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 RE
• 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 RE+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 = $\mathbf\small{m|\vec{g}|h}$
• 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
• Their masses are mA and mB respectively
• The center to center distance is r
2. So we have a system consisting of two masses mA and mB
• 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 mA and mB 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 mB = 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 mA = 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 mA
• 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
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
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 2RE 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 = RE
Mass of the earth = ME
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 2RE = $\mathbf\small{-\frac{GM_Em}{R_E+2R_E}=-\frac{GM_Em}{3R_E}}$
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
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: $\mathbf\small{\frac{Gm^2}{d}=mv^2}$
$\mathbf\small{\Rightarrow \frac{Gm}{d}=v^2}$
$\mathbf\small{\Rightarrow v=\sqrt{\frac{Gm}{d}}}$
• 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 RE
• 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 RE+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 = $\mathbf\small{m|\vec{g}|h}$
• So we have completed a discussion on gravitational potential energy
1. In fig.8.43 below, two bodies A and B are placed at P1 and P2
 |
| Fig.8.43 |
• The center to center distance is r
2. So we have a system consisting of two masses mA and mB
• 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 mA and mB 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 mB = 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 mA = 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 mA
• 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 |
• 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 |
• 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 2RE 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 = RE
Mass of the earth = ME
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 2RE = $\mathbf\small{-\frac{GM_Em}{R_E+2R_E}=-\frac{GM_Em}{3R_E}}$
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 |
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: $\mathbf\small{\frac{Gm^2}{d}=mv^2}$
$\mathbf\small{\Rightarrow \frac{Gm}{d}=v^2}$
$\mathbf\small{\Rightarrow v=\sqrt{\frac{Gm}{d}}}$
• In the next section we will see Escape velocity
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