Chapter 8.2 - Universal Law of Gravitation

In the previous section, we completed a discussion on Kepler's laws. We also saw how the centripetal force helps a body to remain in orbit. In this section we will see Universal law of Gravitation

1. Consider the rotation of the Moon around the Earth. It is shown in fig.8.8 below:

Fig.8.8

• The orbit of the moon is assumed to be a perfect circle. It is indicated by the magenta dashed circle
• The radius of this orbit is R_M. It is measured from the center of the earth
2. What is the speed $\mathbf\small{|\vec{v}|}$ with which the moon moves around the earth?
We can find it in 4 steps:
(i) The Moon takes 27.3 days to complete one rotation around the Earth
(ii) Radius (R_M) of the circular path of the Moon is 3.84 × 10⁸ m
(iii) So the distance traveled in one complete rotation
= Circumference of the circle = 2πR_M = 2π × 3.84 × 10⁸ m
(iv) So the speed $\mathbf\small{|\vec{v}|}$ = ^Distance⁄_Time = $\mathbf\small{\frac{2\pi \times 3.84 \times 10^8}{27.3} \rm{m\;day^{-1}}}$
Note: We can avoid the above 4 steps by using the following formula directly:
$\mathbf\small{|\vec{v}|=\frac{2\pi R}{T}}$
• We derived this formula in an earlier chapter where we discussed 'Angular speed' (Details here) 
3. The next step is to find the centripetal force $\mathbf\small{|\vec{F}_{c(M)}|}$ experienced by the Moon
• We have: $\mathbf\small{|\vec{F}_{c(M)}|=\frac{m_M|\vec{v}|^2}{R_M}}$
(m_M is the mass of the Moon)
• Substituting for the velocity from (2), we get:
$\mathbf\small{|\vec{F}_{c(M)}|=m_M \times \left[\frac{2\pi \times 3.84 \times 10^8}{27.3} \right]^2 \times \frac{1}{R_M}}$
$\mathbf\small{\Rightarrow |\vec{F}_{c(M)}|=m_M \times \left[\frac{2\pi \times 3.84 \times 10^8}{27.3} \right]^2 \times \frac{1}{3.84 \times 10^8}}$
$\mathbf\small{\Rightarrow |\vec{F}_{c(M)}|=m_M \times \left[\frac{4\pi^2 \times 3.84 \times 10^8}{27.3^2} \right]}$
4. Thus we get the centripetal force experienced by the moon
• Based on Newton's second law, we have:
Divide a force by mass, to get the acceleration experienced by that mass
• So, if we divide the centripetal force by the mass, we will get the centripetal acceleration $\mathbf\small{|\vec{a}_{c(M)}|}$ experienced by the moon
• So we divide the result in (3), by the mass:
$\mathbf\small{|\vec{a}_{c(M)}|=m_M \times \left[\frac{4\pi^2 \times 3.84 \times 10^8}{27.3^2} \right]\times \frac{1}{m_M}}$
$\mathbf\small{\Rightarrow |\vec{a}_{c(M)}|=\left[\frac{4\pi^2 \times 3.84 \times 10^8}{27.3^2} \right]\rm{m\;day^{-2}}}$
• We must change the units from m d^-2 to ms^-2
    ♦ 27.3 days = (27.3 × 24 × 60 × 60) seconds       
• Thus we get:
$\mathbf\small{|\vec{a}_{c(M)}|=\left[\frac{4\pi^2 \times 3.84 \times 10^8}{(27.3\times24\times60\times60)^2} \right]=0.002724\;\rm{m\;s^{-2}}}$
5. The centripetal force required by the Moon is supplied by the gravitational force
In other words:
The gravitational pull exerted by the Earth on the Moon = Centripetal force experience by Moon
6. Dividing both sides by 'mass of moon', we get:

$\mathbf{\frac{\text{Gravitational pull exerted by the Earth on the Moon}}{\text{mass of moon}}=\frac{\text{Centripetal force experienced by Moon}}{\text{mass of moon}}}$

$\mathbf{\Rightarrow \frac{\text{Gravitational pull exerted by the Earth on the Moon}}{\text{mass of moon}}=\text{Centripetal acceleration experienced by Moon}}$

• But 'Centripetal acceleration experienced by Moon' is $\mathbf\small{|\vec{a}_{c(M)}|}$ which we calculated in (4) as 0.002724 m s^-2
• So we can write:
$\mathbf{\frac{\text{Gravitational pull exerted by the Earth on the Moon}}{\text{mass of moon}}=0.002724\;\rm{m s^{-2}}}$

