Chapter 8.13 - Gravitational Potential

In the previous section, we saw gravitational potential energy
• We derived Eq.8.15: $\mathbf\small{U_r=-\frac{G\,M_E\,m}{r}}$
• In this section we will see how it is related to our old equation:
Gravitational Potential energy = mgh
• Later in this section we will also see 'gravitational potential'

1. Consider a point mass m
2. Let it be placed on the surface of the earth
• Then it's distance from the center O of the earth is R_E
• So it's potential energy is given by: $\mathbf\small{U_{R_E}=-\frac{G\,M_E\,m}{R_E}}$
3. Let it be taken to a height h above the surface of the earth
• Then it's distance from the center O of the earth is R_E+h
• So now it's potential energy is given by: $\mathbf\small{U_{(R_E+h)}=-\frac{G\,M_E\,m}{(R_E+h)}}$
4. Difference in potential energies = Final energy - Initial energy
= $\mathbf\small{U_{(R_E+h)}-U_{R_E}}$
= $\mathbf\small{-\frac{G\,M_E\,m}{(R_E+h)}-\,-\frac{G\,M_E\,m}{R_E}}$
= $\mathbf\small{-\frac{G\,M_E\,m}{(R_E+h)}+\frac{G\,M_E\,m}{R_E}}$
= $\mathbf\small{G\,M_E\,m \left(\frac{1}{R_E}-\frac{1}{(R_E+h)}\right)}$
= $\mathbf\small{G\,M_E\,m \left(\frac{R_E+h-R_E}{R_E(R_E+h)}\right)}$
= $\mathbf\small{G\,M_E\,m \left(\frac{h}{R_E(R_E+h)}\right)}$
= $\mathbf\small{G\,M_E\,m \left(\frac{h}{R_E^2(1+\frac{h}{R_E})}\right)}$
= $\mathbf\small{\frac{G\,M_E}{R_E^2} \left(\frac{mh}{(1+\frac{h}{R_E})}\right)}$
= $\mathbf\small{\frac{G\,M_E}{R_E^2} \left(\frac{mh}{(1)}\right)}$ (∵ h is small, $\mathbf\small{\frac{h}{R_E}}$ can be ignored)
= $\mathbf\small{m|\vec{g}|h}$ (∵ $\mathbf\small{\frac{G\,M_E}{R_E^2}=|\vec{g}|}$)
5. Thus we get the old relation:
Work done to raise an object of mass m from the surface of earth to a height h = $\mathbf\small{m|\vec{g}|h}$

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• So we have completed a discussion on gravitational potential energy

• Next we have to learn about gravitational potential. We can write about it in steps:
1. In fig.8.43 below, two bodies A and B are placed at P₁ and P₂

Fig.8.43

• Their masses are m_A and m_B respectively
• The center to center distance is r
2. So we have a system consisting of two masses m_A and m_B
• The gravitational potential energy of this system is given by: $\mathbf\small{U=-\frac{G\,m_A\,m_B}{r}}$
3. In normal cases, the masses m_A and m_B do not change
• But the positions can change:
    ♦ A can move away from P₁
    ♦ B can move away from P₂
    ♦ Both A and B can move away from their respective positions P₁ and P₂
• If any of those ‘changes in positions’ happen, the energy of the system will change
• So ‘positions of objects’ is important for finding the potential energy  
4. In the above fig., if m_B = 1 kg, the energy of the system will be equal to $\mathbf\small{U=-\frac{G\,m_A}{r}}$
• We can write: The body A is able to produce a gravitational potential of $\mathbf\small{-\frac{G\,m_A}{r}}$ joules at a distance of r
5. Similarly, if m_A = 1 kg, the energy of the system will be equal to $\mathbf\small{U=-\frac{G\,m_B}{r}}$
• We can write: The body B is able to produce a gravitational potential of $\mathbf\small{-\frac{G\,m_B}{r}}$ joules at a distance of r
6. We can define gravitational field in 5 steps:
(i) Consider a body A. Let it's mass be m_A
• There will be a gravitational field around A
(ii) A mass of 1 kg is initially at infinity
(iii) We want to bring this 1 kg mass into the field of A
• We want to place this 1 kg mass at a distance of r from A
(iv) For that, we have to do a work of $\mathbf\small{-\frac{G\,m_A}{r}}$ joules
• This much work will be stored in that 1 kg mass
• This much work is called gravitational potential (at distance r) created by A
(v) Gravitational potential is denoted by the letter V. Since it is an energy, it is a scalar quantity 
• So we can write:
Eq.8.16: $\mathbf\small{V_A=-\frac{G\,m_A}{r}}$
    ♦ The subscript ‘A’ indicates that, it the gravitational potential created by A 

