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 also. In this section, we will see how graphs can be used for representing the variations in those quantities
First we will see the variation of 'acceleration due to gravity'
We derived 3 equations related to this quantity (Details here)
Eq.8.7:
$\mathbf\small{|\vec{g}|}$ on the surface of the earth is given by: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
Eq.8.8:
$\mathbf\small{|\vec{g}_h|}$ at height h above the surface of the earth: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
Eq.8.11:
$\mathbf\small{|\vec{g}_d|}$ at depth d below the surface of the earth: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
• We know that Eq.8.7 will give a constant value equal to 9.81 m s-2
• A 'constant value' means that, there is no variation of $\mathbf\small{|\vec{g}|}$. So we need not draw any graph
• We will consider the next case: Variation of $\mathbf\small{|\vec{g}_h|}$ with height
• We will write it in steps:
1. We have Eq.8.8
• Substituting the known values in that equation, we get:
$\mathbf\small{|\vec{g}_h|=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^{6}+h)^2}=\frac{3.983\times 10^{14}}{(6.371\times 10^{6}+h)^2}}$
2. By putting various values of h, we can obtain the corresponding $\mathbf\small{|\vec{g}_h|}$
• A table is thus prepared below:
3. The above points can be plotted on a graph. This is shown in fig.8.32 below. The coordinates of some points are also written for easy reference
4. We see that, the graph has a peculiar curved shape. We often come across similar curves in science and engineering
• The graph in fig.8.32 is similar to the graph of $\mathbf\small{y=\frac{1}{x^2}}$
• The exponent is '2' and it is in the denominator
• Note that, in Eq.8.8 also, the exponent is '2' and it is in the denominator
■ Clearly, the $\mathbf\small{|\vec{g}_h|}$ is decreasing with increase in r
5. We get this peculiar curve only if we choose high values (like RE, 2RE, 3RE etc.,) for h
• For low values of h (that is., heights near the surface of the earth), the graph will be nearly a horizontal line. This is clear from the table 8.2 below and it's corresponding graph
■ It is clear that, near the surface of the earth, even if heights are different, the body will be experiencing the same $\mathbf\small{|\vec{g}|}$
1. We have Eq.8.11
• Substituting the known values, we get:
$\mathbf\small{|\vec{g}_d|=9.81\left(1- \frac{d}{6.371\times 10^{6}}\right)}$
2. By putting various values of d, we can obtain the corresponding $\mathbf\small{|\vec{g}_d|}$
• A table is thus prepared below:
3. The above points can be plotted on a graph. This is shown in fig.8.34 below. The coordinates of some points are also written for easy reference
4. We see that, the graph is a straight line. It is sloping downwards towards the right
First we will see the variation of 'acceleration due to gravity'
We derived 3 equations related to this quantity (Details here)
Eq.8.7:
$\mathbf\small{|\vec{g}|}$ on the surface of the earth is given by: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
Eq.8.8:
$\mathbf\small{|\vec{g}_h|}$ at height h above the surface of the earth: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
Eq.8.11:
$\mathbf\small{|\vec{g}_d|}$ at depth d below the surface of the earth: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
• We know that Eq.8.7 will give a constant value equal to 9.81 m s-2
• A 'constant value' means that, there is no variation of $\mathbf\small{|\vec{g}|}$. So we need not draw any graph
• We will consider the next case: Variation of $\mathbf\small{|\vec{g}_h|}$ with height
• We will write it in steps:
1. We have Eq.8.8
• Substituting the known values in that equation, we get:
$\mathbf\small{|\vec{g}_h|=\frac{6.67\times 10^{-11}\times 5.972\times 10^{24}}{(6.371\times 10^{6}+h)^2}=\frac{3.983\times 10^{14}}{(6.371\times 10^{6}+h)^2}}$
2. By putting various values of h, we can obtain the corresponding $\mathbf\small{|\vec{g}_h|}$
• A table is thus prepared below:
 |
| Table 8.1 |
 |
| Fig.8.32 |
• The graph in fig.8.32 is similar to the graph of $\mathbf\small{y=\frac{1}{x^2}}$
• The exponent is '2' and it is in the denominator
• Note that, in Eq.8.8 also, the exponent is '2' and it is in the denominator
■ Clearly, the $\mathbf\small{|\vec{g}_h|}$ is decreasing with increase in r
5. We get this peculiar curve only if we choose high values (like RE, 2RE, 3RE etc.,) for h
• For low values of h (that is., heights near the surface of the earth), the graph will be nearly a horizontal line. This is clear from the table 8.2 below and it's corresponding graph
 |
| Table 8.2 |
 |
| Fig.8.33 |
• Next we will consider the variation of $\mathbf\small{|\vec{g}_d|}$ with depth. We will write it in steps:
• Substituting the known values, we get:
$\mathbf\small{|\vec{g}_d|=9.81\left(1- \frac{d}{6.371\times 10^{6}}\right)}$
2. By putting various values of d, we can obtain the corresponding $\mathbf\small{|\vec{g}_d|}$
• A table is thus prepared below:
 |
| Table 8.3 |
 |
| Fig.8.34 |
• So we can write:
