In the previous section, we completed a discussion on the basics of gravitational force, gravity and acceleration due to gravity. We saw some solved examples and graphs also. In this section, we will see gravitational potential energy
• We know that, potential energy is that energy which is stored in a body at it’s given position
• If the body moves to a new position, it’s potential energy will change
• The ‘change in potential energy’ is equal to the amount of work done by the ‘force which causes the change in position’ on the body (Details here)
• We saw the definition of conservative forces also
♦ If the work done by a force is independent of the path, it is called a conservative force
♦ Gravitational force is a conservative force
We will now write a detailed analysis of the potential energy. In this analysis, we will include Newton's laws of gravitation also. We will write it in steps:
1. Consider a point mass mA situated on the exterior of the earth
• Let it be at a distance of r from the center of the earth
2. We know that, the force of attraction experienced by that body will be given by: $\mathbf\small{|\vec{F}_G|=\frac{G\,M_E\,m_A}{r^2}}$
• Since r is in the denominator, it is clear that, the force will decrease with increase in the distance r
♦ We will see this same fact if we draw the graph of $\mathbf\small{|\vec{F}_G|}$
♦ This graph will be similar to the graph in fig.8.32 that we saw in the previous section
• A demonstration of the 'decreasing force' can be given in 3 steps:
(i) In fig.8.37(a) below, mA is at a distance of r1 from the center O of the earth
• In fig.b, the same mA is at a distance of r2 from O
(ii) In fig.a, the force experienced by mA will be $\mathbf\small{\frac{G\,M_E\,m_A}{r_1^2}}$
• In fig.b, the force experienced by mA will be $\mathbf\small{\frac{G\,M_E\,m_A}{r_2^2}}$
(iii) We see that, r2 is greater than r1
• Since r1 and r2 are in the denominators, the force in fig.b will be less than the force in fig.a
■ So we can write:
If the same body is taken further away from the earth, it will be experiencing a lesser force of attraction
3. Is there any possibility for the forces in both figs.8.36 (a) and (b) to be the same?
Let us check:
• We know that r can be split up as (RE+h)
• So we get two more figs.: (c) and (d)
♦ In fig.c, mA is at a height h1 above the surface of the earth
♦ In fig.d, mA is at a height h2 above the surface of the earth
• Note that:
♦ Fig.a is same as fig.c. The only difference is that, the distance r1 is split into RE and h1
♦ Fig.b is same as fig.d. The only difference is that, the distance r2 is split into RE and h2
4. We know that, the force in fig.d will be less than that in fig.c
• Force acting on mA in fig.c = $\mathbf\small{\frac{G\,M_E\,m_A}{(R_E+h_1)^2}}$
• Force acting on mA in fig.d = $\mathbf\small{\frac{G\,M_E\,m_A}{(R_E+h_2)^2}}$
5. Suppose that, both h1 and h2 are very small when compared to RE. We have seen that, in such cases, those heights can be ignored
We can write:
• Force acting on mA in fig.c = $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^2}}$
• Force acting on mA in fig.d = $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^2}}$
■ Both forces are the same
6. So we can write:
• If the heights attained by a body are very close to the surface of the earth, the same force will be acting on it
• This force is given by: $\mathbf\small{|\vec{F}_G|=\frac{G\,M_E\,m_A}{R_E^2}}$
7. In the right side of the above expression, all quantities (except mA) are constants
• So $\mathbf\small{\frac{G\,M_E}{R_E^2}}$ is a constant
• We can calculate the value of this constant as:
$\mathbf\small{\frac{G\,M_E}{R_E^2}=\frac{6.6742 \times 10^{-11}\rm{(m^3\,kg^{-1}\,s^{-2})}\times5.972 \times 10^{24}\rm{(kg)}}{(6.3781 \times 10^{6})^2\rm{(m^2)}}=9.7979\;\rm{m\,s^{-2}}}$
8. So the result in (6) becomes:
• If the heights attained by a body are very close to the surface of the earth, the same force will be acting on it
• This force is given by: $\mathbf\small{|\vec{F}_G|=9.7979\;\rm{(m\,s^{-2}})}\times m_A\,\rm{(kg)}$
9. We see that, a constant force is acting at smaller heights
• If we divide this force by mass, we will get the acceleration experienced by mA
• We can write: $\mathbf\small{|\vec{a}|=\frac{|\vec{F}_G|}{m_A}=9.7979\;\rm{(m\,s^{-2}})}$
• This $\mathbf\small{|\vec{a}|}$ is the same 'acceleration due to gravity' that we denote as $\mathbf\small{|\vec{g}|}$
• So we can write: $\mathbf\small{|\vec{g}|=9.7979\;\rm{(m\,s^{-2})}}$
■ We see that:
When the mass mA is at smaller heights, it experiences the same acceleration of 9.7979 ms-2 at all those heights
10. The reader must be able to appreciate the following fact:
■ If the heights attained are not very small (when compared to RE), we will not get the same acceleration 9.7979 ms-2 at those heights
• If the reader is not able to appreciate it, he/she must examine the steps (1) to (9) again
• Also recall that, when the heights attained are not very small (when compared to RE), the graph of $\mathbf\small{|\vec{g}_h|}$ is a horizontal line.
♦ A horizontal line indicates a constant value
♦ We saw this graph in the fig.8.33 of the previous section
11. Now we can write every thing in terms of acceleration:
• At smaller heights:
♦ A constant acceleration $\mathbf\small{|\vec{g}|}$ is acting on the body
♦ Hence, a constant force $\mathbf\small{m_A|\vec{g}|}$ is acting on the body
12. Since the force is constant, we can easily calculate the work done
Work done in lifting the body from h1 to h2 = Force × displacement = $\mathbf\small{m|\vec{g}|(h_2-h_1)}$
13. Now is a good time to write a note about datum
• We can consider any horizontal line as the 'datum line'
• Let us consider the surface of the earth as the datum line. This is shown in the fig.8.38 below:
■ We cannot say this:
At the surface of the earth, mA has zero potential energy
• Why we cannot say that?
The answer can be written in 4 steps:
(i) Consider mA to be resting on the surface of the earth
(ii) Let a pit of depth d be made on the surface
(iii) If mA falls freely into that pit, some energy will be released
(iv) So mA has some potential energy even when it is at the datum
14. So what is the 'amount of potential energy that mA possess when it is at the datum'?
• The 'amount of potential energy that mA possess when it is at the datum' does not really matter
• We can write the reason in 5 steps:
(i) Let the 'amount of potential energy that mA possesses when it is at the datum' be W0
(ii) When mA is lifted from the datum to P1, a work of mgh1 will be done on it
• Then we can write:
Amount of potential energy that mA possesses when it is at P1 = mgh1 +W0
(iii) When mA is lifted from the datum to P2, a work of mgh2 will be done on it
• Then we can write:
Amount of potential energy that mA possesses when it is at P2 = mgh2 + W0
(iv) Difference in the potential energies
= (Potential energy at P2 - Potential energy at P1)
= [(mgh2 + W0) - (mgh1 + W0)]
= [mgh2 - mgh1] = mg(h2-h1)
(v) This is the same result obtained in (12)
• W0 gets cancelled
• The difference in potential energies at two points P1 and P2 is equal to the work done in moving the body from P1 to P2
• So it is the initial and final energies that matter. It is not necessary to know the potential energy at the datum
14. Let us make a graphical representation:
• We know that, the work done is equal to the area enclosed in the force-displacement graph (Details here) The fig.8.39 below shows such a graph:
The following points may be noted:
(i) In our present case, the force is a constant. So the graph representing force will be a horizontal line
• This is indicated by the yellow horizontal line
(ii) The distance between this yellow line and the x-axis will be the magnitude of the force
• Also, this distance is the width of the blue rectangle
(iii) The green vertical lines indicate the two heights h1 and h2
♦ The first green line is at a distance of h1 from the origin
♦ The second green line is at a distance of h2 from the origin
• So the distance between the green lines is (h2-h1)
• Also this distance is the length of the blue rectangle
(iv) So area of the blue rectangle = $\mathbf\small{m_A|\vec{g}|(h_2-h_1)}$
• This area is the work done by gravity in moving the mass from P1 to P2
1. Let h1 and h2 in the fig.8.38 above, be very large when compared to RE
This can be represented as in fig.8.40(a) below:
2. We have: r1 = (RE+h1) and r2 = (RE + h2)
• Since h1 and h2 are very large, we cannot ignore them. That is., we cannot write: r1 = r2 = RE
3. Also, since h1 and h2 are very large when compared to RE, there is no need for splitting the distances as (RE+h1) and (RE+h2)
• We can simply write:
♦ P1 is at a distance of r1 from the center of the earth
♦ P2 is at a distance of r2 from the center of the earth
4. We want the work done against gravity when mA is moved from P1 to P2
• As usual, we need to use the formula: Work = Force × displacement
5. But the force is not a constant. It changes as we move from P1 to P2
• So what 'value of force' will we use?
