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 8 - Gravitation

In the previous section, we completed a discussion on rotational motion. In this chapter we will see gravitation.

• In our day to day life, we continuously feel the effects of gravitational force. Let us see some examples:
    ♦ Objects let go from a height always fall downwards
    ♦ It is more difficult to lift a heavier mass than a lighter mass
    ♦ Going uphill is more difficult than going downhill
• Through ages, scientists tried to find proper explanations for such phenomena
• The sixteenth century scientist Galileo Galilei, through his experiments, found out that, all falling objects are subjected to acceleration. This can be explained in 5 steps as follows:
1. We know that, a falling object is moving towards the surface of the earth
• So that falling object is in motion
2. Since it is in motion, we can say: It has a velocity
3. But that velocity is continuously increasing
• That means, the velocity is continuously changing
4. Since there is ‘change in velocity’, we can say: The body is subjected to acceleration
5. Galilio found out that, all bodies are subjected to the same acceleration
• That means:
    ♦ Some bodies may be having a greater mass (we call them heavy bodies)
    ♦ Some bodies may be having a lesser mass (we call them light bodies)
    ♦ Whatever be the mass, the acceleration experienced by all bodies will be the same

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• Next we will see another notable discovery made by scientists of the seventeenth century. We will write it in 5 steps:

1. Observe a planet
2. Record it’s position
3. Note down the exact date and time at which the observation is made
4. Wait for one year
5. At the exact date and time, observe the planet
■ It will be located at the same exact position recorded in (2)

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• Danish scientist Tycho Brahe received funds from various Kings and rulers. Those Funds were granted to him, for carrying out extensive scientific research works. So he was able to spend a major part of his life time recording such observations of the planets and stars

• The data recorded by Brahe was later analysed by his assistant Johannes Kepler
• Kepler formulated three laws, which are now known as Kepler’s laws
• Kepler’s laws was the starting point of the works done by Sir Isaac Newton
    ♦ Those works led to the discovery of Universal law of Gravitation
• So let us discuss about the Kepler’s laws in some detail

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Kepler’s first law
All planets move in elliptical orbits with the sun situated at one of the foci of the ellipse
• This law is also known as Law of Orbits

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• To fully understand this law, we must know some of the 'salient features of ellipse'

• Learning a simple method to 'draw an ellipse' will help us to appreciate those salient features
• The method of drawing can be written in 3 steps:
1. Fix two pins at any ‘convenient distance apart’ on a drawing board 
• This is shown in fig.8.1(a) below:

Fig.8.1

• Name the positions of the pins as F₁ and F₂
2. Tie one end of an inextensible string to the pin at F₁
• Tie the other end to the pin at F₂
3. Using the tip of a pencil, stretch the string taut and draw a curve
• The string must be taut all the while
• The closed curve thus obtained is called the ellipse

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Let us see the important features of an ellipse:

1. F₁ and F₂ are called the foci
2. Consider any point T on the ellipse. There will be two distances F₁T and F₂T
• Point T was obtained while keeping the string taut
• So (F₁T + F₂T) = Length of the string
• This is true for any point on the string
■ So we can write:
For any point on the string, the sum of the distances from the foci is the same
3. Draw a line connecting F₁ and F₂
• Extend this line towards either sides so that, it meets the ellipse at P and A (fig.b)
• Mark the midpoint of PA as O
• This O is the center of the ellipse
4. PA is called the major axis of the ellipse
• OP or OA, which is half of the major axis is called semi major axis
5. An interesting case:
• Move F₂ towards the left
• Let it coincide with F₁
• Draw the curve as before
• We will get a circle This is shown in fig.8.2(a) below

Fig.8.2

Now we will see how the 'ellipse and it's foci' are related to 'planetary motion'

1. The sun is present at the focus F₁
• It is marked as S in the above fig.8.2(b)
2. The planet orbits around the sun
• The motion of the planet is along the ellipse
3. Consider the position P
• PS is the smallest distance possible between the planet and the sun
■ The position P is called Perihelion
4. Consider the position A
• AS is the largest distance possible between the planet and the sun
■ The position A is called Aphelion

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Kepler’s second law
The line that joins any planet to the sun sweeps equal areas in equal intervals of time
• This law is also known as Law of Areas

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• To fully understand this law, we can make use of fig.8.3 below:

