Chapter 8.17 - Energy of an Orbiting Satellite

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, M_E, m, R_E 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: m_A = m_B = 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:
$\mathbf\small{U_A=-X_A=-\frac{G\,M_E\,m}{2R}}$
$\mathbf\small{U_B=-X_B=-\frac{G\,M_E\,m}{8R}}$
⇒ U_A:U_B = 4:1
3. Similarly:
$\mathbf\small{K_A=\frac{1}{2}X_A=\frac{G\,M_E\,m}{4R}}$
$\mathbf\small{K_B=\frac{1}{2}X_B=\frac{G\,M_E\,m}{16R}}$
⇒ K_A:K_B = 16:4 = 4:1
4. Similarly:
$\mathbf\small{E_A=-\frac{1}{2}X_A=-\frac{G\,M_E\,m}{4R}}$
$\mathbf\small{E_B=-\frac{1}{2}X_B=-\frac{G\,M_E\,m}{16R}}$
⇒ E_A:E_B = 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 2R_E about the earth. How much energy is required to transfer it to a circular orbit of radius 4R_E? 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 K_i and K_f
• The second term is the difference of U_i and U_f
(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

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• In the next section we will see Geostationary satellites

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Chapter 8.16 - Earth Satellites

In the previous section, we completed a discussion on escape velocity
• In this section we will see Earth satellites

First we will find the speed with which satellites move around the earth
1. Consider a satellite moving around the earth in a circular orbit
• Let it's mass be m
• Let it's speed be V
• Let it be at a height h above the surface of the earth
    ♦ So the radius r of the circular orbit will be equal to (RE + h)
2. Any object moving in a circular path requires centripetal force 
• We know that, for our present case, the centripetal force will be equal to $\mathbf\small{\frac{mV^2}{R_E+h}}$
3. This centripetal force is provided by the gravitational force between the earth and the satellite
• We know that, the gravitational force will be equal to $\mathbf\small{\frac{G\,M_E\,m}{(R_E+h)^2}}$
4. Equating the results in (2) and (3), we get: $\mathbf\small{\frac{mV^2}{R_E+h}=\frac{G\,M_E\,m}{(R_E+h)^2}}$
$\mathbf\small{\Rightarrow \frac{V^2}{R_E+h}=\frac{G\,M_E}{(R_E+h)^2}}$
$\mathbf\small{\Rightarrow V^2=\frac{G\,M_E}{(R_E+h)}}$
Thus we get:
Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
5. In the above equation, there is no m
• All quantities except h are constants
• The h is in the denominator
• So we can write: 
    ♦ The speed of a satellite does not depend on it's mass
    ♦ When h increases, speed decreases 
6. If the satellite is very close to the surface of the earth, (R_E+h) can be taken approximately equal to R_E
• In such cases, we can write a separate equation:
Speed of satellites very close to the surface of the earth is given by:
Eq.8.23: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{R_E}}}$
7. In the above equation 8.23, let us multiply both numerator and denominator by ✓R_E
• We get: $\mathbf\small{V=\sqrt{\frac{G\,M_E\,R_E}{R_E^2}}}$
• Thus we get:
Speed of satellites very close to the surface of the earth is given by:
Eq.8.24: $\mathbf\small{V=\sqrt{g\,R_E}}$
(∵ $\mathbf\small{g={\frac{G\,M_E}{R_E^2}}}$)

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Next we want the time period T of a satellite