The following solved example will help us to fully understand this concept
Solved example 8.29
Four equal masses m are placed at the four corners of a square ABCD. The side of the square is a. What is the gravitational potential energy of the system? Also find the gravitational potential at the center of the system
Solution:
Part (i):
1. Fig.8.44(a) below shows the arrangement

Fig.8.44

2. We have to consider one pair at a time
• Consider the pair A-B
• The potential energy due to this pair is given by: $\mathbf\small{U_{A,B}=-\frac{Gm^2}{a}}$
3. Along the periphery, there are 3 more pairs like this. All the four pairs along the periphery are identical. So we can write:
Total gravitational potential energy of the pairs along the periphery
= $\mathbf\small{U_{AB}\;+U_{B,C}\;+U_{C,D}\;+U_{D,A}}$
= $\mathbf\small{-\frac{Gm^2}{a}\;+-\frac{Gm^2}{a}\;+-\frac{Gm^2}{a}\;+-\frac{Gm^2}{a}}$
= $\mathbf\small{-\frac{4Gm^2}{a}}$
4. Next we consider the pair A-C along the diagonal
• The distance between the two masses in this pair = √2 a
    ♦ This is shown in fig.b
• So the potential energy due to this pair is given by: $\mathbf\small{U_{A,C}=-\frac{Gm^2}{\sqrt{2}\,a}}$
5. There is one more diagonal pair B-D like this. Both the diagonal pairs are identical. So we can write:
Total gravitational potential energy of the diagonal pairs
= $\mathbf\small{U_{A,C}\;+U_{B,D}}$
= $\mathbf\small{-\frac{Gm^2}{\sqrt{2}\,a}\;-\frac{Gm^2}{\sqrt{2}\,a}}$
= $\mathbf\small{-\frac{2Gm^2}{\sqrt{2}\,a}\;=-\frac{\sqrt{2}Gm^2}{\,a}}$
6. Thus we get:
Total gravitational potential energy of the system = $\mathbf\small{-\frac{4Gm^2}{a}\;+-\frac{\sqrt{2}Gm^2}{\,a}}$
= $\mathbf\small{-\frac{Gm^2}{a}(4+\sqrt{2})=-5.41\frac{Gm^2}{a}}$
Part (ii):
1. In this part we calculate the gravitational potential at the center
• We have to consider the potential created by each mass
• First we consider the mass at A
• It is at a distance of $\mathbf\small{\frac{a}{\sqrt{2}}}$ from the center O
    ♦ This is shown in fig.c
• So we get: $\mathbf\small{V_A=-\frac{Gm}{\frac{a}{\sqrt{2}}}=-\frac{\sqrt{2}Gm}{a}}$
2. There are three more masses. All of them are at the same distance of $\mathbf\small{\frac{a}{\sqrt{2}}}$ from the center. So all four masses will create the same potential
• Thus we get:
Total potential at the center of the square
= $\mathbf\small{V_A\;+V_B\;+V_C\;+V_D}$
= $\mathbf\small{-\frac{\sqrt{2}Gm}{a}\;+-\frac{\sqrt{2}Gm}{a}\;+-\frac{\sqrt{2}Gm}{a}\;+-\frac{\sqrt{2}Gm}{a}}$
= $\mathbf\small{-\frac{4\sqrt{2}Gm}{a}}$

Solved example 8.30
Three particles of masses m, 2m and 4m are placed at the corners of an equilateral triangle of side a
(i) Calculate the potential energy of the system
(ii) Work done on the system if all the sides are changed from a to 2a
Assume that the potential energy is zero when the sides are infinity
Solution:
Part (i):
1. Fig.8.45 below shows the arrangement