7. Now consider an object of mass m_O
• Let it be situated on the surface of the Earth
• We know that, just like the moon, the object is also experiencing a pull by the earth
• If we divide the 'magnitude of that pull' by m_O, what do we get?
• We will obviously get the familiar 9.8 m s^-2
■ We can write:
$\mathbf{\frac{\text{Gravitational pull exerted by the Earth on an 'object on the surface of the Earth'}}{\text{mass of that object}}=9.8\;\rm{m s^{-2}}}$
8. Compare the results in (6) and (7)
• We see that the result in (7) is (^9.8⁄_0.002724) = 3597 times greater
That means: Near the surface of the earth,  'the acceleration due to gravity' is 3597 times greater  
• Is this because the denominator (which is mass of object) is very low in (7) ?
• Let us check:
    ♦ Mass of the moon is 7.35 × 10²² kg
    ♦ Mass of the object can be about 5 or 7 kg
    ♦ So the mass of the object is about 10²² times lesser than the mass of moon
    ♦ This will not make the acceleration 'greater by just 3597 times'
• So obviously, the numerator is playing a big role here
• That is: The numerator in (6) is far less than the numerator in (7)
9. Sir Isaac Newton found out that, the 'gravitational pull' is inversely proportional to the 'square of the distance between two objects'
■ He also found out that, the 'gravitational pull' is directly proportional to the 'product of the masses of the two objects'
• We will now apply those findings to our present case:
• Let:
    ♦ $\mathbf\small{|\vec{F}_{G(O)}|}$ denote the gravitational pull exerted by the Earth on the object on surface
    ♦ $\mathbf\small{|\vec{F}_{G(M)}|}$ denote the gravitational pull exerted by the Earth on the Moon
    ♦ R_E denote the radius of the Earth. It is equal to 6.378 × 10⁶ m  
    ♦ R_M denote the distance from the center of the Earth to the center of the Moon
       It is equal to 3.84 × 10⁸ m
    ♦ m_O denote the mass of the object
    ♦ m_M denote the mass of the Moon
    ♦ $\mathbf\small{|\vec{a}_{G(O)}|}$ denote the acceleration due to 'gravitational pull from the Earth' experienced by the object on the surface
    ♦ $\mathbf\small{|\vec{a}_{G(M)}|}$ denote the acceleration due to 'gravitational pull from the Earth' experienced by the Moon
We can write:
(i) $\mathbf\small{|\vec{F}_{G(O)}|\propto \frac{1}{(R_E)^2}}$
Also $\mathbf\small{|\vec{F}_{G(O)}|\propto (m_O \times m_E)}$
$\mathbf\small{\Rightarrow|\vec{F}_{G(O)}|=G\times \frac{(m_O \times m_E)}{(R_E)^2}}$
(ii) $\mathbf\small{|\vec{F}_{G(M)}|\propto \frac{1}{(R_M)^2}}$
Also $\mathbf\small{|\vec{F}_{G(M)}|\propto (m_M \times m_E)}$
$\mathbf\small{\Rightarrow|\vec{F}_{G(M)}|=G\times \frac{(m_M \times m_E)}{(R_M)^2}}$ 
(Here G is the constant of proportionality)
10. Dividing (i) by (ii), we get:
$\mathbf\small{\frac{|\vec{F}_{G(O)}|}{|\vec{F}_{G(M)}|}= \frac{(R_M)^2}{(R_E)^2}}$
• Substituting the values of R_M and R_E, we get:
$\mathbf\small{\frac{|\vec{F}_{G(O)}|}{|\vec{F}_{G(M)}|}= \frac{(3.84\times 10^8)^2}{(6.378\times 10^6)^2}=3624.88}$
$\mathbf\small{\Rightarrow|\vec{F}_{G(O)}|=3624.88\times|\vec{F}_{G(M)}|}$
■ That is:
Gravitational pull by the Earth on a surface object
is 3624.88 times greater than
the gravitational pull by the Earth on the moon
11. Now we will find such a relation between accelerations:
We have:
(i) $\mathbf\small{|\vec{a}_{G(O)}|=\frac{|\vec{F}_{G(O)}|}{m_O}}$
Substituting for $\mathbf\small{|\vec{F}_{G(O)}|}$ from (9),we get:
$\mathbf\small{|\vec{a}_{G(O)}|=G\times \frac{(m_O \times m_E)}{(R_E)^2}\times \frac{1}{m_O}}$
$\mathbf\small{\Rightarrow |\vec{a}_{G(O)}|=G\times \frac{(m_E)}{(R_E)^2}}$
(ii) $\mathbf\small{|\vec{a}_{G(M)}|=\frac{|\vec{F}_{G(M)}|}{m_M}}$
Substituting for $\mathbf\small{|\vec{F}_{G(M)}|}$ from (9),we get:
$\mathbf\small{|\vec{a}_{G(M)}|=G\times \frac{(m_M \times m_E)}{(R_M)^2}\times \frac{1}{m_M}}$
$\mathbf\small{\Rightarrow |\vec{a}_{G(M)}|=G\times \frac{(m_E)}{(R_M)^2}}$
12. Dividing (i) by (ii), we get:
$\mathbf\small{\frac{|\vec{a}_{G(O)}|}{|\vec{a}_{G(M)}|}=\frac{(R_M)^2}{(R_E)^2}}$
$\mathbf\small{\Rightarrow |\vec{a}_{G(O)}|=\frac{(R_M)^2}{(R_E)^2}\times |\vec{a}_{G(M)}|}$
Substituting the values, we get:
$\mathbf\small{\Rightarrow |\vec{a}_{G(O)}|=\frac{(3.84\times 10^8)^2}{(6.378\times 10^6)^2}\times |\vec{a}_{G(M)}|}$
$\mathbf\small{\Rightarrow |\vec{a}_{G(O)}|=3624.88\times |\vec{a}_{G(M)}|}$
• This result is similar to the result in (10)
■ That is:
Acceleration (due to Earth's gravitational pull) experienced by the surface object
is also 3624.88 times greater than
the acceleration (due to Earth's gravitational pull) experienced by the Moon
13. Consider the acceleration experienced by the Moon $\mathbf\small{|\vec{a}_{G(M)}|}$ 
• It is the same centripetal acceleration $\mathbf\small{|\vec{a}_{c(M)}|}$ experienced by the moon
    ♦ We wrote this in (6)
• We have calculated it's value as 0.002724 ms^-2
14. Substituting this in (12), we get:
• $\mathbf\small{|\vec{a}_O|=3624.88\times 0.002724}$ = 9.874 ms^-2  
• This is very close to the value of g that we use today
• So the calculations carried out by Newton were correct

