In the previous section, we completed a discussion on speed and time period of earth satellites
• In this section we will see energy of an orbiting satellite
1. A satellite will always be at a constant height h from the surface of the earth. So it will be having a constant potential energy
• We know that, this potential energy is given by $\mathbf\small{U=-\frac{G\,M_E\,m}{(R_E+h)}}$
2. But the satellite is in constant motion also. It has a constant speed V
• So it will have a kinetic energy of $\mathbf\small{\frac{1}{2}m\,V^2}$
3. In the previous section we saw Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• So the kinetic energy will be given by: $\mathbf\small{K=\frac{1}{2}m\,\left(\sqrt{\frac{G\,M_E}{(R_E+h)}}\right)^2}$
• Thus we get Eq.8.28: $\mathbf\small{K=\frac{1}{2}\frac{G\,M_E\,m}{(R_E+h)}}$
4. If we add the results in (2) and (3), we will get the total energy E
• That is., E = U + K
• So we can write Eq.8.29: $\mathbf\small{E=-\frac{G\,M_E\,m}{(R_E+h)}+\frac{1}{2}\frac{G\,M_E\,m}{(R_E+h)}}$
5. We see that, on the right side there are some common items
• If we put $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$, we will get:
Eq.8.30: $\mathbf\small{E=-X+\frac{1}{2}X=-\frac{1}{2}X=-\frac{G\,M_E\,m}{2(R_E+h)}}$
6. We see that $\mathbf\small{U=-X}$ and $\mathbf\small{K=\frac{1}{2}X}$
• Taking ratios, we get: $\mathbf\small{\frac{U}{K}=\frac{-X}{\frac{1}{2}X}=-2}$
• Thus we get Eq.8.31: $\mathbf\small{U=-2K}$ or $\mathbf\small{K=-\frac{U}{2}}$
7. We note another interesting information:
• From Eq.8.30, we have: $\mathbf\small{E=-\frac{G\,M_E\,m}{2(R_E+h)}}$
• The items on the right side are G, ME, m, RE and h. All items are positive quantities
• A negative sign is already present in Eq.8.30
• So from Eq.8.30, we will never get a positive value for E
• That means, the total mechanical energy of an earth satellite will always be negative
(Mechanical energy = Kinetic energy + Potential energy)
• This is indeed expected. The reason can be written in 5 steps:
(i) We know that, as height of a satellite increases, it's energy increases
(ii) We also know that, when the height approaches infinity, the energy must approach zero
(iii) When the height is infinity, the energy must become zero (The height h is in the denominator)
(iv) All this is possible only if the energy is negative
(v) If the energy is zero or positive, it would mean that, the satellite is at infinity. It is no longer bound to earth. Such a satellite will escape away from earth. It will not rotate around the earth
Now we will see some solved examples
Solved example 8.49
Two satellites A and B rotates in two different orbits around the earth. The masses of A and B are 3m and m respectively. The radii of the orbits are r and 4r respectively. If E is the mechanical energy of A, calculate the mechanical energy of B
Solution:
1. We have Eq.8.30: $\mathbf\small{E=-X+\frac{1}{2}X=-\frac{1}{2}X=-\frac{G\,M_E\,m}{2(R_E+h)}}$
Substituting the values, we get:
$\mathbf\small{E_A=-\frac{G\,M_E\,(3m)}{2(r)}}$
$\mathbf\small{E_B=\frac{G\,M_E\,(m)}{2(4r)}}$
2. Taking ratios, we get: $\mathbf\small{\frac{E_A}{E_B}=-\frac{G\,M_E\,(3m)}{2(r)}\times \frac{2(4r)}{G\,M_E\,(m)}=12}$
$\mathbf\small{\Rightarrow E_B=\frac{E_A}{12}=\frac{E}{12}}$
Solved example 8.50
A satellite moving around the earth has a total mechanical energy of E. What is it's kinetic energy ?
Solution:
1. From Eq.8.31, we have: U = -2K
2. So E = (U + K) = (-2K + K) = -K
• Thus we get: Kinetic energy (K) of the satellite = -E
• Note that, E will be a negative quantity. So -E will be positive
Solved example 8.51
Two identical satellites are orbiting at distances R and 7R from the surface of the earth. R is the radius of the earth. What is the ratio of their kinetic energies? What is the ratio of their potential energies? What is the ratio of their total energies?