Fig.8.45

2. We have to consider one pair at a time
• Consider the pair A-B
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{A,B}=-\frac{2Gm^2}{a}}$
• Consider the pair B-C
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{B,C}=-\frac{8Gm^2}{a}}$
• Consider the pair C-A
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{C,A}=-\frac{4Gm^2}{a}}$
3. Total gravitational potential energy of the system
= $\mathbf\small{U_{A,B}\;+U_{B,C}\;+U_{C,A}}$
= $\mathbf\small{-\frac{2Gm^2}{a}\;+-\frac{8Gm^2}{a}\;+-\frac{4Gm^2}{a}}$
= $\mathbf\small{-\frac{14Gm^2}{a}}$
Part (ii):
When the separation is 2a
1. We have to consider one pair at a time
• Consider the pair A-B
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{A,B}=-\frac{2Gm^2}{2a}}$
• Consider the pair B-C
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{B,C}=-\frac{8Gm^2}{2a}}$
• Consider the pair C-A
    ♦ The potential energy due to this pair is given by: $\mathbf\small{U_{C,A}=-\frac{4Gm^2}{2a}}$
2. Total gravitational potential energy of the system
= $\mathbf\small{U_{A,B}\;+U_{B,C}\;+U_{C,A}}$
= $\mathbf\small{-\frac{2Gm^2}{2a}\;+-\frac{8Gm^2}{a}\;+-\frac{4Gm^2}{2a}}$
= $\mathbf\small{-\frac{7Gm^2}{a}}$
3. To find the work done:
(i) Initially the particles are infinite distance apart. Then the energy of the system will be zero
(ii) Energy (U₁) of the system when the separation is a = $\mathbf\small{-\frac{14Gm^2}{a}}$
(iii) Energy (U₂) of the system when the separation is 2a = $\mathbf\small{-\frac{7Gm^2}{a}}$
(iv) Work done
= Change in energy
= Final energy - Initial energy
= U₂-U₁ = $\mathbf\small{-\frac{7Gm^2}{a}\;--\frac{14Gm^2}{a}}$
= $\mathbf\small{-\frac{7Gm^2}{a}\;+\frac{14Gm^2}{a}}$
= $\mathbf\small{\frac{7Gm^2}{a}}$

Solved example 8.31
An object is dropped from a height of 2R_E from the surface of the earth. Find the speed with which it will hit the surface of the earth. Neglect the effect of air resistance
Radius of the earth = R_E
Mass of the earth = M_E
Solution:
1. Let m be the mass of the object
• Potential energy of the object when it is at the surface of the earth = $\mathbf\small{-\frac{GM_Em}{R_E}}$
• Potential energy of the object when it is at a height of 2R_E = $\mathbf\small{-\frac{GM_Em}{R_E+2R_E}=-\frac{GM_Em}{3R_E}}$
2. So difference in potential energy = $\mathbf\small{-\frac{GM_Em}{3R_E}--\frac{GM_Em}{R_E}}$
= $\mathbf\small{\frac{GM_Em}{R_E}-\frac{GM_Em}{3R_E}=\frac{2GM_Em}{3R_E}}$
3. Since air resistance is neglected, we can write:
• The difference in potential energy will be converted into kinetic energy
4. Let v be the speed with which the object hits the ground
• Then it's kinetic energy at the instant of impact = $\mathbf\small{\frac{1}{2}mv^2}$
5. Equating the results in (2) and (4), we get: $\mathbf\small{\frac{1}{2}mv^2=\frac{2GM_Em}{3R_E}}$
$\mathbf\small{\Rightarrow v^2=\frac{4GM_E}{3R_E}}$
$\mathbf\small{\Rightarrow v=\sqrt{\frac{4GM_E}{3R_E}}=2\sqrt{\frac{GM_E}{3R_E}}}$

Solved example 8.32
In fig.8.46 below, two identical particles, each of mass m, are kept at rest at a distance d apart. They are allowed to move under the influence of their mutual gravitational force of attraction. What will be the speed of each when the distance between them is 0.5d

