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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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 = $\mathbf\small{m|\vec{g}|h}$

---

• 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 $\mathbf\small{|\vec{g}|}$

---

• 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

---

■ 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

---

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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