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■ Newton proposed the Universal Law of Gravitation. It states that:

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them
• We can elaborate this in 8 steps:
1. Consider two bodies having masses m₁ and m₂
2. They are situated at a distance of r apart as shown in fig.8.9 below:

Fig.8.9

3. Body of mass m₁ attracts the other body of mass m₂
4. The magnitude of this attractive force $\mathbf\small{|\vec{F}_G|}$ is proportional to the product (m₁m₂)
• Mathematically this can be written as: $\mathbf\small{|\vec{F}_G|\propto(m_1\times m_2)}$  
5. Also, the magnitude of this attractive force is inversely proportional to the square of the distance r
• Mathematically this can be written as: $\mathbf\small{|\vec{F}_G|\propto\frac{1}{r^2}}$
6. Combining the results in (4) and (5), we get: $\mathbf\small{|\vec{F}_G|\propto\frac{(m_1 \times m_2)}{r^2}}$
7. Introducing a constant of proportionality, we can write: $\mathbf\small{|\vec{F}_G|=G\frac{m_1\; m_2}{r^2}}$
• Where G is the Universal Gravitational Constant
(Some basics about 'proportionality constant' can be seen here)
• In the a later section, we will see how scientists calculated the exact value of G
8. The force $\mathbf\small{|\vec{F}_G|}$ obtained in (7), is the pulling force exerted by m₁ on m₂
• But for every force, there exists an equal and opposite reaction force
• So m₂ exerts a pulling force of the same magnitude $\mathbf\small{|\vec{F}_G|}$ on m₁.

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• The above 8 steps help us to calculate the magnitude of the gravitational force between two bodies
• But we want the direction also. In the next section, we will see details about the direction. Before that, we will see a few solved examples related to magnitude

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Solved example 8.5

Explain how the Universal law of gravitation can be used to prove Kepler's third law
Solution:
1. Consider any planet say Earth
• The gravitational force of attraction between Earth and sun will be equal to $\mathbf\small{G\frac{m_E\,m_S}{a_E^2}}$
• Where 
    ♦ m_E and m_S are the masses of the Earth and the Sun respectively
    ♦ a_E is the radius of the Earth's orbit 
• This force provides the centripetal force which keeps Earth in it's orbit
2. Centripetal force acting on Earth is $\mathbf\small{\frac{4m_E\, \pi^2\,a_E}{T_E^2}}$
• Where T_E is the time period of the Earth to complete one revolution around the sun
3. Equating the results in (1) and (2), we get: $\mathbf\small{G\frac{m_E\,m_S}{a_E^2}=\frac{4m_E\, \pi^2\,a_E}{T_E^2}}$
$\mathbf\small{\Rightarrow G\frac{\,m_S}{a_E^2}=\frac{4\, \pi^2\,a_E}{T_E^2}}$
$\mathbf\small{\Rightarrow \frac{T_E^2}{a_E^3}=\frac{4\, \pi^2}{G\,m_S}}$
4. Consider any other planet say Jupiter
• The gravitational force of attraction between Jupiter and sun will be equal to $\mathbf\small{G\frac{m_J\,m_S}{a_J^2}}$
• Where 
    ♦ m_J and m_S are the masses of the Jupiter and the Sun respectively
    ♦ a_J is the radius of the Jupiter's orbit 
• This force provides the centripetal force which keeps Jupiter in it's orbit
5. Centripetal force acting on Jupiter is $\mathbf\small{\frac{4m_J\, \pi^2\,a_J}{T_J^2}}$
• Where T_J is the time period of the Jupiter to complete one revolution around the sun
6. Equating the results in (4) and (5), we get: $\mathbf\small{G\frac{m_J\,m_S}{a_J^2}=\frac{4m_J\, \pi^2\,a_J}{T_J^2}}$
$\mathbf\small{\Rightarrow G\frac{\,m_S}{a_J^2}=\frac{4\, \pi^2\,a_J}{T_J^2}}$
$\mathbf\small{\Rightarrow \frac{T_J^2}{a_J^3}=\frac{4\, \pi^2}{G\,m_S}}$
7. Right side of the result in (6) is same as the right side of the result in (3) 
• We will get the same right side for any planet of the solar system
• So we can write: $\mathbf\small{\frac{T^2}{a^3}}$ = A constant
• This is Kepler's third law