1. Let the time period of an earth satellite be T
• That means., T seconds are required by that satellite to complete one rotation around the earth
2. Obviously, during those T seconds, the satellite will travel a distance equal to the 'circumference of it's orbit'
• This circumference is equal to $\mathbf\small{2\pi(R_E+h)}$
3. When we divide 'distance traveled' by time, we get the speed
• So we can write: $\mathbf\small{V={\frac{2\pi(R_E+h)}{T}}}$
4. But we have already calculated V
• Putting that value, the result in (3) becomes:
$\mathbf\small{\sqrt{\frac{G\,M_E}{(R_E+h)}}={\frac{2\pi(R_E+h)}{T}}}$
• Squaring both sides, we get: $\mathbf\small{\frac{G\,M_E}{(R_E+h)}={\frac{4\pi^2(R_E+h)^2}{T^2}}}$
$\mathbf\small{\Rightarrow T^2={\frac{4\pi^2(R_E+h)^3}{G\,M_E}}}$
$\mathbf\small{\Rightarrow T^2={\frac{4\pi^2}{G\,M_E}}(R_E+h)^3}$
5. $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ is a constant. So we can write:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
Where k = $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ = a constant
• So we can write:
The square of the 'time period of an earth satellite' is proportional to the cube of the 'distance of that planet from the center of the earth'
• Thus it is clear that, earth satellites obey Kepler's third law
6. From the result in (4), we can obtain an expression for the time period:
Eq.8.26: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
7. If the satellite is very close to the surface of the earth, (R_E+h) can be taken approximately equal to R_E
• Then we can rearrange Eq.8.26:
$\mathbf\small{T={\frac{2\pi(R_E)^{3/2}}{\sqrt{G\,M_E}}}}$
• Squaring both sides, we get: $\mathbf\small{T^2={\frac{4\pi^2(R_E)^{3}}{G\,M_E}}}$
$\mathbf\small{\Rightarrow T^2=4\pi^2 \left(\frac{R_E^{2}}{G\,M_E}\right)R_E}$
• But $\mathbf\small{\left(\frac{R_E^{2}}{G\,M_E}\right)}$ is $\mathbf\small{\frac{1}{g}}$
• So we get: $\mathbf\small{T^2=4\pi^2\frac{R_E}{g}}$
• Thus we get:
Time period of satellites very close to the surface of the earth is given by:
Eq.8.27: $\mathbf\small{T=2\pi\sqrt{\frac{R_E}{g}}}$
8. Let us put the known values in Eq.8.26. We get: $\mathbf\small{T=2\pi\sqrt{\frac{6.4\times 10^6}{9.8}}}$
• This works out to approximately 85 minutes
• So we can write:
Satellites which are close to the earth will have a time period of approximately 85 minutes

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So we have seen speed (V) and time period (T). Many other properties of celestial bodies can be calculated based on these two items. Some solved examples given below will demonstrate this concept:

Solved example 8.43
An artificial satellite very close to the surface of the earth, revolves with a speed v. What will be the speed of another artificial satellite, whose height from the surface is 0.5R_E ? 
Solution:
1. Given that, the satellite is very close to the surface of the earth
• So we can use Eq.8.23: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{R_E}}}$
• Substituting the given value 'v', we get: $\mathbf\small{v=\sqrt{\frac{G\,M_E}{R_E}}}$
2. Now we want the speed of another satellite whose h is 0.5RE
• We can use Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• Let v' be the speed of this satellite
• Substituting the values, we get: $\mathbf\small{v'=\sqrt{\frac{G\,M_E}{(1.5R_E)}}}$
3. Taking ratios, we get:
$\mathbf\small{\frac{v}{v'}=\sqrt{\frac{G\,M_E}{R_E}}\times \sqrt{\frac{1.5R_E}{G\,M_E}}=\sqrt{1.5}}$
• Thus we get: $\mathbf\small{v'=\frac{v}{\sqrt{1.5}}}$

Solved example 8.44
Two satellites A and B revolve around a planet in orbits of radii 4R and R respectively. If the speed of the satellite A is 3v, what is the speed of B?
Solution:
1. We can use Eq.8.22: $\mathbf\small{V=\sqrt{\frac{G\,M_E}{(R_E+h)}}}$
• Substituting the values we get:
$\mathbf\small{V_A=\sqrt{\frac{G\,M_E}{4R}}}$
$\mathbf\small{V_B=\sqrt{\frac{G\,M_E}{R}}}$
2. Taking ratios, we get:
$\mathbf\small{\frac{V_A}{V_B}=\sqrt{\frac{G\,M_E}{4R}}\times \sqrt{\frac{R}{G\,M_E}}=\sqrt{\frac{1}{4}}=\frac{1}{2}}$
3. But given that V_A = 3v
• So we get: $\mathbf\small{V_B=\sqrt{2}\,V_A=3\times 2\,v=6v}$

Solved example 8.45
The radii of the orbits of two satellites A and B are in the ratio 1:4. Calculate T_A : T_B
Solution:
1. We can use Kepler's law:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
$\mathbf\small{\Rightarrow T^2=k\;r^3}$
• Where (R_E+h) = r = distance from the center of the planet = radius of the orbit
2. Substituting the values, we get:
$\mathbf\small{T_A^2=k\;r_A^3}$
$\mathbf\small{T_B^2=k\;r_B^3}$
3. Taking ratios. we get:
$\mathbf\small{\frac{T_A^2}{T_B^2}=\frac{r_A^3}{r_B^3}\Rightarrow \left(\frac{T_A}{T_B}\right)^2=\left(\frac{r_A}{r_B}\right)^3}$
$\mathbf\small{\Rightarrow \left(\frac{T_A}{T_B}\right)^2=\left(\frac{1}{4}\right)^3=\frac{1}{64}}$
$\mathbf\small{\Rightarrow \frac{T_A}{T_B}=\frac{1}{8}}$