6. Let us try the graphical method. Fig.8.41 below shows the force-displacement graph:
• The graph (the yellow curve) representing force is not a horizontal line
• It is not even a line. It is a curve
• How do we plot this curve on the graph paper?
The answer can be written in 3 steps:
(i) We have: $\mathbf\small{|\vec{F}|=\frac{G\,M_E\,m_A}{r^2}}$
(ii) This is same as: $\mathbf\small{y=K\left(\frac{1}{x^2}\right)}$
Where:
♦ $\mathbf\small{y=|\vec{F}|}$
♦ $\mathbf\small{K=G\,M_E\,m_A}$
♦ $\mathbf\small{x=r}$
(iii) That means, the equation of the yellow curve is: $\mathbf\small{y=K\left(\frac{1}{x^2}\right)}$
• Putting, various values of x, we will get the corresponding values of y
• Thus the curve can be plotted
(However, we do not need the actual plotting at present. We just need to confirm that, the 'graph of the force' is a curve)
7. Consider the blue region. It is bounded by four items:
(i) The x-axis
(ii) The vertical green line at r1
(iii) The vertical green line at r2
(iv) The yellow curve
8. Using calculus, we can find the area of such complicated shapes
■ For this particular case, calculus gives us this result:
Area of the blue region = $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$
9. But area of the blue region is equal to the work done in moving mA from P1 to P2
So we can write:
Eq.8.14:
Work done in moving mA from P1 to P2 = $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$
10. How do we verify that the right side of Eq.8.14, is indeed a 'work'?
It can be done in 3 steps:
(i) $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$ can be written as: $\mathbf\small{G\,M_E\,m_A\left(\frac{r_1-r_2}{r_1r_2}\right)}$
(ii) Inputting the units, we get: $\mathbf\small{(N\,m^2\,kg^{-2})(kg)(kg)\left(\frac{(m)}{(m)(m)}\right)=N\,m}$
(iii) $\mathbf\small{N\,m}$ corresponds to 'force × displacement', which is 'work done'
11. Using Eq.8.14, we can calculate the work done for moving the body between any two points P1 and P2
• Let us consider two important points (see fig.8.40.b above):
♦ First point P1 is at a distance of r from the center of the earth
♦ Second point P2 is at a distance of infinity (∞) from the center of the earth
• Substituting these values in Eq.8.14, we get:
Work done in moving mA from ∞ to P1
= $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{\infty }-\frac{1}{r}\right)}$ = $\mathbf\small{G\,M_E\,m_A\left(0-\frac{1}{r}\right)}$ = $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$
(∵Any quantity divided by ∞ gives zero)
12. ∞ is such a very large distance. At that distance, there will not be any action of earth's gravity
• Gravitational potential energy comes into effect only if there is an action of gravity
• But at ∞ distance from earth, there is no action of earth's gravity
■ So the gravitational potential energy at ∞ is zero
13. We moved mA from a 'point of zero potential' to P1
• For that, we did a work of $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$ joules
• So the potential energy stored in the body when it is at P1 is $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$ joules
14. This is the case for a body A of mass mA. We can write the general case:
Eq.8.15: $\mathbf\small{U_r=-\frac{G\,M_E\,m}{r}}$
• Where $\mathbf\small{U_r}$ is the gravitational potential energy stored in a body of mass m, when it is at a distance of r from the center of the earth