Fig.8.46

Solution:
1. Initially, the particles are at rest. So the only energy available initially is the potential energy which is equal to $\mathbf\small{-\frac{Gm^2}{d}}$
2. When the particles begin to move, there will be both potential energy and kinetic energy
3. Let v be the velocity of the particles at the instant when the distance between them is 0.5d
• Then the kinetic energy of the system at that instant = $\mathbf\small{\frac{1}{2}mv^2+\frac{1}{2}mv^2=mv^2}$ 
4. Potential energy of the system at that instant = $\mathbf\small{-\frac{Gm^2}{0.5d}=-\frac{2Gm^2}{d}}$  
5. So loss in potential energy
= Final potential energy - Initial potential energy
= $\mathbf\small{-\frac{2Gm^2}{d}--\frac{Gm^2}{d}}$
= $\mathbf\small{\frac{Gm^2}{d}-\frac{2Gm^2}{d}=-\frac{Gm^2}{d}}$
• The negative sign indicates that energy is lost 
6. 'Magnitude of this loss in potential energy' is equal to the 'kinetic energy of the system at that instant'
So equating the results in (3) and (5), we get: $\mathbf\small{\frac{Gm^2}{d}=mv^2}$
$\mathbf\small{\Rightarrow \frac{Gm}{d}=v^2}$
$\mathbf\small{\Rightarrow v=\sqrt{\frac{Gm}{d}}}$

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• In the next section we will see Escape velocity

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Chapter 8.11 - Graph of Acceleration due to Gravity

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:

Table 8.1

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

Fig.8.32

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 R_E, 2R_E, 3R_E 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

■ It is clear that, near the surface of the earth, even if heights are different, the body will be experiencing the same $\mathbf\small{|\vec{g}|}$

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• Next we will consider the variation of $\mathbf\small{|\vec{g}_d|}$ with depth. We will write it in steps:

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:

Table 8.3

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

Fig.8.34

4. We see that, the graph is a straight line. It is sloping downwards 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

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■ 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 m_A 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 = R_E - d
    ♦ R_E = Radius of the earth
    ♦ M_E = Mass of the earth
2. If we divide this force by m_A, we will get the acceleration experienced by that m_A
• 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

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:
$\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$

• 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

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

Fig.8.35

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 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 R_E 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 R_E
14. So we must use two equations for drawing the graph:
• When 0 ≤ r ≤ R_E, we must use $\mathbf\small{|\vec{g}_r|=\frac{G\,M_E}{R_E^3}r}$
• When r > R_E, 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:

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 M_E and radius R_E
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

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 < R_E, 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 R_E

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Let us write a summary of the above graphs:

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

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

■ 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 m_A, we get force

(ii) When force is divided by the second mass m_B, we get the field created by the first mass m_A

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• In the next section, we will see gravitational potential energy

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Chapter 8.9 - Solved examples on Gravity

In the previous section, we completed a discussion on the basics of gravitational force, gravity and acceleration due to gravity. We saw a solved example also. In this section, we will see a few more solved examples

Solved example 8.15
The acceleration due to gravitational force at the surface of the moon is 1.67 ms^-2. If the radius of the moon is 1.74 × 10⁶ m, calculate the mass of the moon
Solution:
1. For earth, we have: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
• We derived this formula by assuming that the earth is made up of concentric shells
• This assumption can be applied to any planet or heavenly body having a spherical shape
2. So for the moon, we can write: $\mathbf\small{|\vec{g}_M|=\frac{GM_M}{R_M^2}}$
Substituting the values, we get:
$\mathbf\small{1.67\;\rm{(m\;s^{-2})}=\frac{6.67\times 10^{-11}\rm{(N\;m^{2}\;kg^{-2})}M_M}{(1.74\times 10^{6})^2\rm{(m^2)}}}$
$\mathbf\small{\Rightarrow 1.67\;\rm{(m\;s^{-2})}=2.203\times 10^{-23}M_M\rm{(N\;kg^{-2})}}$
$\mathbf\small{\Rightarrow M_M=7.58 \times 10^{22}\frac{\rm{(m\;s^{-2})}}{\rm{N\,kg^{-2}}}}$
• We can put 'kg m s^-2' instead of 'N'. So we get: 
$\mathbf\small{M_M=7.58 \times 10^{22}\frac{\rm{(m\;s^{-2})}}{\rm{kg\,m\;s^{-2}\,kg^{-2}}}}$
$\mathbf\small{\Rightarrow M_M=7.58 \times 10^{22}\;\rm{kg}}$

