass B appears to be placed 'eccentrically' when compared to A. But that should not be a problem. We must be able to find the force $\mathbf\small{\vec{F}_{G(A,B)}}$ regardless of the position of B
5. In fig.b, the center of A is marked as O
• A white dashed line joins O and B
6. In fig.c, a 'random point mass' P is marked on A
• The force $\mathbf\small{\vec{F}_{G(P,B)}}$ is also shown
• We know that, this force will be directed from B towards P. This is indicated by the white dotted line
7. This force $\mathbf\small{\vec{F}_{G(P,B)}}$ can be resolved into two perpendicular components. This is shown in fig.8.19(a) below
• One component acts along OA. It is shown in red color
• The other component acts perpendicular to OA. It is shown in yellow color
8. Once a vector is resolved into it's two perpendicular components, we need to consider those two components only
• The 'combined effect produced by the two components' is same as the 'effect produced by $\mathbf\small{\vec{F}_{G(P,B)}}$'
• That means, we can ignore $\mathbf\small{\vec{F}_{G(P,B)}}$ and do calculations with the two components alone
9. In fig.8.19(b), another point mass P' is marked
• This P' has some peculiarities:
♦ P' and P are symmetric points. The axis of symmetry is OB
♦ P' is the one and only symmetric point of P (when the axis OB is considered)
10. Now we repeat all the works on P':
It can be summarized in 3 steps:
(i) $\mathbf\small{\vec{F}_{G(P',B)}}$ is the force exerted by P' on B
(ii) The red vector in fig.8.19(b) is the parallel component of $\mathbf\small{\vec{F}_{G(P',B)}}$
(iii) The yellow vector in fig.8.19(b) is the perpendicular component of $\mathbf\small{\vec{F}_{G(P',B)}}$
11. Now compare figs (a) and (b):
• Distances PB and P'B are equal. So the following two magnitudes are equal:
(i) Magnitude of $\mathbf\small{\vec{F}_{G(P,B)}}$
(ii) Magnitude of $\mathbf\small{\vec{F}_{G(P',B)}}$
• PB and P'B make the same angle with OB. So the following two magnitudes are also equal:
(i) Magnitude of the red vector in fig.a
(ii) Magnitude of the red vector in fig.b
• The following two magnitudes are also equal:
(i) Magnitude of the yellow vector in fig.a
(ii) Magnitude of the yellow vector in fig.b
12. We see something of interest:
• The red vectors in figs. (a) and (b) act in the same direction
♦ That means, parallel components add up
• The yellow vectors in figs. (a) and (b) act in the opposite directions
♦ That means, perpendicular components cancel each other
♦ Remember that, the magnitudes of the two yellow vectors are equal
13. Now, like P, we can consider any other point
• It will have a unique symmetric point on the other side of OB
• For those two points also, the parallel components will add up and the perpendicular components will cancel each other
• So in effect, there will be no force perpendicular to OB
■ All forces will be acting along OB
14. OB is the line joining the center of A to B
■ So we can write:
The point mass B will experience an attractive force towards the spherical shell A. This force will be appear as if all the mass of A is concentrated at it's centre O
15. The result written in (14) can be used in general:
■ If we want to find the 'force exerted by a spherical shell' on a 'point mass situated outside the shell', assume that, all the mass of the shell is concentrated at it's center, and then do the calculations
1. Fig.8.20(a) below shows a spherical shell. We will call it: A
• Note that, it is spherical in shape. And it is hollow inside. So in a 2D drawing, it will appear as a circle
2. A point mass is kept inside A. We will call the point mass: B
3. So A is an extended object and B is a point mass
• We want to find the 'gravitational force of attraction' $\mathbf\small{\vec{F}_{G(A,B)}}$ exerted by A on B
4. The point mass B appears to be placed 'eccentrically' inside A. But that should not be a problem. We must be able to find the force $\mathbf\small{\vec{F}_{G(A,B)}}$ regardless of the position of B
5. In fig.b, the center of A is marked as O
• A white dashed line joins O and B
6. In fig.b, a 'random point mass' is marked on A. It is shown in green color
• The force of attraction between the green point and B is also shown
♦ This force is indicated by the green vector
7. Now consider fig.8.21(a) below:
• The red dashed line is a chord of the circle
• This chord is drawn perpendicular to OB
8. The red dashed chord separates the circle into two parts
• The upper part is smaller
• The lower part is larger
9. A 'random point mass' is marked on the upper part. It is shown in yellow color
• The force of attraction between the yellow point and B is also shown
♦ This force is indicated by the yellow vector
10. Let us compare the magnitudes of the green and yellow vectors:
• Both act between 'a point mass on A' and B
• But for the yellow vector, the point mass is closer to B
• So the yellow vector will have a greater magnitude than the green vector