15. In Eq.8.15, the quantity on the right side is given a negative sign
• The reason can be written in steps:
(i) In the expression, since r is in the denominator, it is clear that when the distance r from the center of the earth increases, the potential energy decreases
(ii) Consider the 3 conditions:
• When r approaches ∞, the potential energy must approach zero
• When r becomes equal to ∞, the potential energy must become equal to zero
• Also, when r increases, the potential energy must increase. This is because, when height increases, the potential energy naturally increases
(iii) So we must satisfy 3 conditions mentioned in (ii)
■ This is possible only if a negative sign is given to the quantity on the right side
(iv) This case is similar to a number line shown in fig.8.42 below:
• In this number line, the values of Ur always lies on the left of zero
• As we move from far left towards the right, the absolute value of Ur decreases. But the actual value of Ur is increases
15. We have to note an important point here. It can be written in steps:
(i) The body moves from infinity to P1
(ii) We get the impression that:
• The body falls from infinity to P1, and so, the work done for the motion will be provided by gravity
(iii) But this is not the case
• If the body is allowed to fall freely, some of the energy will be converted to kinetic energy
• We do not want such a 'conversion to kinetic energy'. Because, we want to find the potential energy only
(iv) So, we must ensure that, the body moves (from ∞ to P1) with out acceleration
• For that, we must do work against gravity. This work is stored as potential energy
• We know that, potential energy is that energy which is stored in a body at it’s given position
• If the body moves to a new position, it’s potential energy will change
• The ‘change in potential energy’ is equal to the amount of work done by the ‘force which causes the change in position’ on the body (Details here)
• We saw the definition of conservative forces also
♦ If the work done by a force is independent of the path, it is called a conservative force
♦ Gravitational force is a conservative force
We will now write a detailed analysis of the potential energy. In this analysis, we will include Newton's laws of gravitation also. We will write it in steps:
1. Consider a point mass mA situated on the exterior of the earth
• Let it be at a distance of r from the center of the earth
2. We know that, the force of attraction experienced by that body will be given by: $\mathbf\small{|\vec{F}_G|=\frac{G\,M_E\,m_A}{r^2}}$
• Since r is in the denominator, it is clear that, the force will decrease with increase in the distance r
♦ We will see this same fact if we draw the graph of $\mathbf\small{|\vec{F}_G|}$
♦ This graph will be similar to the graph in fig.8.32 that we saw in the previous section
• A demonstration of the 'decreasing force' can be given in 3 steps:
(i) In fig.8.37(a) below, mA is at a distance of r1 from the center O of the earth
• In fig.b, the same mA is at a distance of r2 from O
 |
| Fig.8.37 |
• In fig.b, the force experienced by mA will be $\mathbf\small{\frac{G\,M_E\,m_A}{r_2^2}}$
(iii) We see that, r2 is greater than r1
• Since r1 and r2 are in the denominators, the force in fig.b will be less than the force in fig.a
■ So we can write:
If the same body is taken further away from the earth, it will be experiencing a lesser force of attraction