Solved example 8.16
At a particular height h₁ above the surface of the earth, the value of acceleration due to gravity is 1/64 of $\mathbf\small{|\vec{g}|}$. What is the value of h₁?
Solution:
1. Given that: $\mathbf\small{|\vec{g}_{h_1}|=\frac{|\vec{g}|}{64}}$
2. We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
3. So we get: $\mathbf\small{\frac{|\vec{g}|}{64}=\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow 64=\left(1+\frac{h_1}{R_E}\right)^2}$
$\mathbf\small{\Rightarrow 8=\left(1+\frac{h_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{h_1}{R_E}=7}$
$\mathbf\small{\Rightarrow h_1=7R_E}$ = 7 × 6.378 × 10⁶ m = 44.65 × 10⁶ m

Solved example 8.17
Assume that, earth is made up of lead. What would be the value of $\mathbf\small{|\vec{g}|}$?
Density of lead = 11.34 × 10³ kg m^-3
Solution:
1. We have Eq.8.7: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
2. If RE is the radius of the earth and $\mathbf\small{\rho}$ the density of lead, we have: $\mathbf\small{M_E=\frac{4\pi R_E^3 \rho}{3}}$
• Substituting this in (1), we get: $\mathbf\small{|\vec{g}|=\frac{G}{R_E^2}\times \frac{4\pi R_E^3 \rho}{3}}$
$\mathbf\small{\Rightarrow |\vec{g}|=\frac{4G\pi R_E \rho}{3}}$
3. Substituting the values, we get:
$\mathbf\small{|\vec{g}|=\frac{4\times 6.67\times 10^{-11}\rm{(N\;m^{2}\;kg^{-2})}\times \pi \times 6.4 \times 10^6 \rm{(m)} \times 11.34\times 10^{3}\rm{(kg\;m^{-3})}}{3}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\rm{(N\;m^{2}\;kg^{-2})}\rm{(m)}\rm{(kg\;m^{-3})}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\rm{(kg\,m\,s^{-2}\;m^{2}\;kg^{-2})}\rm{(m)}\rm{(kg\;m^{-3})}}$
$\mathbf\small{\Rightarrow |\vec{g}|=20.27\;\rm{m\,s^{-2}}}$

Solved example 8.18
On the surface of the earth, the acceleration due to gravity is $\mathbf\small{|\vec{g}|}$. At what depth d₁ from the surface will the acceleration become $\mathbf\small{0.5|\vec{g}|}$?
Solution:
1. Given that: $\mathbf\small{|\vec{g}_{d_1}|=0.5|\vec{g}|}$
2. We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
3. So we get: $\mathbf\small{0.5|\vec{g}|=|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow 0.5=\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{d_1}{R_E}=0.5}$
$\mathbf\small{\Rightarrow d_1=0.5R_E}$

Solved example 8.19
A body weighs 100 N on the surface of the earth. It is taken down to a depth of 32 km below the surface. What will be it's new weight? Take radius of the earth as 6400 km
Solution:
1.  We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
2. So we get: $\mathbf\small{|\vec{g}_{32}|=|\vec{g}|\left(1- \frac{32000}{6400000}\right)}$
$\mathbf\small{\Rightarrow |\vec{g}_{32}|=|\vec{g}|\left(1- \frac{1}{200}\right)=0.995|\vec{g}|}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{100}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{100}{|\vec{g}|}\times 0.995|\vec{g}|=99.5\;N}$

Solved example 8.20
At a particular height h₁, the acceleration due to gravity is lesser by $\mathbf\small{|\Delta \vec{g}_{h_1}|}$. At a particular depth d₁, the acceleration due to gravity is lesser by $\mathbf\small{|\Delta \vec{g}_{d_1}|}$. If $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\Delta \vec{g}_{d_1}|}$, what is the relation between d₁ and h₁?
Solution:
1. We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
2. We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
3. Given that $\mathbf\small{|\vec{g}|-|\vec{g}_{h_1}|=|\Delta \vec{g}_{h_1}|}$
• So we get: $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\vec{g}|-\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}}$
4. Given that $\mathbf\small{|\vec{g}|-|\vec{g}_{d_1}|=|\Delta \vec{g}_{d_1}|}$
• So we get: $\mathbf\small{|\Delta \vec{g}_{d_1}|=|\vec{g}|-|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
5. Given that: $\mathbf\small{|\Delta \vec{g}_{h_1}|=|\Delta \vec{g}_{d_1}|}$
• So equating the results in (3) and (4), we get:
$\mathbf\small{|\vec{g}|-\frac{|\vec{g}|}{\left(1+\frac{h_1}{R_E}\right)^2}=|\vec{g}|-|\vec{g}|\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow 1-\frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=1-\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=\left(1- \frac{d_1}{R_E}\right)}$
6. Consider the left side of the above result. We have seen it's binomial expansion after discarding higher powers: $\mathbf\small{\frac{1}{\left(1+\frac{h_1}{R_E}\right)^2}=\left(1- \frac{2h_1}{R_E}\right)}$
• So the result in (5) becomes: $\mathbf\small{\left(1- \frac{2h_1}{R_E}\right)=\left(1- \frac{d_1}{R_E}\right)}$
$\mathbf\small{\Rightarrow \frac{2h_1}{R_E}=\frac{d_1}{R_E}}$
$\mathbf\small{\Rightarrow d_1 = 2h_1}$