Solved example 8.6
Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸  m. Show that the mass of Jupiter is about one-thousandth that of the sun
Solution:
1. Consider the rotation of Io around Jupiter
Equating the gravitational force and the centripetal force, we get:
$\mathbf\small{G\frac{m_I\,m_J}{a_I^2}=\frac{4m_I\, \pi^2\,a_I}{T_I^2}}$
• Where 
    ♦ m_I and m_J are the masses of Io and Jupiter respectively

    ♦ a_I is the radius of Io's orbit 
    ♦ T_I is the time in which Io completes one revolution around Jupiter
$\mathbf\small{\Rightarrow G\frac{\,m_J}{a_I^2}=\frac{4\, \pi^2\,a_I}{T_I^2}}$
$\mathbf\small{\Rightarrow \frac{T_I^2}{a_I^3}=\frac{4\, \pi^2}{G\,m_J}}$
$\mathbf\small{\Rightarrow m_J=\left(\frac{4\, \pi^2}{G} \right)\frac{a_I^3}{T_I^2}}$
2. Substituting the known values, we get:
$\mathbf\small{m_J=\left(\frac{4\, \pi^2}{G} \right)\frac{(4.22 \times 10^8)^3}{1.769^2}}$
3. Consider the rotation of Earth around Sun
Equating the gravitational force and the centripetal force, we get:
$\mathbf\small{G\frac{m_E\,m_S}{a_E^2}=\frac{4m_E\, \pi^2\,a_E}{T_E^2}}$
$\mathbf\small{\Rightarrow G\frac{\,m_S}{a_E^2}=\frac{4\, \pi^2\,a_E}{T_E^2}}$
$\mathbf\small{\Rightarrow \frac{T_E^2}{a_E^3}=\frac{4\, \pi^2}{G\,m_S}}$
$\mathbf\small{\Rightarrow m_S=\left(\frac{4\, \pi^2}{G} \right)\frac{a_E^3}{T_E^2}}$
4. Substituting the known values, we get:
$\mathbf\small{\Rightarrow m_S=\left(\frac{4\, \pi^2}{G} \right)\frac{(1.496 \times 10^{11})^3}{365.25^2}}$
5. Dividing (2) by (4), we get:
$\mathbf\small{\frac{m_J}{m_S}=\left(\frac{(4.22 \times 10^8)^3}{1.769^2}\right)\div \left(\frac{(1.496 \times 10^{11})^3}{365.25^2}\right)}$
$\mathbf\small{\Rightarrow \frac{m_J}{m_S}=\left(\frac{(4.22 \times 10^8)^3}{1.769^2}\right)\times \left(\frac{365.25^2}{(1.496 \times 10^{11})^3}\right)}$ = 0.000955
$\mathbf\small{\Rightarrow \frac{m_J}{m_S}=0.000957}$
$\mathbf\small{\Rightarrow m_S= \frac{m_J}{0.000957}}$

$\mathbf\small{\Rightarrow m_S=1045\;m_J}$

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In the next section, we will see direction

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Chapter 8.1 - Kepler's Third law

In the previous section, we saw Kepler's first and second laws. In this section we will see Kepler's third law. Later in this section we will recap the 'significance of centripetal force in rotational motion'

Kepler’s third law
The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet
• This law is also known as Law of periods

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• To fully understand this law, we must carry out some calculations. They can be done in steps:

Step 1:
1. Consider any planet say, Venus
• Venus has it's own ellipse
2. Write down the following distance:
• The distance from center (O) of that ellipse to the perihelion(P)/aphelion(A)
    ♦ Obviously, it is the semi-major axis of the ellipse
    ♦ Let us denote this distance as a
    ♦ Then, for Venus, it will be a_Venus
    ♦ So a_Venus is the semi-major axis of Venus
    ♦ We can use the unit 'm' to write this distance 
3. Write down the following time:
• The time required by Venus to complete one revolution around the Sun
    ♦ Let us denote this time as T
    ♦ Then for Venus, it will be T_Venus 
    ♦ We can use the unit 'year' to write this time
4. Calculate the ratio: $\mathbf\small{\frac{(T_{\text{Venus}})^2}{(a_{\text{Venus}})^3}}$