11. Now let us compare the directions of the green and yellow vectors:
• The green vector tries to pull B towards the lower portion of the shell
• The yellow vector tries to pull B towards the upper portion of the shell
12. So the green and yellow vectors are acting in somewhat opposite directions
• Also we know that, the yellow vector has a greater magnitude
• So we would expect the point mass B to move towards the upper portion of the shell
13. But when we take a closer look, we find some thing of interest:
♦ We can mark numerous green points. Because the lower arc is larger
♦ But we cannot mark that many yellow points. Because the upper arc is smaller
• This is shown in fig.b
■ That means:
♦ There will be numerous green vectors
♦ There will not be that many yellow vectors
14. We can write:
♦ The green vectors, though smaller in strengths, are greater in numbers
♦ The yellow vectors, though greater in strengths, are smaller in numbers
15. That means:
♦ The net upward force will be equal to the net downward force
♦ The two net forces will cancel each other
• We will see the precise mathematical proof in higher classes
• At present, all we need to know is this:
The net force on B is zero
16. The result written in (15) can be used in general:
■ The 'force exerted by a spherical shell' on a 'point mass situated inside the shell', is zero
1. Consider an extended object like a block of wood. We will call it: A
2. A point mass is kept in the vicinity of A. We will call the point mass: B
3. So A is an extended object and B is a point mass
• We want to find the 'gravitational force of attraction' $\mathbf\small{\vec{F}_{G(A,B)}}$ exerted by A on B
4. In this case, each point mass in A will be exerting an attractive force on B
• So we will see a large number of force vectors
♦ All of them will be starting from B
♦ All of them will be pointing towards A
☆ But all of them will not be pointing towards a single point in A
☆ Different vectors will be pointing towards different points in A
5. So we can write:
■ The tail ends of all the force vectors will be the same point. But the head ends will be different
• That means, all the force vectors will be having different directions
6. Now, to obtain the resultant force on B, we must add all the force vectors
• It must be a vector addition
• Such a vector addition of different vectors can be done easily using the principles of calculus. We will see it in higher classes
7. However, for some 'simple extended objects', we can obtain the resultant, without the application of calculus
We will now see the case of a spherical shell, which is a 'simple extended object'. We will write it in steps
• Note that, it is spherical in shape. And it is hollow inside. So in a 2D drawing, it will appear as a circle
 |
| Fig.8.18 |
3. So A is an extended object and B is a point mass
• We want to find the 'gravitational force of attraction' $\mathbf\small{\vec{F}_{G(A,B)}}$ exerted by A on B
4. The point mass B appears to be placed 'eccentrically' when compared to A. But that should not be a problem. We must be able to find the force $\mathbf\small{\vec{F}_{G(A,B)}}$ regardless of the position of B
5. In fig.b, the center of A is marked as O
• A white dashed line joins O and B
6. In fig.c, a 'random point mass' P is marked on A
• The force $\mathbf\small{\vec{F}_{G(P,B)}}$ is also shown
• We know that, this force will be directed from B towards P. This is indicated by the white dotted line
7. This force $\mathbf\small{\vec{F}_{G(P,B)}}$ can be resolved into two perpendicular components. This is shown in fig.8.19(a) below
 |
| Fig.8.19 |
• The other component acts perpendicular to OA. It is shown in yellow color
8. Once a vector is resolved into it's two perpendicular components, we need to consider those two components only
• The 'combined effect produced by the two components' is same as the 'effect produced by $\mathbf\small{\vec{F}_{G(P,B)}}$'
• That means, we can ignore $\mathbf\small{\vec{F}_{G(P,B)}}$ and do calculations with the two components alone
9. In fig.8.19(b), another point mass P' is marked
• This P' has some peculiarities:
♦ P' and P are symmetric points. The axis of symmetry is OB
♦ P' is the one and only symmetric point of P (when the axis OB is considered)
10. Now we repeat all the works on P':
It can be summarized in 3 steps:
(i) $\mathbf\small{\vec{F}_{G(P',B)}}$ is the force exerted by P' on B
(ii) The red vector in fig.8.19(b) is the parallel component of $\mathbf\small{\vec{F}_{G(P',B)}}$
(iii) The yellow vector in fig.8.19(b) is the perpendicular component of $\mathbf\small{\vec{F}_{G(P',B)}}$
11. Now compare figs (a) and (b):
• Distances PB and P'B are equal. So the following two magnitudes are equal:
(i) Magnitude of $\mathbf\small{\vec{F}_{G(P,B)}}$
(ii) Magnitude of $\mathbf\small{\vec{F}_{G(P',B)}}$
• PB and P'B make the same angle with OB. So the following two magnitudes are also equal:
(i) Magnitude of the red vector in fig.a