Solved example 8.21
A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth?
Solution:
1.  We have Eq.8.9 that we derived in the previous section: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
2. So we get: $\mathbf\small{|\vec{g}_{(0.5R_E)}|=\frac{|\vec{g}|}{\left(1+\frac{0.5R_E}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow |\vec{g}_{(0.5R_E)}|=\frac{|\vec{g}|}{\left(1+0.5\right)^2}=\frac{|\vec{g}|}{1.5^2}=\frac{|\vec{g}|}{2.25}}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{63}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{63}{|\vec{g}|}\times \frac{|\vec{g}|}{2.25}=28\, \rm{N}}$

Solved example 2.22
Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the center of the earth, if it weighed 250 N on the surface?
Solution:
1.  We have Eq.8.10 that we derived in the previous section: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
2. So we get: $\mathbf\small{|\vec{g}_{(0.5R_E)}|=|\vec{g}|\left(1- \frac{0.5R_E}{R_E}\right)=0.5|\vec{g}|}$
3. Actual mass of the body = $\mathbf\small{\frac{m|\vec{g}|}{|\vec{g}|}=\frac{250}{|\vec{g}|}}$
4. So new weight = $\mathbf\small{\frac{250}{|\vec{g}|}\times 0.5|\vec{g}|=125\;\rm{N}}$

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• In the next section, we will see gravitational field and it's intensity

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Chapter 8.8 - Acceleration due to Gravity

In the previous section, we saw the force exerted by earth at three important positions. We will write a summary:

Eq.8.3:
Magnitude of the gravity on a point mass m_A at any distance r:
$\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$

Eq.8.4:
Magnitude of the gravity on a point mass m_A situated on the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
Eq.8.5:
Magnitude of the gravity on a point mass m_A situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
OR
Eq.8.6:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$

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• Our next aim is to find the acceleration experienced by the point mass at the three positions

• For that we apply Newton's second law
That is., When we divide force by mass, we get acceleration
• In vector form, the magnitude of 'acceleration due to gravity' is expressed as $\mathbf\small{|\vec{g}|}$
• We will now find $\mathbf\small{|\vec{g}|}$ at the three positions:

Position 1: On the surface of the earth
1. Eq.8.4, gives the force. We must divide this force by mass. Thus we get:
Acceleration (due to gravity) experienced by a point mass m_A situated on the surface of the earth
= $\mathbf\small{|\vec{g}|= \frac{GM_E\,m_A}{R_E^2}\times \frac{1}{m_A}=\frac{GM_E}{R_E^2}}$
2. Thus we can write:
Eq.8.7: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
3. $\mathbf\small{|\vec{g}|}$ can be determined independently by means of experiments. In such a situation, M_E is the only unknown quantity in Eq.8.7. So we can easily calculate M_E
• Thus a popular statement came into existence: Cavendish weighed the Earth

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■ We wrote: Acceleration (due to gravity) experienced by a point mass m_A situated on the surface of the earth

(i) A mass on the surface of the earth may be stationary. But still, gravity is acting on it
(ii) It is stationary because, the reaction from the ground cancels the force of gravity
(iii) If that ground is removed, the mass will begin to travel towards the center of the earth with an acceleration of $\mathbf\small{|\vec{g}|}$
(iv) Similarly, if a mass is dropped from a moderate height above the surface of the earth, it will fall with an acceleration of $\mathbf\small{|\vec{g}|}$
• The above steps are helpful to get the idea about 'acceleration on surface of earth' 