When depth increases, g decreases
5. But why is it a straight line?
The answer can be written in 4 steps:
(i) Consider Eq.8.11: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
• It can be rearranged as: $\mathbf\small{|\vec{g}_d|=|\vec{g}|- \left(\frac{|\vec{g}|}{R_E}\right)d}$
$\mathbf\small{\Rightarrow|\vec{g}_d|=- \left(\frac{|\vec{g}|}{R_E}\right)d+|\vec{g}|}$
(ii) This is of the form: $\mathbf\small{y=- \left(m\right)x+c}$
• So it is a straight line
(iii) Note that, the slope is negative. So the line slopes downwards towards the right
• Also, the y intercept c is $\mathbf\small{|\vec{g}|}$
(iv) We can write:
$\mathbf\small{|\vec{g}_d|}$ is proportional to d
■ The two graphs in figs 8.32 and 8.34 are important graphs
• The first one shows the variation of $\mathbf\small{|\vec{g}_h|}$ when we go upwards from the surface of the earth
• The second one shows the variation of $\mathbf\small{|\vec{g}_d|}$ when we go downwards from the surface of the earth
■ So for both the graphs, the starting point is the surface of the earth. That is why, we obtain 9.81 m s-2 as the first point in both the cases
■ What if we start from the center of the earth?
• Then, instead of h or d, we will be using r. We will write it in steps:
1. First, we will consider the interior (Details here)
• We have Eq.8.6:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ ME = Mass of the earth
2. If we divide this force by mA, we will get the acceleration experienced by that mA
• So we can write: $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
3. We need not consider depth below surface. What we need is the distance r from O
• Substituting the known values on the right side, we get: $\mathbf\small{|\vec{g}_r|=(1.54\times 10^{-6})r}$
4. By putting various values of r, we can obtain the corresponding $\mathbf\small{|\vec{g}_r|}$
• A table is thus prepared below:
5. The above table 8.4 gives the values in the interior. Next we want the values in the exterior
• We can use Eq.8.8 again:
$\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
7. By putting various values of r, we can obtain the corresponding $\mathbf\small{|\vec{g}_r|}$ values
• The table is given below:
8. The points from the above two tables can be plotted on a graph. This is shown in fig.8.35 below. The coordinates of some points are also written for easy reference
9. We see that, the first portion of the graph is a straight line. It is shown in magenta color. It is sloping upwards towards the right
• So we can write:
When depth increases, g decreases
5. But why is it a straight line?
The answer can be written in 4 steps:
(i) Consider Eq.8.11: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
• It can be rearranged as: $\mathbf\small{|\vec{g}_d|=|\vec{g}|- \left(\frac{|\vec{g}|}{R_E}\right)d}$
$\mathbf\small{\Rightarrow|\vec{g}_d|=- \left(\frac{|\vec{g}|}{R_E}\right)d+|\vec{g}|}$
(ii) This is of the form: $\mathbf\small{y=- \left(m\right)x+c}$
• So it is a straight line
(iii) Note that, the slope is negative. So the line slopes downwards towards the right
• Also, the y intercept c is $\mathbf\small{|\vec{g}|}$
(iv) We can write:
$\mathbf\small{|\vec{g}_d|}$ is proportional to d
• The first one shows the variation of $\mathbf\small{|\vec{g}_h|}$ when we go upwards from the surface of the earth
• The second one shows the variation of $\mathbf\small{|\vec{g}_d|}$ when we go downwards from the surface of the earth
■ So for both the graphs, the starting point is the surface of the earth. That is why, we obtain 9.81 m s-2 as the first point in both the cases
■ What if we start from the center of the earth?