3. Is there any possibility for the forces in both figs.8.36 (a) and (b) to be the same?
Let us check:
• We know that r can be split up as (RE+h)
• So we get two more figs.: (c) and (d)
♦ In fig.c, mA is at a height h1 above the surface of the earth
♦ In fig.d, mA is at a height h2 above the surface of the earth
• Note that:
♦ Fig.a is same as fig.c. The only difference is that, the distance r1 is split into RE and h1
♦ Fig.b is same as fig.d. The only difference is that, the distance r2 is split into RE and h2
4. We know that, the force in fig.d will be less than that in fig.c
• Force acting on mA in fig.c = $\mathbf\small{\frac{G\,M_E\,m_A}{(R_E+h_1)^2}}$
• Force acting on mA in fig.d = $\mathbf\small{\frac{G\,M_E\,m_A}{(R_E+h_2)^2}}$
5. Suppose that, both h1 and h2 are very small when compared to RE. We have seen that, in such cases, those heights can be ignored
We can write:
• Force acting on mA in fig.c = $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^2}}$
• Force acting on mA in fig.d = $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^2}}$
■ Both forces are the same
6. So we can write:
• If the heights attained by a body are very close to the surface of the earth, the same force will be acting on it
• This force is given by: $\mathbf\small{|\vec{F}_G|=\frac{G\,M_E\,m_A}{R_E^2}}$
7. In the right side of the above expression, all quantities (except mA) are constants
• So $\mathbf\small{\frac{G\,M_E}{R_E^2}}$ is a constant
• We can calculate the value of this constant as:
$\mathbf\small{\frac{G\,M_E}{R_E^2}=\frac{6.6742 \times 10^{-11}\rm{(m^3\,kg^{-1}\,s^{-2})}\times5.972 \times 10^{24}\rm{(kg)}}{(6.3781 \times 10^{6})^2\rm{(m^2)}}=9.7979\;\rm{m\,s^{-2}}}$
8. So the result in (6) becomes:
• If the heights attained by a body are very close to the surface of the earth, the same force will be acting on it
• This force is given by: $\mathbf\small{|\vec{F}_G|=9.7979\;\rm{(m\,s^{-2}})}\times m_A\,\rm{(kg)}$
9. We see that, a constant force is acting at smaller heights
• If we divide this force by mass, we will get the acceleration experienced by mA
• We can write: $\mathbf\small{|\vec{a}|=\frac{|\vec{F}_G|}{m_A}=9.7979\;\rm{(m\,s^{-2}})}$
• This $\mathbf\small{|\vec{a}|}$ is the same 'acceleration due to gravity' that we denote as $\mathbf\small{|\vec{g}|}$
• So we can write: $\mathbf\small{|\vec{g}|=9.7979\;\rm{(m\,s^{-2})}}$
■ We see that:
When the mass mA is at smaller heights, it experiences the same acceleration of 9.7979 ms-2 at all those heights
10. The reader must be able to appreciate the following fact:
■ If the heights attained are not very small (when compared to RE), we will not get the same acceleration 9.7979 ms-2 at those heights
• If the reader is not able to appreciate it, he/she must examine the steps (1) to (9) again
• Also recall that, when the heights attained are not very small (when compared to RE), the graph of $\mathbf\small{|\vec{g}_h|}$ is a horizontal line.
♦ A horizontal line indicates a constant value
♦ We saw this graph in the fig.8.33 of the previous section
11. Now we can write every thing in terms of acceleration:
• At smaller heights:
♦ A constant acceleration $\mathbf\small{|\vec{g}|}$ is acting on the body
♦ Hence, a constant force $\mathbf\small{m_A|\vec{g}|}$ is acting on the body
12. Since the force is constant, we can easily calculate the work done
Work done in lifting the body from h1 to h2 = Force × displacement = $\mathbf\small{m|\vec{g}|(h_2-h_1)}$
13. Now is a good time to write a note about datum
• We can consider any horizontal line as the 'datum line'
• Let us consider the surface of the earth as the datum line. This is shown in the fig.8.38 below:
 |
| Fig.8.38 |
At the surface of the earth, mA has zero potential energy
• Why we cannot say that?