Step 2:
1. Consider any other planet say, Mars
• Mars has it's own ellipse
2. Write down the following distance:
• The distance from center (O) of that ellipse to the perihelion(P)/aphelion(A)
    ♦ Obviously, it is the semi-major axis of the ellipse
    ♦ For Mars, it will be a_Mars
    ♦ So a_Mars is the semi-major axis of Mars
    ♦ We can use the unit 'm' to write this distance 
3. Write down the following time:
• The time required by Mars to complete one revolution around the Sun
    ♦ For Mars, it will be T_Mars 
    ♦ We can use the unit 'year' to write this time
4. Calculate the ratio: $\mathbf\small{\frac{(T_{\text{Mars}})^2}{(a_{\text{Mars}})^3}}$

Step 3:
1. Consider any other planet say, Jupiter
• Jupiter has it's own ellipse
2. Write down the following distance:
• The distance from center (O) of that ellipse to the perihelion(P)/aphelion(A)
    ♦ Obviously, it is the semi-major axis of the ellipse
    ♦ For Jupiter, it will be a_Jupiter
    ♦ So a_Jupiter is the semi-major axis of Jupiter
    ♦ We can use the unit 'm' to write this distance 
3. Write down the following time:
• The time required by Jupiter to complete one revolution around the Sun
    ♦ For Jupiter, it will be T_Jupiter 
    ♦ We can use the unit 'year' to write this time
4. Calculate the ratio: $\mathbf\small{\frac{(T_{\text{Jupiter}})^2}{(a_{\text{Jupiter}})^3}}$

Step 4:
• Compare the ratios. We get a surprise
• We see that, all ratios are the same
• That is: $\mathbf\small{\frac{(T_{\text{Venus}})^2}{(a_{\text{Venus}})^3}=\frac{(T_{\text{Mars}})^2}{(a_{\text{Mars}})^3}=\frac{(T_{\text{Jupiter}})^2}{(a_{\text{Jupiter}})^3}}$

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• The steps are complete

• Remember that, we chose planets at random. So the ratio must be the same for all the eight planets
• It is indeed so. The calculations for all the planets can be seen here
• Kepler was astonished to see the same result for all the planets
• From the table, we have: $\mathbf\small{\frac{T^2}{a^3}=\text{A constant}(\approx 3\times10^{-34}\;\rm{y^2 m^{-3}})}$
• From this, we get:
Eq.8.1: $\mathbf\small{T^2=\text{(A constant)}\times a^3}$
    ♦ That means, $\mathbf\small{T^2}$ is proportional to $\mathbf\small{a^3}$
    ♦ This is the mathematical representation of the third law
■ The second law is related to 'equality of some items' 
    ♦ Those items are related to any one planet
■ The third law is also related to 'equality of some items'
    ♦ But those items are related to all the eight planets
■ Scientists thus received hints about the 'presence of a common factor' among the eight planets
In a later section, we will see how Sir Isaac Newton gave a satisfactory explanation for this 'common factor'

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Now we will see some solved examples:

Solved example 8.1
A planet moving around the sun sweeps areas A₁ in 2 days, A₂ in 3 days and A₃ in 6 days. Find the relation between A₁, A₂ and A₃
Solution:
1. From Kepler's second law, we have:
A planet sweeps equal areas in equal intervals of time
2. In our present case, the planet sweeps A₁ in 2 days
• So in 1 day, the planet sweeps ¹⁄₂ A₁
3. Thus we get:
• In 3 days, the planet will sweep (3 × ¹⁄₂ A₁) = ³⁄₂A₁
• But given that, the planet sweeps A₂ in 3 days
• So we get: ³⁄₂A₁ = A₂
⇒ A₁ = ²⁄₃ A₂
4. Similarly we get:
• In 6 days, the planet will sweep (6 × ¹⁄₂A₁) = 3A₁
• But given that, the planet sweeps A₃ in 6 days

• So we get: 3A₁ = A₃
A₁ = ¹⁄₃ A₃
5. Equating the 3 items, we get:
A₁ = ²⁄₃ A₂ = ¹⁄₃ A₃
• Multiplying by 3, we get: 3A₁ = 2A₂ = A₃

Solved example 8.2
A satellite is orbiting around the earth at a distance of 6R from the surface of the earth. It takes 24 hr to complete one revolution. How much time will another satellite take to make one revolution, if it is orbiting at a distance of 2.5 R from the surface of the earth?
Solution:
1. Radius of the first satellite from center of the earth = (R+6R) = 7R
• Radius of the second satellite from center of the earth = (R+2.5R) = 3.5R
2. From Kepler's third law, we have: $\mathbf\small{\frac{T^2}{a^3}=\text{A constant}}$
• So we can write: $\mathbf\small{\frac{T_1^2}{a_1^3}=\frac{T_2^2}{a_2^3}}$
• Substituting the values, we get: $\mathbf\small{\frac{24^2}{(7R)^3}=\frac{T_2^2}{(3.5R)^3}}$
$\mathbf\small{\Rightarrow \frac{24^2\times (3.5R)^3}{(7R)^3}=\frac{T_2^2}{1}}$
• Thus we get: T₂ = √(72) = 6√2 hr