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Position 2: At a height h above the surface of the earth
1. Eq.8.3 gives the gravity at any point above the surface of the earth
• If the point mass is at a height h, it's distance r from the center of the earth O will be given by:
r = (R_E+h)
2. We have Eq.8.3: $\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$ 
3. Now we divide this force by mass. We will get the acceleration experienced by that body at height h
• We will denote this acceleration as $\mathbf\small{|\vec{g}_h|}$ 
• So we can write: $\mathbf\small{|\vec{g}_h|=\frac{GM_E\,m_A}{(R_E+h)^2}\times \frac{1}{m_A}}$
4. Thus we get:
Eq.8.8: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{(R_E+h)^2}}$
5. Compare Eqs.8.7 and 8.8
• The numerators are the same
• But in Eq.8.8, the denominator is larger. So we can write:
At a height h, the acceleration due to gravity is lesser than that on the surface of the earth
6. Now let us write a comparison between h and R_E
(i) The radius (R_E) of the earth is 6378.1 km
(ii) The fig. below shows the various layers of the atmosphere. The fig. is obtained from wikimedia commons. The link is given below:
https://commons.wikimedia.org/wiki/File:Atmosphere_layers.jpg

(iii) We see that the maximum height of the atmosphere is 90 km
• This is far less than the radius R_E
(iv) So in normal cases, we can write: $\mathbf\small{h<<R_E}$
• In such cases, the ratio $\mathbf\small{\frac{h}{R_E}}$ will be much less than 1
    ♦ An example:
    ♦ Let us consider a normal case. The case of a weather balloon at 40 km above the earth's surface
    ♦ The ratio $\mathbf\small{\frac{h}{R_E}=\frac{40}{6378.1}=0.0063}$ 
• So we can write: For normal cases, $\mathbf\small{\frac{h}{R_E}<<1}$
7. Consider the denominator in Eq.8.8: $\mathbf\small{(R_E+h)^2}$
• This can be written as: $\mathbf\small{\left[R_E \left(1+\frac{h}{R_E}\right)\right]^2}$
= $\mathbf\small{R_E^2 \left(1+\frac{h}{R_E}\right)^2}$
• So Eq.8.8 becomes: $\mathbf\small{|\vec{g}_h|=\frac{GM_E}{R_E^2 \left(1+\frac{h}{R_E}\right)^2}=\left[\frac{GM_E}{R_E^2}\times \frac{1}{\left(1+\frac{h}{R_E}\right)^2} \right]}$
8. But from Eq.8.7, $\mathbf\small{\frac{GM_E}{R_E^2}}$ is $\mathbf\small{|\vec{g}|}$ on the surface of the earth
• So the result in (7) becomes: 
Eq.8.9: $\mathbf\small{|\vec{g}_h|=\frac{|\vec{g}|}{\left(1+\frac{h}{R_E}\right)^2}}$
$\mathbf\small{\Rightarrow |\vec{g}_h|=|\vec{g}|\left(1+\frac{h}{R_E}\right)^{-2}}$
9. The exponent is '-2'. So we have to use binomial expansion
• We saw that, $\mathbf\small{\frac{h}{R_E}}$ will be small. So $\mathbf\small{\frac{h^2}{R_E^2}}$ will be very small
• For example, 0.001 is a small quantity. It's square (0.001² = 0.000001) is a very small quantity
• So we can discard $\mathbf\small{\frac{h^2}{R_E^2}}$ and higher powers
10. So the result in (8) becomes:
Eq.8.10: $\mathbf\small{|\vec{g}_h|=|\vec{g}|\left(1-\frac{2h}{R_E}\right)}$
11. For small values of h, the value of $\mathbf\small{\frac{2h}{R_E}}$ will be less than 1
• So the value of $\mathbf\small{\left(1-\frac{2h}{R_E}\right)}$ will also be less than 1 but greater than zero
• So the value of $\mathbf\small{|\vec{g}|\left(1-\frac{2h}{R_E}\right)}$ will be less than $\mathbf\small{|\vec{g}|}$ 
• That means: the value of $\mathbf\small{|\vec{g}_h|}$ will be less than the value of $\mathbf\small{|\vec{g}|}$
■ We can also write:
For small heights h, the acceleration due to gravity decreases by a factor $\mathbf\small{\left(1-\frac{2h}{R_E}\right)}$