The answer can be written in 4 steps:
(i) Consider mA to be resting on the surface of the earth
(ii) Let a pit of depth d be made on the surface
(iii) If mA falls freely into that pit, some energy will be released
(iv) So mA has some potential energy even when it is at the datum
14. So what is the 'amount of potential energy that mA possess when it is at the datum'?
• The 'amount of potential energy that mA possess when it is at the datum' does not really matter
• We can write the reason in 5 steps:
(i) Let the 'amount of potential energy that mA possesses when it is at the datum' be W0
(ii) When mA is lifted from the datum to P1, a work of mgh1 will be done on it
• Then we can write:
Amount of potential energy that mA possesses when it is at P1 = mgh1 +W0
(iii) When mA is lifted from the datum to P2, a work of mgh2 will be done on it
• Then we can write:
Amount of potential energy that mA possesses when it is at P2 = mgh2 + W0
(iv) Difference in the potential energies
= (Potential energy at P2 - Potential energy at P1)
= [(mgh2 + W0) - (mgh1 + W0)]
= [mgh2 - mgh1] = mg(h2-h1)
(v) This is the same result obtained in (12)
• W0 gets cancelled
• The difference in potential energies at two points P1 and P2 is equal to the work done in moving the body from P1 to P2
• So it is the initial and final energies that matter. It is not necessary to know the potential energy at the datum
14. Let us make a graphical representation:
• We know that, the work done is equal to the area enclosed in the force-displacement graph (Details here) The fig.8.39 below shows such a graph:
 |
| Fig.8.39 |
(i) In our present case, the force is a constant. So the graph representing force will be a horizontal line
• This is indicated by the yellow horizontal line
(ii) The distance between this yellow line and the x-axis will be the magnitude of the force
• Also, this distance is the width of the blue rectangle
(iii) The green vertical lines indicate the two heights h1 and h2
♦ The first green line is at a distance of h1 from the origin
♦ The second green line is at a distance of h2 from the origin
• So the distance between the green lines is (h2-h1)
• Also this distance is the length of the blue rectangle
(iv) So area of the blue rectangle = $\mathbf\small{m_A|\vec{g}|(h_2-h_1)}$
• This area is the work done by gravity in moving the mass from P1 to P2
• If the force is not constant, we will not get a horizontal yellow line. So we will not get a rectangle
• In such cases, the area cannot be calculated easily
• We will now write a detailed analysis on such cases:
• In such cases, the area cannot be calculated easily
• We will now write a detailed analysis on such cases:
This can be represented as in fig.8.40(a) below:
 |
| Fug.8.40 |
• Since h1 and h2 are very large, we cannot ignore them. That is., we cannot write: r1 = r2 = RE
3. Also, since h1 and h2 are very large when compared to RE, there is no need for splitting the distances as (RE+h1) and (RE+h2)
• We can simply write:
♦ P1 is at a distance of r1 from the center of the earth
♦ P2 is at a distance of r2 from the center of the earth
4. We want the work done against gravity when mA is moved from P1 to P2
• As usual, we need to use the formula: Work = Force × displacement
5. But the force is not a constant. It changes as we move from P1 to P2
• So what 'value of force' will we use?
6. Let us try the graphical method. Fig.8.41 below shows the force-displacement graph:
 |
| Fig.8.41 |
• It is not even a line. It is a curve
• How do we plot this curve on the graph paper?
The answer can be written in 3 steps:
(i) We have: $\mathbf\small{|\vec{F}|=\frac{G\,M_E\,m_A}{r^2}}$
(ii) This is same as: $\mathbf\small{y=K\left(\frac{1}{x^2}\right)}$
Where:
♦ $\mathbf\small{y=|\vec{F}|}$
♦ $\mathbf\small{K=G\,M_E\,m_A}$
♦ $\mathbf\small{x=r}$
(iii) That means, the equation of the yellow curve is: $\mathbf\small{y=K\left(\frac{1}{x^2}\right)}$
• Putting, various values of x, we will get the corresponding values of y
• Thus the curve can be plotted
(However, we do not need the actual plotting at present. We just need to confirm that, the 'graph of the force' is a curve)
7. Consider the blue region. It is bounded by four items:
(i) The x-axis
(ii) The vertical green line at r1
(iii) The vertical green line at r2
(iv) The yellow curve
8. Using calculus, we can find the area of such complicated shapes
■ For this particular case, calculus gives us this result:
Area of the blue region = $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$
9. But area of the blue region is equal to the work done in moving mA from P1 to P2
So we can write:
Eq.8.14:
Work done in moving mA from P1 to P2 = $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$