Solved example 8.3
Period of revolution of two planets A and B around the sun are: T_A = T and T_B = 8T
What is the relation between their distances a_A and a_B from the sun?
Solution:
1. We have:$\mathbf\small{\frac{T_1^2}{a_1^3}=\frac{T_2^2}{a_2^3}}$
• Substituting the values, we get: $\mathbf\small{\frac{T^2}{a_A^3}=\frac{(8T)^2}{a_B^3}}$
$\mathbf\small{\Rightarrow \frac{1}{a_A^3}=\frac{(8)^2}{a_B^3}=\frac{(2^3)^2}{a_B^3}=\frac{(2^2)^3}{a_B^3}=\frac{(4)^3}{a_B^3}}$
$\mathbf\small{\Rightarrow \frac{a_B^3}{a_A^3}=\frac{(4)^3}{1}}$
$\mathbf\small{\Rightarrow \frac{a_B}{a_A}=4}$
2. So we can write: 
Distance of B from the sun = 4 × Distance of A from the sun

Solved example 8.4
Suppose there existed a planet that went around the sun twice as fast as the Earth. What would be it's orbital size as compared to that of the earth?

Solution:
1. If the planet travels with double speed, the time required will be half
2. We have:$\mathbf\small{\frac{T_1^2}{a_1^3}=\frac{T_2^2}{a_2^3}}$
• Substituting the values, we get: $\mathbf\small{\frac{T^2}{a_E^3}=\frac{(0.5T)^2}{a_P^3}}$
$\mathbf\small{\Rightarrow \frac{1}{a_E^3}=\frac{(0.5)^2}{a_P^3}=\frac{0.25}{a_P^3}}$
$\mathbf\small{\Rightarrow \frac{a_P^3}{a_E^3}=0.25}$
$\mathbf\small{\Rightarrow \frac{a_P}{a_E}=(0.25)^{\frac{1}{3}}=0.63}$
3. So we can write: 
Distance of the planet from the sun = 0.63 × Distance of Earth from the sun

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Next, we have to learn about the Universal law of Gravitation. But before that, we will write the answer for a fundamental question:

■ How does centripetal force help a body to keep revolving, with out falling down?
(We have seen the basics about centripetal force here)
• In the animation in fig.8.6 below, a block attached to a string is revolving around a vertical axis

Fig.8.6

    ♦ The vertical axis is shown in blue color
    ♦ The string is shown in yellow color
    ♦ The block is shown in red color

• A two dimensional view is shown in fig.8.7 below:

Fig.8.7

• Fig.8.7(a) shows the normal position, where the block is at rest
• When the rotation starts, the block is raised to a higher level. This is shown in fig.b
• In this situation, the string is making an angle of θ with the horizontal
• If the speed of rotation is increased, the block will be raised to an even higher level
• Can we increase the speed to such a high value that, the string becomes perfectly horizontal?
We will write the answer in steps:
1. Fig.c shows the Free body diagram of the block
• Recall that, a free body diagram will show only those forces which are exerted on the body. It will not show forces which are exerted by the body
2. The tension $\mathbf\small{|\vec{T}|}$ in the string is resolved into horizontal and vertical components
3. Considering equilibrium in the horizontal direction, we get: $\mathbf\small{|\vec{T}|\cos \theta=|\vec{F}_c|}$
• Where $\mathbf\small{|\vec{F}_c|}$ is the magnitude of the centripetal force
4. Considering the equilibrium in the vertical direction, we get: $\mathbf\small{|\vec{T}|\sin \theta=m|\vec{g}|}$
5. Dividing (4) by (3), we get: $\mathbf\small{\tan \theta=\frac{m|\vec{g}|}{|\vec{F}_c|}}$
6. But $\mathbf\small{|\vec{F}_c|=\frac{m|\vec{v}|^2}{R}}$
• Where:
    ♦ $\mathbf\small{|\vec{v}|}$ is the speed of the block
    ♦ R is the radius of the circular path of the block
7. So the result in (5) becomes: $\mathbf\small{\tan \theta=\frac{m|\vec{g}|}{\frac{m|\vec{v}|^2}{R}}}$
$\mathbf\small{\Rightarrow \tan \theta=\frac{R|\vec{g}|}{|\vec{v}|^2}}$
8. $\mathbf\small{|\vec{v}|}$ is in the denominator
• So when $\mathbf\small{|\vec{v}|}$ increases, tan θ decreases
• When tan θ decreases, θ decreases
• That means, the angle which the string makes with the horizontal decreases
9. So we can write:
As the speed $\mathbf\small{|\vec{v}|}$ of rotation increases, the string becomes more and more horizontal
10. If the speed increase to such a level that, the string becomes perfectly horizontal, the vertical component $\mathbf\small{|\vec{T}|\sin \theta}$ will vanish  
• In such a situation, there will not be any force to oppose $\mathbf\small{m|\vec{g}|}$
• So the block will fall down wards
• But just as it falls, a small θ will appear again. As a result the vertical component $\mathbf\small{|\vec{T}|\sin \theta}$ will appear again
• Thus the block will continue to rotate
11. Suppose that, the following two points are true:
(i) There is no air resistance
(ii) There is no friction at the anchor point indicated by the small green sphere
• If the two points are true, the block will continue to rotate for ever
12. In the case of planets, we have the 'gravitational force' in place of $\mathbf\small{|\vec{T}|}$ 
• Since there are no external forces, the planets continue to revolve forever
• Thus we can understand that, the gravitational force supplies the necessary centripetal force for the revolution of planets around the sun