Solved example 8.46
The planet Mars has two moons, Phobos and Delmos. (i) Phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 10³ km. Calculate the mass of mars. (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days ?
Solution:
Part (i):
1. We are given the time period. So we will use an equation connecting T and mass
• We have Eq.8.26 for an earth satellite: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
2. For a Mars satellite, we can write: $\mathbf\small{T={\frac{2\pi(R_M+h)^{3/2}}{\sqrt{G\,M_M}}}}$
• Substituting the values, we get:
$\mathbf\small{\left[(7)(60)+39\right](60)={\frac{2\pi\left[(9.4)(10)^3 (10)^3\right]^{3/2}}{\sqrt{(6.67)(10^{-11})\,M_M}}}}$
3. The mass is the only unknown quantity. So we get: M_M = 6.48 × 10²³ kg

Part(ii):
1. We can use Kepler's law:
Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
$\mathbf\small{\Rightarrow T^2=k\;r^3}$
• Where (R_E+h) = r = distance from the center of the planet = radius of the orbit
2. Substituting the values, we get:
$\mathbf\small{T_E^2=k\;r_E^3}$
$\mathbf\small{T_M^2=k\;r_M^3}$
3. Taking ratios. we get:
$\mathbf\small{\frac{T_E^2}{T_M^2}=\frac{r_E^3}{r_M^3}\Rightarrow \left(\frac{T_E}{T_M}\right)^2=\left(\frac{r_E}{r_M}\right)^3}$
$\mathbf\small{\Rightarrow \left(\frac{T_E}{T_M}\right)^2=\left(\frac{r_E}{1.52\,r_E}\right)^3=\frac{1}{1.52^3}}$
4. But T_E = 365 days. So we get:
$\mathbf\small{\frac{365^2}{T_M^2}=\frac{1}{1.52^3}}$
• Thus we get: T_M = 684 days  

Solved example 8.47
You are given the following data: g = 9.81 m s^-2, R_E = 6.37 × 10⁶ m, the distance to the moon R = 3.84 × 10⁸ m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth M_E in two different ways. 
Solution:
Method 1:
We will use an equation which connects mass and force
1. Consider a body of mass m resting on the surface of the earth
• The gravitational force of attraction acting on it towards the center of the earth is $\mathbf\small{\frac{G\,M_E\,m}{R_E^2}}$
2. But this force is the weight mg of the body
3. Equating the two, we get: $\mathbf\small{mg=\frac{G\,M_E\,m}{R_E^2}}$
$\mathbf\small{\Rightarrow g=\frac{G\,M_E}{R_E^2}}$
4. Substituting the known values, we get: $\mathbf\small{9.81=\frac{(6.67 \times 10^{-11})\,M_E}{(6.37 \times 10^{6})^2}}$
• M_E is the only unknown quantity. So we get:
M_E = 5.97 × 10²⁴ kg

Method 2:
• We will use an equation which connects mass and time period T
• We have Eq.8.26: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
• Substituting the values, we get: $\mathbf\small{(27.3)(24\times 60\times 60)={\frac{2\pi(3.84\times 10^8)^{3/2}}{\sqrt{(6.67\times 10^{-11})\,M_E}}}}$
• M_E is the only unknown quantity. So we get:
M_E = 6.024 × 10²⁴ kg
■ The mass obtained by the two methods are approximately equal

Solved example 8.48
Express the constant k of Eq. (8.25) in days and km. Given k = 10^-13 s² m^-3. The moon is at a distance of 3.84 × 10⁵ km from the earth. Obtain its time-period of revolution in days.
Solution:
Part (i):
1. We have Eq.8.25: $\mathbf\small{T^2=k(R_E+h)^3}$
• Where k = $\mathbf\small{\frac{4\pi^2}{G\,M_E}}$ = a constant
2. The equation can be rearranged as $\mathbf\small{k=\frac{T^2}{(R_E+h)^3}}$
3. In SI system, the unit of time is s and the unit of distance is m
• So the units of k can be calculated as: $\mathbf\small{k=\frac{s^2}{m^3}}$
4. We want time in 'terms of days' and distance in 'terms of km'
• 1 s = $\mathbf\small{\frac{1}{24 \times 60 \times 60}=\frac{1}{86400}}$  days
• 1 m = 10^-3 km
5. So 1 s² m^-3 = $\mathbf\small{\frac{(\frac{1}{86400})^2}{(10^{-3})^3}}$
• So 10^-13 s² m^-3 = $\mathbf\small{10^{-13}\times \frac{(\frac{1}{86400})^2}{(10^{-3})^3}}$ = 1.33 × 10^-14 days² km^-3
Part (ii):
• Using Eq.8.25, we get: $\mathbf\small{1.33 \times 10^{-14}\times 3.84 \times 10^5}$ = 753.08
• Thus T = ✓(753.08) = 27.3 days

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• In the next section we will see energy of satellites

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