Position 3: At a depth d below the surface of the earth
1. Eq.8.6 gives the gravity at any point below the surface of the earth
• If the point mass is at a depth d, it's distance r from the center of the earth O will be given by:
r = (R_E-d)
2. We have Eq.8.6: $\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
3. Now we divide this force by mass. We will get the acceleration experienced by that body at depth d
• We will denote this acceleration as $\mathbf\small{|\vec{g}_d|}$ 
• So we can write: $\mathbf\small{|\vec{g}_d|=\frac{G\,M_E\,m_A}{R_E^3}(R_E-d)\times \frac{1}{m_A}}$
$\mathbf\small{\Rightarrow |\vec{g}_d|=\frac{G\,M_E}{R_E^3}(R_E-d)}$
4. The above result in (3) can be written as: $\mathbf\small{|\vec{g}_d|=\frac{G\,M_E}{R_E^2}\times\frac{(R_E-d)}{R_E}}$
• But from Eq.8.7, $\mathbf\small{\frac{GM_E}{R_E^2}}$ is $\mathbf\small{|\vec{g}|}$ on the surface of the earth
• So the result in (3) becomes: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\frac{(R_E-d)}{R_E}=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
5. So we get:
Eq.8.11: $\mathbf\small{|\vec{g}_d|=|\vec{g}|\left(1- \frac{d}{R_E}\right)}$
6. d will be always less than R_E. So the value of $\mathbf\small{\frac{d}{R_E}}$ will be less than 1
• So the value of $\mathbf\small{\left(1-\frac{d}{R_E}\right)}$ will also be less than 1 but greater than zero
• So the value of $\mathbf\small{|\vec{g}|\left(1-\frac{d}{R_E}\right)}$ will be less than $\mathbf\small{|\vec{g}|}$ 
• That means: the value of $\mathbf\small{|\vec{g}_d|}$ will be less than the value of $\mathbf\small{|\vec{g}|}$
■ We can also write:
At a depth d, the acceleration due to gravity decreases by a factor $\mathbf\small{\left(1-\frac{d}{R_E}\right)}$

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■ Let us compare the two results:

(i) Result in position 2 (11)
(ii) Result in Position 3 (6)
• We see that, in both cases, there will be a decrease in the acceleration. That is:
    ♦ $\mathbf\small{|\vec{g}_d|}$ is less than the $\mathbf\small{|\vec{g}|}$ at the surface
    ♦ $\mathbf\small{|\vec{g}_h|}$ is also less than the $\mathbf\small{|\vec{g}|}$ at the surface
■ We can write:
• The 'acceleration due to gravity of the earth' is maximum on the surface of the earth
• The acceleration decreases when we go higher up above the surface
• The acceleration decreases when we go lower down below the surface

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Now we will see a solved example

Solved example 8.14
The mass of a planet P is 4 times the mass of the earth. It's radius is also 4 times the radius of the earth. What is the relation between the following two items:
(i) $\mathbf\small{|\vec{g}|}$ (ii) Acceleration (due to gravitational force) experienced by a point mass on the surface of P
Solution:
1. $\mathbf\small{|\vec{g}|}$ is simply the acceleration (due to gravitational force) experienced by a point mass on the surface of the earth
• The formula is: $\mathbf\small{|\vec{g}|=\frac{GM_E}{R_E^2}}$
2. We derived this formula by assuming that the earth is made up of concentric shells
• This assumption can be applied to any planet having a spherical shape
• So for the given planet, we can write: $\mathbf\small{|\vec{g}_P|=\frac{GM_P}{R_P^2}}$
3. Taking ratios, we get: $\mathbf\small{\frac{|\vec{g}|}{|\vec{g}_P|}=\frac{GM_E}{R_E^2}\times\frac{R_P^2}{GM_P}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=\frac{M_E}{M_P}\times\frac{R_P^2}{R_E^2}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=\frac{M_E}{4M_E}\times\frac{(4R_E)^2}{R_E^2}}$
$\mathbf\small{\Rightarrow \frac{|\vec{g}|}{|\vec{g}_P|}=4}$
$\mathbf\small{\Rightarrow |\vec{g}_P|=\frac{|\vec{g}|}{4}}$
4. So we can write:
Acceleration (due to gravitational force) experienced by a point mass on the surface of P is one fourth of $\mathbf\small{|\vec{g}|}$

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• We have seen the basics about gravitational force, gravity and acceleration due to gravity
• In the next section, we will see a few more solved examples

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