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• In the next section, we will Universal law of Gravitation

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Chapter 6.17 - Examples in Vertical Circular Motion

In the previous section, we saw that the critical velocity at the topmost point of the vertical circle is $\mathbf\small{\sqrt{rg}}$. We also saw that the direction of the reaction at the top most point can be upwards or downwards. In this section we will see some practical cases.

In fig.6.53 below, a sphere of mass 1.2 kg rolls down a track shown in yellow color. 

Fig.6.53

It starts rolling from the point A. It reaches the lowest point B and then rolls upwards on the other side. We must ensure that the sphere does not fly off while passing through the highest point C on the other side. How can we ensure that? Assume that the surfaces are smooth. [g = 10 ms^-2]
Let us analyze:
1. Let the sphere roll down from a height 'h'
• That is., the point 'A' is at a height of 'h' from the datum
• When it reaches the lowest point 'B', it will have a large velocity 
2. But when it rolls upwards on the other side, the velocity will begin to decrease.
• The velocity will reach the smallest possible value when it reaches 'C'
• However, since 'C' is at a lower level than 'A', v_C will have a significant value
3. If v_C is very large, the sphere will fly off at 'C'
• We have to find the safe velocity at C so that 'flying off' is avoided
4. For that, we will draw the FBD of the sphere at C
• It is shown in fig.6.54 below:

Fig.6.54

• The normal reaction from the track is F_N(C)
■ The sign convention is: • Forces directed towards the center of the circle are positive • Forces directed away from the center of the circle are negative
5. So the net force acting on the sphere is (mg-F_N(C))
• This net force must be equal to the centripetal force
• So we can write: $\mathbf\small{mg-F_{N(C)}=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
6. If the normal reaction F_N(C) is zero, then it would mean that, the sphere is not touching the track
• That is., if the normal reaction F_N(C) is zero, the sphere is about to fly off
• Then the equation in (5) becomes: $\mathbf\small{0=m \left(g-\frac{v_C^2}{r}\right)}$
7. Obviously, for the left side to become zero, $\mathbf\small{\frac{v_C^2}{r}}$ must be equal to g
• So we get: $\mathbf\small{v_C=\sqrt{rg}}$
• We can write: If v_C is equal to $\mathbf\small{\sqrt{rg}}$, the sphere will be at the 'point of flying off'
8. We can prevent such a situation by reducing the speed
• That is., v_C must be less than $\mathbf\small{\sqrt{rg}}$
9. In our present case, 
• If the elevation of point A is high, v_C will be high
• If the elevation of point A is low, v_C will be low
10. Let us find the limiting height:
• For that, we use the Law of conservation of energy
• Let us apply it to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write E_A = E_C
11. Let us write the various energies:
(i) Total energy E_A:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh}$ 
■ So total energy E_A = $\mathbf\small{mgh}$
(ii) Total energy E_C
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mg \times 3}$ 
■ So total energy E_C = $\mathbf\small{\frac{m\,v_C^2}{2}+3mg}$
12. Equating the two energies we get:
$\mathbf\small{mgh=\frac{m\,v_C^2}{2}+3mg}$
• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{2gh=v_C^2+6g}$
$\mathbf\small{\Longrightarrow 20h=v_C^2+60}$
13. But v_C must not exceed $\mathbf\small{\sqrt{rg}}$
• That is., v_C must not exceed $\mathbf\small{\sqrt{1.5 \times 10} = \sqrt{15}}$     
• Substituting this in (12), we get:
$\mathbf\small{20h=15+60=75}$
Thus we get: h = ⁷⁵⁄₂₀ = 3.75 m
14. If the sphere is released from a greater height, the speed v_C will exceed $\mathbf\small{\sqrt{15}}$ the and it will fly off at C
• So we must release it only from a height less than 3.75 m

Another example:
In fig.6.55 below, a sphere of mass 1.2 kg rolls down a track shown in yellow color. 

