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
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
• 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 aVenus
♦ So aVenus 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 TVenus
♦ 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 aMars
♦ So aMars 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 TMars
♦ 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 aJupiter
♦ So aJupiter 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 TJupiter
♦ 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: $\mathbf\small{\frac{(T_{\text{Venus}})^2}{(a_{\text{Venus}})^3}=\frac{(T_{\text{Mars}})^2}{(a_{\text{Mars}})^3}=\frac{(T_{\text{Jupiter}})^2}{(a_{\text{Jupiter}})^3}}$
• 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'
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: TA = T and TB = 8T
What is the relation between their distances aA and aB 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
■ 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
♦ 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(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
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 aVenus
♦ So aVenus 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 TVenus
♦ 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 aMars
♦ So aMars 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 TMars
♦ 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 aJupiter
♦ So aJupiter 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 TJupiter
♦ 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: $\mathbf\small{\frac{(T_{\text{Venus}})^2}{(a_{\text{Venus}})^3}=\frac{(T_{\text{Mars}})^2}{(a_{\text{Mars}})^3}=\frac{(T_{\text{Jupiter}})^2}{(a_{\text{Jupiter}})^3}}$
• The steps are complete
• 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'
Now we will see some solved examples:
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: TA = T and TB = 8T
What is the relation between their distances aA and aB 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
Next, we have to learn about the Universal law of Gravitation. But before that, we will write the answer for a fundamental question:
(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 string is shown in yellow color
♦ The block is shown in red color
 |
| Fig.8.7 |
• 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
• In the next section, we will Universal law of Gravitation
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