(ii) Magnitude of the red vector in fig.b
• The following two magnitudes are also equal:
(i) Magnitude of the yellow vector in fig.a
(ii) Magnitude of the yellow vector in fig.b
12. We see something of interest:
• The red vectors in figs. (a) and (b) act in the same direction
♦ That means, parallel components add up
• The yellow vectors in figs. (a) and (b) act in the opposite directions
♦ That means, perpendicular components cancel each other
♦ Remember that, the magnitudes of the two yellow vectors are equal
13. Now, like P, we can consider any other point
• It will have a unique symmetric point on the other side of OB
• For those two points also, the parallel components will add up and the perpendicular components will cancel each other
• So in effect, there will be no force perpendicular to OB
■ All forces will be acting along OB
14. OB is the line joining the center of A to B
■ So we can write:
The point mass B will experience an attractive force towards the spherical shell A. This force will be appear as if all the mass of A is concentrated at it's centre O
15. The result written in (14) can be used in general:
■ If we want to find the 'force exerted by a spherical shell' on a 'point mass situated outside the shell', assume that, all the mass of the shell is concentrated at it's center, and then do the calculations
In the above discussion, the point mass B was outside the shell. Now we will consider the case when B is inside the shell. We will write it in steps
• Note that, it is spherical in shape. And it is hollow inside. So in a 2D drawing, it will appear as a circle
 |
| Fig.8.20 |
3. So A is an extended object and B is a point mass
• We want to find the 'gravitational force of attraction' $\mathbf\small{\vec{F}_{G(A,B)}}$ exerted by A on B
4. The point mass B appears to be placed 'eccentrically' inside A. But that should not be a problem. We must be able to find the force $\mathbf\small{\vec{F}_{G(A,B)}}$ regardless of the position of B
5. In fig.b, the center of A is marked as O
• A white dashed line joins O and B
6. In fig.b, a 'random point mass' is marked on A. It is shown in green color
• The force of attraction between the green point and B is also shown
♦ This force is indicated by the green vector
7. Now consider fig.8.21(a) below:
 |
| Fig.8.21 |
• This chord is drawn perpendicular to OB
8. The red dashed chord separates the circle into two parts
• The upper part is smaller
• The lower part is larger
9. A 'random point mass' is marked on the upper part. It is shown in yellow color
• The force of attraction between the yellow point and B is also shown
♦ This force is indicated by the yellow vector
10. Let us compare the magnitudes of the green and yellow vectors:
• Both act between 'a point mass on A' and B
• But for the yellow vector, the point mass is closer to B
• So the yellow vector will have a greater magnitude than the green vector
11. Now let us compare the directions of the green and yellow vectors:
• The green vector tries to pull B towards the lower portion of the shell
• The yellow vector tries to pull B towards the upper portion of the shell
12. So the green and yellow vectors are acting in somewhat opposite directions
• Also we know that, the yellow vector has a greater magnitude
• So we would expect the point mass B to move towards the upper portion of the shell
13. But when we take a closer look, we find some thing of interest:
♦ We can mark numerous green points. Because the lower arc is larger
♦ But we cannot mark that many yellow points. Because the upper arc is smaller
• This is shown in fig.b
■ That means:
♦ There will be numerous green vectors
♦ There will not be that many yellow vectors
14. We can write:
♦ The green vectors, though smaller in strengths, are greater in numbers
♦ The yellow vectors, though greater in strengths, are smaller in numbers
15. That means:
♦ The net upward force will be equal to the net downward force
♦ The two net forces will cancel each other
• We will see the precise mathematical proof in higher classes
• At present, all we need to know is this:
The net force on B is zero
16. The result written in (15) can be used in general:
■ The 'force exerted by a spherical shell' on a 'point mass situated inside the shell', is zero
• So we have seen the effects of a 'simple extended object'
• We chose a spherical shell to be the 'simple extended object'
• In later sections, we will see the applications of the results that we derived in this discussion
• In the next section, we will see an experiment which helps to obtain the value of the Universal gravitational constant G
• We chose a spherical shell to be the 'simple extended object'
• In later sections, we will see the applications of the results that we derived in this discussion
• In the next section, we will see an experiment which helps to obtain the value of the Universal gravitational constant G
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