Solution:
Given that the satellites are identical. So we can write: mA = mB = m
1. First we calculate X using the equation: $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$
Substituting the values, we get:
$\mathbf\small{X_A=\frac{G\,M_E\,m}{(R+R)}=\frac{G\,M_E\,m}{2R}}$
$\mathbf\small{X_B=\frac{G\,M_E\,m}{(R+7R)}=\frac{G\,M_E\,m}{8R}}$
2. Thus we get:
$\mathbf\small{U_A=-X_A=-\frac{G\,M_E\,m}{2R}}$
$\mathbf\small{U_B=-X_B=-\frac{G\,M_E\,m}{8R}}$
⇒ UA:UB = 4:1
3. Similarly:
$\mathbf\small{K_A=\frac{1}{2}X_A=\frac{G\,M_E\,m}{4R}}$
$\mathbf\small{K_B=\frac{1}{2}X_B=\frac{G\,M_E\,m}{16R}}$
⇒ KA:KB = 16:4 = 4:1
4. Similarly:
$\mathbf\small{E_A=-\frac{1}{2}X_A=-\frac{G\,M_E\,m}{4R}}$
$\mathbf\small{E_B=-\frac{1}{2}X_B=-\frac{G\,M_E\,m}{16R}}$
⇒ EA:EB = 16:4 = 4:1
Solved example 8.52
What is the energy required to launch a m kg satellite from the earth's surface to an orbit of radius 8R
Solution:
1. When the satellite is on the surface of the earth, it has no kinetic energy
• It's energy is completely potential. It is equal to $\mathbf\small{-\frac{G\,M_E\,m}{R}}$
2. When the satellite is in the orbit of radius 8R, it has both kinetic and potential energies
• The total energy is given by:
Eq.8.30: $\mathbf\small{E=-\frac{G\,M_E\,m}{2(8R)}=-\frac{G\,M_E\,m}{16R}}$
3. Difference in energies = $\mathbf\small{-\frac{G\,M_E\,m}{16R}--\frac{G\,M_E\,m}{R}}$
= $\mathbf\small{\frac{G\,M_E\,m}{R}-\frac{G\,M_E\,m}{16R}=\frac{15G\,M_E\,m}{16R}}$
Solved example 8.53
A 400 kg satellite is in a circular orbit of radius 2RE about the earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in kinetic and potential energies?
Solution:
1. Let $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$
• Then we get:
♦ Initial potential energy = $\mathbf\small{U_i=-X_i=-\frac{G\,M_E\,(400)}{2R_E}}$
♦ Initial kinetic energy = $\mathbf\small{K_i=\frac{1}{2}X_i=\frac{G\,M_E\,(400)}{4R_E}=\frac{G\,M_E\,(200)}{2R_E}}$
2. So total initial energy = $\mathbf\small{U_i+K_i=-\frac{G\,M_E\,(100)}{R_E}}$
3. Also we get:
♦ Final potential energy = $\mathbf\small{U_f=-X_f=-\frac{G\,M_E\,(400)}{4R_E}}$
♦ Final kinetic energy = $\mathbf\small{K_f=\frac{1}{2}X_f=\frac{G\,M_E\,(400)}{8R_E}=\frac{G\,M_E\,(200)}{4R_E}}$
4. So total final energy = $\mathbf\small{U_f+K_f=-\frac{G\,M_E\,(50)}{R_E}}$
5. So energy required = Total final energy - Total initial energy
= $\mathbf\small{-\frac{G\,M_E\,(50)}{R_E}--\frac{G\,M_E\,(100)}{R_E}=\frac{G\,M_E\,(50)}{R_E}}$
• Substituting the values, we get:
Energy required = $\mathbf\small{\frac{G\,M_E\,(50)}{R_E}=\frac{G\,M_E\,(50)R_E}{R_E^2}=g(50)R_E=(9.81)(50)(6.37\times 10^6)}$ = 3.13 × 109 J
6. Change in kinetic energy = $\mathbf\small{K_f-K_i=\frac{G\,M_E\,(200)}{4R_E}-\frac{G\,M_E\,(200)}{2R_E}=-\frac{G\,M_E\,(50)}{R_E}}$
= $\mathbf\small{-\frac{G\,M_E\,(50)R_E}{R_E^2}=-g(50)R_E=-(9.81)(50)(6.37\times 10^6)}$ = -3.13 × 109 J
7. Change in potential energy = $\mathbf\small{U_f-U_i=-\frac{G\,M_E\,(400)}{4R_E}--\frac{G\,M_E\,(400)}{2R_E}=\frac{G\,M_E\,(100)}{R_E}}$
= $\mathbf\small{\frac{G\,M_E\,(100)R_E}{R_E^2}=g(100)R_E=(9.81)(100)(6.37\times 10^6)}$ = -6.25 × 109 J
An interesting result:
(i) We have: $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$
• $\mathbf\small{E_i=-X_i+\frac{X_i}{2}}$
• $\mathbf\small{E_f=-X_f+\frac{X_f}{2}}$
(ii) $\mathbf\small{\Delta E=E_f-E_i=(-X_f+\frac{X_f}{2})-(-X_i+\frac{X_i}{2})}$
$\mathbf\small{\Rightarrow \Delta E=(X_i-X_f)+\frac{(X_f-X_i)}{2}}$
(iii) Note the two terms on the right side. We see that:
• Absolute value of the first term
Is equal to
• Twice the absolute value of the second term
(iv) The first term is the difference of Ki and Kf
• The second term is the difference of Ui and Uf
(v) So we can write: |ΔK| = 2|ΔU|
Solved example 8.54
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite = 200 kg; mass of earth = 6 × 1024 kg; radius of earth = 6.4 × 106 m; G = 6.67 × 10-11 N m2 kg-2
Solution:
1. Let X = $\mathbf\small{\frac{G\,M_E\,m}{(R_E+h)}}$
• Then we get:
Initial potential energy = $\mathbf\small{U_i=-X_i=-\frac{G\,M_E\,(200)}{R_E+400000}}$
= $\mathbf\small{-\frac{(6.67\times 10^{-11})\,(6\times 10^{24})\,(200)}{(6.4\times 10^6)+400000}}$
= -11.77 × 109 J
• Initial kinetic energy = $\mathbf\small{K_i=\frac{1}{2}X_i}$ = (11.77 × 109 ➗ 2) = 5.9 × 109 J
2. So total initial energy
= $\mathbf\small{U_i+K_i}$ = (-11.77 × 109 + 5.9 × 109) = -5.9 × 109 J
3. The final potential energy will be zero because, when the satellite is out of the influence of the earth, there is no gravitational force. So there is no gravitational potential energy
• The final kinetic energy will also be zero. This is because, we want the satellite to 'just escape' from the influence of the earth. We do not want it to move with any velocity after escaping. This way, we will get the minimum required energy
• So the total final energy = 0
4. So the energy required = (0 - -5.9 × 109) = 5.9 × 109 J
5. The energy obtained in (2) is called binding energy of the satellite. The satellite remains bound to the earth because of this energy
• In this section we will see energy of an orbiting satellite
1. A satellite will always be at a constant height h from the surface of the earth. So it will be having a constant potential energy
• We know that, this potential energy is given by $\mathbf\small{U=-\frac{G\,M_E\,m}{(R_E+h)}}$
2. But the satellite is in constant motion also. It has a constant speed V
• So it will have a kinetic energy of $\mathbf\small{\frac{1}{2}m\,V^2}$
3. In the previous section we saw Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• So the kinetic energy will be given by: $\mathbf\small{K=\frac{1}{2}m\,\left(\sqrt{\frac{G\,M_E}{(R_E+h)}}\right)^2}$
• Thus we get Eq.8.28: $\mathbf\small{K=\frac{1}{2}\frac{G\,M_E\,m}{(R_E+h)}}$
4. If we add the results in (2) and (3), we will get the total energy E
• That is., E = U + K
• So we can write Eq.8.29: $\mathbf\small{E=-\frac{G\,M_E\,m}{(R_E+h)}+\frac{1}{2}\frac{G\,M_E\,m}{(R_E+h)}}$
5. We see that, on the right side there are some common items
• If we put $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$, we will get:
Eq.8.30: $\mathbf\small{E=-X+\frac{1}{2}X=-\frac{1}{2}X=-\frac{G\,M_E\,m}{2(R_E+h)}}$
6. We see that $\mathbf\small{U=-X}$ and $\mathbf\small{K=\frac{1}{2}X}$
• Taking ratios, we get: $\mathbf\small{\frac{U}{K}=\frac{-X}{\frac{1}{2}X}=-2}$
• Thus we get Eq.8.31: $\mathbf\small{U=-2K}$ or $\mathbf\small{K=-\frac{U}{2}}$
7. We note another interesting information:
• From Eq.8.30, we have: $\mathbf\small{E=-\frac{G\,M_E\,m}{2(R_E+h)}}$
• The items on the right side are G, ME, m, RE and h. All items are positive quantities
• A negative sign is already present in Eq.8.30
• So from Eq.8.30, we will never get a positive value for E
• That means, the total mechanical energy of an earth satellite will always be negative
(Mechanical energy = Kinetic energy + Potential energy)
• This is indeed expected. The reason can be written in 5 steps:
(i) We know that, as height of a satellite increases, it's energy increases
(ii) We also know that, when the height approaches infinity, the energy must approach zero
(iii) When the height is infinity, the energy must become zero (The height h is in the denominator)
(iv) All this is possible only if the energy is negative
(v) If the energy is zero or positive, it would mean that, the satellite is at infinity. It is no longer bound to earth. Such a satellite will escape away from earth. It will not rotate around the earth
Now we will see some solved examples
Solved example 8.49
Two satellites A and B rotates in two different orbits around the earth. The masses of A and B are 3m and m respectively. The radii of the orbits are r and 4r respectively. If E is the mechanical energy of A, calculate the mechanical energy of B
Solution:
1. We have Eq.8.30: $\mathbf\small{E=-X+\frac{1}{2}X=-\frac{1}{2}X=-\frac{G\,M_E\,m}{2(R_E+h)}}$
Substituting the values, we get:
$\mathbf\small{E_A=-\frac{G\,M_E\,(3m)}{2(r)}}$
$\mathbf\small{E_B=\frac{G\,M_E\,(m)}{2(4r)}}$
2. Taking ratios, we get: $\mathbf\small{\frac{E_A}{E_B}=-\frac{G\,M_E\,(3m)}{2(r)}\times \frac{2(4r)}{G\,M_E\,(m)}=12}$
$\mathbf\small{\Rightarrow E_B=\frac{E_A}{12}=\frac{E}{12}}$
Solved example 8.50
A satellite moving around the earth has a total mechanical energy of E. What is it's kinetic energy ?
Solution:
1. From Eq.8.31, we have: U = -2K
2. So E = (U + K) = (-2K + K) = -K
• Thus we get: Kinetic energy (K) of the satellite = -E
• Note that, E will be a negative quantity. So -E will be positive
Solved example 8.51
Two identical satellites are orbiting at distances R and 7R from the surface of the earth. R is the radius of the earth. What is the ratio of their kinetic energies? What is the ratio of their potential energies? What is the ratio of their total energies?
Solution:
Given that the satellites are identical. So we can write: mA = mB = m
1. First we calculate X using the equation: $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$
Substituting the values, we get:
$\mathbf\small{X_A=\frac{G\,M_E\,m}{(R+R)}=\frac{G\,M_E\,m}{2R}}$
$\mathbf\small{X_B=\frac{G\,M_E\,m}{(R+7R)}=\frac{G\,M_E\,m}{8R}}$
2. Thus we get:
$\mathbf\small{U_A=-X_A=-\frac{G\,M_E\,m}{2R}}$
$\mathbf\small{U_B=-X_B=-\frac{G\,M_E\,m}{8R}}$
⇒ UA:UB = 4:1
3. Similarly:
$\mathbf\small{K_A=\frac{1}{2}X_A=\frac{G\,M_E\,m}{4R}}$
$\mathbf\small{K_B=\frac{1}{2}X_B=\frac{G\,M_E\,m}{16R}}$
⇒ KA:KB = 16:4 = 4:1
4. Similarly:
$\mathbf\small{E_A=-\frac{1}{2}X_A=-\frac{G\,M_E\,m}{4R}}$
$\mathbf\small{E_B=-\frac{1}{2}X_B=-\frac{G\,M_E\,m}{16R}}$
⇒ EA:EB = 16:4 = 4:1
Solved example 8.52
What is the energy required to launch a m kg satellite from the earth's surface to an orbit of radius 8R
Solution:
1. When the satellite is on the surface of the earth, it has no kinetic energy
• It's energy is completely potential. It is equal to $\mathbf\small{-\frac{G\,M_E\,m}{R}}$
2. When the satellite is in the orbit of radius 8R, it has both kinetic and potential energies
• The total energy is given by:
Eq.8.30: $\mathbf\small{E=-\frac{G\,M_E\,m}{2(8R)}=-\frac{G\,M_E\,m}{16R}}$
3. Difference in energies = $\mathbf\small{-\frac{G\,M_E\,m}{16R}--\frac{G\,M_E\,m}{R}}$
= $\mathbf\small{\frac{G\,M_E\,m}{R}-\frac{G\,M_E\,m}{16R}=\frac{15G\,M_E\,m}{16R}}$
Solved example 8.53
A 400 kg satellite is in a circular orbit of radius 2RE about the earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in kinetic and potential energies?