Fig.6.55

It starts rolling from the point A. It reaches the lowest point B and then rolls upwards into a vertical circular track on the other side. We must ensure that the sphere does not fall while passing through the highest point C on the circular track. How can we ensure that? Assume that the surfaces are smooth. [g = 10 ms^-2]
Let us analyze:
1. Let the sphere roll down from a height 'h'
• That is., the point 'A' is at a height of 'h' from the datum
• When it reaches the lowest point B, it will have a large velocity 
2. But when it rolls upwards on vertical circular track the other side, the velocity will begin to decrease.
• The velocity will reach the smallest possible value when it reaches 'C'
• However, since 'C' is at a lower level than 'A', v_C will have a significant value
3. Even then, if v_C is very small, the sphere will fall down at 'C'
• We have to find the safe velocity at C
4. For that, we will draw the FBD of the sphere at C
• It is shown in fig.6.56 below:

Fig.6.56

• The normal reaction from the track is F_{N(C)
■ The sign convention is: • Forces directed towards the center of the circle are positive • Forces directed away from the center of the circle are negative}
5. So the net force acting on the sphere is (mg+F_N(C)) 
• This net force must be equal to the centripetal force
• So we can write: $\mathbf\small{mg+F_{N(C)}=\frac{mv_C^2}{r}}$
$\mathbf\small{\Longrightarrow F_{N(C)}=m \left(\frac{v_C^2}{r}-g\right)}$
6. If the normal reaction F_N(C) is zero, then it would mean that, the sphere is not touching the track
• That is., if the normal reaction F_N(C) is zero, the sphere will fall down
• Then the equation in (5) becomes: $\mathbf\small{0=m \left(\frac{v_C^2}{r}-g\right)}$
7. Obviously, for the left side to become zero, $\mathbf\small{\frac{v_C^2}{r}}$ must be equal to g
• So we get: $\mathbf\small{v_C=\sqrt{rg}}$
• We can write: If v_C is equal to $\mathbf\small{\sqrt{rg}}$, the sphere will fall down
8. We can prevent such a situation by increasing the speed
• That is., v_C must be greater than $\mathbf\small{\sqrt{rg}}$
9. In our present case, 
• If the elevation of point A is high, v_C will be high
• If the elevation of point A is low, v_C will be low
10. Let us find the limiting height:
• For that, we use the Law of conservation of energy
• Let us apply it to the points A and C
■ By the Law of conservation of energy, the total energy must be the same at those two points
• We can write E_A = E_C
11. Let us write the various energies:
(i) Total energy E_A:
• Kinetic energy = 0
• Gravitational potential energy = $\mathbf\small{mgh}$ 
■ So total energy E_A = $\mathbf\small{mgh}$
(ii) Total energy E_C
• Kinetic energy = $\mathbf\small{\frac{m\,v_C^2}{2}}$
• Gravitational potential energy = $\mathbf\small{mg \times 3}$ 
■ So total energy E_C = $\mathbf\small{\frac{m\,v_C^2}{2}+3mg}$
12. Equating the two energies we get:
$\mathbf\small{mgh=\frac{m\,v_C^2}{2}+3mg}$

• Multiplying both sides by $\mathbf\small{\frac{2}{m}}$, we get:
$\mathbf\small{2gh=v_C^2+6g}$
$\mathbf\small{\Longrightarrow 20h=v_C^2+60}$
13. But v_C must exceed $\mathbf\small{\sqrt{rg}}$
• That is., v_C must exceed $\mathbf\small{\sqrt{1.5 \times 10} = \sqrt{15}}$     
• Substituting this in (12), we get:
$\mathbf\small{20h=15+60=75}$
Thus we get: h = ⁷⁵⁄₂₀ = 3.75 m
14. If the sphere is released from a height lesser than 3.75 m, the speed v_C will be lesser than $\mathbf\small{\sqrt{15}}$ and so the sphere will fall at point C
• So we must release it only from a height greater than 3.75 m

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We can write a summary based on the above two examples
■ In example 1, the normal reaction from the track is given by: $\mathbf\small{F_{N(C)}=m \left(g-\frac{v_C^2}{r}\right)}$
    ♦ If F_N(C) is not to become zero, v_C must be less than $\mathbf\small{\sqrt{rg}}$
    ♦ This is because, $\mathbf\small{\frac{v_C^2}{r}}$ is being subtracted from g 
■ In example 2, the normal reaction from the track is given by: $\mathbf\small{F_{N(C)}=m \left(\frac{v_C^2}{r}-g\right)}$
    ♦ If F_N(C) is not to become zero, v_C must be greater than $\mathbf\small{\sqrt{rg}}$
    ♦ This is because, g is being subtracted from $\mathbf\small{\frac{v_C^2}{r}}$

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So we have completed a discussion on conservation of mechanical energy. Next we will see the law of conservation in other forms of energy such as heat energy, sound energy, nuclear energy etc.,   
Before that, we will see a few more solved examples in the next section.

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