• Then, instead of h or d, we will be using r. We will write it in steps:
1. First, we will consider the interior (Details here)
• We have Eq.8.6:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ ME = Mass of the earth
2. If we divide this force by mA, we will get the acceleration experienced by that mA
• So we can write: $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
3. We need not consider depth below surface. What we need is the distance r from O
• Substituting the known values on the right side, we get: $\mathbf\small{|\vec{g}_r|=(1.54\times 10^{-6})r}$
4. By putting various values of r, we can obtain the corresponding $\mathbf\small{|\vec{g}_r|}$
• A table is thus prepared below:
 |
| Table 8.4 |
• We can use Eq.8.8 again:
$\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
• But this time, we put r instead of (RE+h)
• So we can write:
$\mathbf\small{|\vec{g}_r|=\frac{GM_E}{r^2}}$
6. Substituting the known values, we get:
$\mathbf\small{|\vec{g}_r|=\frac{3.983 \times 10^{14}}{r^2}}$7. By putting various values of r, we can obtain the corresponding $\mathbf\small{|\vec{g}_r|}$ values
• The table is given below:
 |
| Table 8.5 |
 |
| Fig.8.35 |
• So we can write:
When r increases, $\mathbf\small{|\vec{g}_r|}$ also increases
10. But why is it a straight line?
The answer can be written in 4 steps:
(i) Consider the equation that we wrote in (2): $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
(ii) This is of the form: $\mathbf\small{y=mx}$
• So it is a straight line
(iii) Note that, the slope is positive. So the line slopes upwards towards the right
• Also, the y intercept c is not present. So the line passes through the origin
(iv) We can write:
In the interior of the earth, $\mathbf\small{|\vec{g}_r|}$ is proportional to r
11. Note that, in the previous graph in fig.8.34, the slope is downwards. But in the above fig.8.35, in the initial portion, the slope is upwards. Why is that so?
The answer can be written in 5 steps:
(i) Consider two points:
♦ The center O of the earth
♦ Any one point on the surface
(ii) A person can move between those two points in two ways:
♦ He can start from O and move towards the surface point
♦ He can start from the surface point and move towards O
(iii) If he starts from O and move upwards towards the surface point, he will feel that, acceleration due to gravity acting on him is increasing
• If he starts from the surface point and move downwards towards O, he will feel that, acceleration due to gravity acting on him is decreasing
■ We saw the reason when we analysed the 'gravity in the interior' in a previous section:
The shells outside the point under consideration have no effect
(iv) Thus we get upward and downward slopes in the two different graphs
12. Now we consider the exterior portion
• Here we have a green curve
• This is similar to the green curve in the previous fig.8.32. Both are related to the acceleration above the surface
13. But we see a difference:
• In fig.8.32, the green curve starts from the y-axis. But in fig.8.35, the green curve starts at a distance of RE away from the y-axis
• The reason can be written in steps:
(i) The graph in fig.8.32 is related to exterior only
• The graph in fig.8.35 is related to both interior and exterior
(ii) In fig.8.32, the starting point is the surface of the earth. It coincides with the y-axis
• In fig.8.35, the starting point is the center O. So this O coincides with the y-axis
(iii) The green curve is related to the exterior. It can begin only when r becomes RE
14. So we must use two equations for drawing the graph:
• When 0 ≤ r ≤ RE, we must use $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
• When r > RE, we must use $\mathbf\small{|\vec{g}_r|=\frac{GM_E}{r^2}}$
1. Consider a sphere having the same mass and radius as the earth
• But it's inside is hollow
• So it is a spherical shell of mass ME and radius RE
2. We want to draw the graph of $\mathbf\small{|\vec{g}_r|}$
• It will be the same graph in fig.8.35, but without the magenta line
• So the required graph will be the one shown in fig.8.36 below
3. So why is the magenta line absent?
The answer can be written in 4 steps:
(i) In the interior of a shell, there will not be any gravitational force
(ii) If there is no force, there will not be any acceleration
(iii) So for r < RE, we will not get any y-coordinates
• That means, for r < RE, we will not get any points to plot
(iv) So the graph will begin only when r reaches RE
1. The graph in fig.8.32 shows the variation of $\mathbf\small{|\vec{g}_h|}$. We consider points on and above the surface of the earth