Solution:
1. Let $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$
• Then we get:
♦ Initial potential energy = $\mathbf\small{U_i=-X_i=-\frac{G\,M_E\,(400)}{2R_E}}$
♦ Initial kinetic energy = $\mathbf\small{K_i=\frac{1}{2}X_i=\frac{G\,M_E\,(400)}{4R_E}=\frac{G\,M_E\,(200)}{2R_E}}$
2. So total initial energy = $\mathbf\small{U_i+K_i=-\frac{G\,M_E\,(100)}{R_E}}$
3. Also we get:
♦ Final potential energy = $\mathbf\small{U_f=-X_f=-\frac{G\,M_E\,(400)}{4R_E}}$
♦ Final kinetic energy = $\mathbf\small{K_f=\frac{1}{2}X_f=\frac{G\,M_E\,(400)}{8R_E}=\frac{G\,M_E\,(200)}{4R_E}}$
4. So total final energy = $\mathbf\small{U_f+K_f=-\frac{G\,M_E\,(50)}{R_E}}$
5. So energy required = Total final energy - Total initial energy
= $\mathbf\small{-\frac{G\,M_E\,(50)}{R_E}--\frac{G\,M_E\,(100)}{R_E}=\frac{G\,M_E\,(50)}{R_E}}$
• Substituting the values, we get:
Energy required = $\mathbf\small{\frac{G\,M_E\,(50)}{R_E}=\frac{G\,M_E\,(50)R_E}{R_E^2}=g(50)R_E=(9.81)(50)(6.37\times 10^6)}$ = 3.13 × 109 J
6. Change in kinetic energy = $\mathbf\small{K_f-K_i=\frac{G\,M_E\,(200)}{4R_E}-\frac{G\,M_E\,(200)}{2R_E}=-\frac{G\,M_E\,(50)}{R_E}}$
= $\mathbf\small{-\frac{G\,M_E\,(50)R_E}{R_E^2}=-g(50)R_E=-(9.81)(50)(6.37\times 10^6)}$ = -3.13 × 109 J
7. Change in potential energy = $\mathbf\small{U_f-U_i=-\frac{G\,M_E\,(400)}{4R_E}--\frac{G\,M_E\,(400)}{2R_E}=\frac{G\,M_E\,(100)}{R_E}}$
= $\mathbf\small{\frac{G\,M_E\,(100)R_E}{R_E^2}=g(100)R_E=(9.81)(100)(6.37\times 10^6)}$ = -6.25 × 109 J
An interesting result:
(i) We have: $\mathbf\small{X=\frac{G\,M_E\,m}{(R_E+h)}}$
• $\mathbf\small{E_i=-X_i+\frac{X_i}{2}}$
• $\mathbf\small{E_f=-X_f+\frac{X_f}{2}}$
(ii) $\mathbf\small{\Delta E=E_f-E_i=(-X_f+\frac{X_f}{2})-(-X_i+\frac{X_i}{2})}$
$\mathbf\small{\Rightarrow \Delta E=(X_i-X_f)+\frac{(X_f-X_i)}{2}}$
(iii) Note the two terms on the right side. We see that:
• Absolute value of the first term
Is equal to
• Twice the absolute value of the second term
(iv) The first term is the difference of Ki and Kf
• The second term is the difference of Ui and Uf
(v) So we can write: |ΔK| = 2|ΔU|
Solved example 8.54
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite = 200 kg; mass of earth = 6 × 1024 kg; radius of earth = 6.4 × 106 m; G = 6.67 × 10-11 N m2 kg-2
Solution:
1. Let X = $\mathbf\small{\frac{G\,M_E\,m}{(R_E+h)}}$
• Then we get:
Initial potential energy = $\mathbf\small{U_i=-X_i=-\frac{G\,M_E\,(200)}{R_E+400000}}$
= $\mathbf\small{-\frac{(6.67\times 10^{-11})\,(6\times 10^{24})\,(200)}{(6.4\times 10^6)+400000}}$
= -11.77 × 109 J
• Initial kinetic energy = $\mathbf\small{K_i=\frac{1}{2}X_i}$ = (11.77 × 109 ➗ 2) = 5.9 × 109 J
2. So total initial energy
= $\mathbf\small{U_i+K_i}$ = (-11.77 × 109 + 5.9 × 109) = -5.9 × 109 J
3. The final potential energy will be zero because, when the satellite is out of the influence of the earth, there is no gravitational force. So there is no gravitational potential energy
• The final kinetic energy will also be zero. This is because, we want the satellite to 'just escape' from the influence of the earth. We do not want it to move with any velocity after escaping. This way, we will get the minimum required energy
• So the total final energy = 0
4. So the energy required = (0 - -5.9 × 109) = 5.9 × 109 J
5. The energy obtained in (2) is called binding energy of the satellite. The satellite remains bound to the earth because of this energy
• In the next section we will see Geostationary satellites
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