10. How do we verify that the right side of Eq.8.14, is indeed a 'work'?
It can be done in 3 steps:
(i) $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}$ can be written as: $\mathbf\small{G\,M_E\,m_A\left(\frac{r_1-r_2}{r_1r_2}\right)}$
(ii) Inputting the units, we get: $\mathbf\small{(N\,m^2\,kg^{-2})(kg)(kg)\left(\frac{(m)}{(m)(m)}\right)=N\,m}$
(iii) $\mathbf\small{N\,m}$ corresponds to 'force × displacement', which is 'work done'
11. Using Eq.8.14, we can calculate the work done for moving the body between any two points P1 and P2
• Let us consider two important points (see fig.8.40.b above):
♦ First point P1 is at a distance of r from the center of the earth
♦ Second point P2 is at a distance of infinity (∞) from the center of the earth
• Substituting these values in Eq.8.14, we get:
Work done in moving mA from ∞ to P1
= $\mathbf\small{G\,M_E\,m_A\left(\frac{1}{\infty }-\frac{1}{r}\right)}$ = $\mathbf\small{G\,M_E\,m_A\left(0-\frac{1}{r}\right)}$ = $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$
(∵Any quantity divided by ∞ gives zero)
12. ∞ is such a very large distance. At that distance, there will not be any action of earth's gravity
• Gravitational potential energy comes into effect only if there is an action of gravity
• But at ∞ distance from earth, there is no action of earth's gravity
■ So the gravitational potential energy at ∞ is zero
13. We moved mA from a 'point of zero potential' to P1
• For that, we did a work of $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$ joules
• So the potential energy stored in the body when it is at P1 is $\mathbf\small{\frac{G\,M_E\,m_A}{r}}$ joules
14. This is the case for a body A of mass mA. We can write the general case:
Eq.8.15: $\mathbf\small{U_r=-\frac{G\,M_E\,m}{r}}$
• Where $\mathbf\small{U_r}$ is the gravitational potential energy stored in a body of mass m, when it is at a distance of r from the center of the earth
15. In Eq.8.15, the quantity on the right side is given a negative sign
• The reason can be written in steps:
(i) In the expression, since r is in the denominator, it is clear that when the distance r from the center of the earth increases, the potential energy decreases
(ii) Consider the 3 conditions:
• When r approaches ∞, the potential energy must approach zero
• When r becomes equal to ∞, the potential energy must become equal to zero
• Also, when r increases, the potential energy must increase. This is because, when height increases, the potential energy naturally increases
(iii) So we must satisfy 3 conditions mentioned in (ii)
■ This is possible only if a negative sign is given to the quantity on the right side
(iv) This case is similar to a number line shown in fig.8.42 below:
 |
| Fig.8.42 |
• As we move from far left towards the right, the absolute value of Ur decreases. But the actual value of Ur is increases
15. We have to note an important point here. It can be written in steps:
(i) The body moves from infinity to P1
(ii) We get the impression that:
• The body falls from infinity to P1, and so, the work done for the motion will be provided by gravity
(iii) But this is not the case
• If the body is allowed to fall freely, some of the energy will be converted to kinetic energy
• We do not want such a 'conversion to kinetic energy'. Because, we want to find the potential energy only
(iv) So, we must ensure that, the body moves (from ∞ to P1) with out acceleration
• For that, we must do work against gravity. This work is stored as potential energy
• So using Eq.8.15, we can calculate the potential energy of a mass m at any point (situated at a distance r) from the center of the earth
• How is all this related to the potential energy mgh that we learned in our previous classes?
• We will see it in the next section
• How is all this related to the potential energy mgh that we learned in our previous classes?
• We will see it in the next section
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