2. The graph in fig.8.33 shows that, there is no variation in $\mathbf\small{|\vec{g}_h|}$ if h takes small values
10. But why is it a straight line?
The answer can be written in 4 steps:
(i) Consider the equation that we wrote in (2): $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
(ii) This is of the form: $\mathbf\small{y=mx}$
• So it is a straight line
(iii) Note that, the slope is positive. So the line slopes upwards towards the right
• Also, the y intercept c is not present. So the line passes through the origin
(iv) We can write:
In the interior of the earth, $\mathbf\small{|\vec{g}_r|}$ is proportional to r
11. Note that, in the previous graph in fig.8.34, the slope is downwards. But in the above fig.8.35, in the initial portion, the slope is upwards. Why is that so?
The answer can be written in 5 steps:
(i) Consider two points:
♦ The center O of the earth
♦ Any one point on the surface
(ii) A person can move between those two points in two ways:
♦ He can start from O and move towards the surface point
♦ He can start from the surface point and move towards O
(iii) If he starts from O and move upwards towards the surface point, he will feel that, acceleration due to gravity acting on him is increasing
• If he starts from the surface point and move downwards towards O, he will feel that, acceleration due to gravity acting on him is decreasing
■ We saw the reason when we analysed the 'gravity in the interior' in a previous section:
The shells outside the point under consideration have no effect
(iv) Thus we get upward and downward slopes in the two different graphs
12. Now we consider the exterior portion
• Here we have a green curve
• This is similar to the green curve in the previous fig.8.32. Both are related to the acceleration above the surface
13. But we see a difference:
• In fig.8.32, the green curve starts from the y-axis. But in fig.8.35, the green curve starts at a distance of RE away from the y-axis
• The reason can be written in steps:
(i) The graph in fig.8.32 is related to exterior only
• The graph in fig.8.35 is related to both interior and exterior
(ii) In fig.8.32, the starting point is the surface of the earth. It coincides with the y-axis
• In fig.8.35, the starting point is the center O. So this O coincides with the y-axis
(iii) The green curve is related to the exterior. It can begin only when r becomes RE
14. So we must use two equations for drawing the graph:
• When 0 ≤ r ≤ RE, we must use $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
• When r > RE, we must use $\mathbf\small{|\vec{g}_r|=\frac{GM_E}{r^2}}$
Next we will see an interesting case. We will write it in steps:
• But it's inside is hollow
• So it is a spherical shell of mass ME and radius RE
2. We want to draw the graph of $\mathbf\small{|\vec{g}_r|}$
• It will be the same graph in fig.8.35, but without the magenta line
• So the required graph will be the one shown in fig.8.36 below
 |
| Fig.8.36 |
The answer can be written in 4 steps:
(i) In the interior of a shell, there will not be any gravitational force
(ii) If there is no force, there will not be any acceleration
(iii) So for r < RE, we will not get any y-coordinates
• That means, for r < RE, we will not get any points to plot
(iv) So the graph will begin only when r reaches RE
Let us write a summary of the above graphs:
2. The graph in fig.8.33 shows that, there is no variation in $\mathbf\small{|\vec{g}_h|}$ if h takes small values
3. The graph in fig.8.34 shows the variation of $\mathbf\small{|\vec{g}_d|}$. We consider points on and below the surface of the earth
4. The graph in fig.8.35 shows the variation of $\mathbf\small{|\vec{g}_r|}$. We consider points from the center O of the earth, moving outwards
5. The graph in fig.8.36 shows the variation of $\mathbf\small{|\vec{g}_r|}$ in the case of a spherical shell. We consider points from the center O of the earth, moving outwards
5. The graph in fig.8.36 shows the variation of $\mathbf\small{|\vec{g}_r|}$ in the case of a spherical shell. We consider points from the center O of the earth, moving outwards
■ So there are 5 categories related to acceleration
■ We can prepare graphs related to gravitational force also for all the 5 categories
■ We can prepare graphs related to intensity of gravitational field also for all the 5 categories
• The reader is advised to draw all of them and become familiar with their shapes and types of variations
• Hint:
(i) When acceleration is multiplied by a mass say mA, we get force
(ii) When force is divided by the second mass mB, we get the field created by the first mass mA
• In the next section, we will see gravitational potential energy
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