Fig.8.3

• A planet is moving around the sun S
• We must complete 4 steps:
Step 1:
1. Note down the time t₁ at which the planet is at any convenient point U
2. Note down the time t₂ at which the planet is at any other convenient point V
3. Calculate Δt = (t₂-t₁)
4. Calculate the area A₁ enclosed between the three items:
(i) Line US
(ii) Line VS
(iii) Arc UV
■ A₁ is the area swept by the line US during a time interval of Δt
Step 2:
1. Note down the time t₃ at which the planet is at any other convenient point W
2. Note down the point X at which the planet reaches exactly at time (t₃+Δt)
• Here Δt must be the same Δt calculated in step 1
• This ensures that, the two items are equal:
(i) Time taken to travel from U to V
(ii) Time taken to travel from W to X
3. Calculate the area A₂ enclosed between the three items:
(i) Line WS
(ii) Line XS
(iii) Arc WX
■ A₂ is the area swept by the line WS during a time interval of Δt
Step 3:
1. Note down the time t₄ at which the planet is at any other convenient point Y
2. Note down the point Z at which the planet reaches exactly at time (t₄+Δt)
• Here Δt must be the same Δt calculated in step 1
• This ensures that, the three items are equal:
(i) Time taken to travel from U to V
(ii) Time taken to travel from W to X
(ii) Time taken to travel from Y to Z
3. Calculate the area A₃ enclosed between the three items:
(i) Line YS
(ii) Line ZS
(iii) Arc YZ
■ A₃ is the area swept by the line YS during a time interval of Δt
Step 4:
■ Compare the areas
We will find that: A₁ = A₂ = A₃

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• The above four steps were carried out several times by Kepler
• Thus he arrived at the second law
• If we have precision instruments to observe the planets, we too can perform the 4 steps and verify the law

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• The second law is based on 'analysis of observations'
    ♦ Observations were made by Tycho Brahe
    ♦ Analysis of those observations were done by Kepler

• While doing the analysis, Kepler discovered that, there is an ‘equality in areas’
• But we want to know the ‘cause of such an equality’
• Indeed, scientists were able to find the ‘cause’
• Before discussing the 'explanation given by scientists', let us see an 'immediate inference'. It can be written in 3 steps:
1. Let us assume that, the three areas in fig.8.3 above, are triangles
• For the △SUV near the aphelion, the base is small
• For the △SWX and △SYZ near the perihelion, the bases are larger
2. In spite of the differences in bases, the areas are the same
• This is due to the difference in heights
    ♦ For the △SUV near the aphelion, the height is large
    ♦ For the △SWX and △SYZ near the perihelion, the heights are smaller
3. Remember that, Δt is same in all the three cases
■ So from (1), it is clear that:
• The planet travel with greater speeds when it is near the perihelion
• That is how it is able to cover greater ‘base distance’ in the same interval of time
■ So we can write the ‘immediate inference’:
The ‘equality in areas’ is due to:
    ♦ Greater speeds near perihelion
    ♦ Lesser speeds near aphelion

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But this leads to more questions:

    ♦ Why do planets move faster when they are near the perihelion ?
    ♦ Why do planets move slower when they are near the aphelion ?
The answers can be written in steps:
1. In fig.8.4 below, the elliptical orbit is shown:

Fig.8.4

• The position of the sun is denoted by S
• The position of the planet is denoted by Q
2. Let S be the origin of the coordinate system
• Then the vector joining S and Q can be considered as the position vector $\mathbf\small{\vec{r}}$ of the planet    
3. After a time duration of Δt, the planet reaches Q’
4. So the vector joining Q and Q' is the displacement vector
• Magnitude of the displacement is given by: (velocity × time)
• So the vector from Q to Q' is: $\mathbf\small{\vec{v}(\Delta t)}$
• '$\mathbf\small{\vec{v}(\Delta t)}$' is a vector
    ♦ It's magnitude is: $\mathbf\small{|\vec{v}|\times\Delta t}$
    ♦ It's direction is same as that of $\mathbf\small{\vec{v}}$
    ♦ In effect, it is the displacement vector from Q to Q' 
5. Consider the △SQQ'
• One side of this triangle is $\mathbf\small{\vec{r}}$
• The other side is $\mathbf\small{\vec{v}(\Delta t)}$
■ Then we have:
• Area of △SQQ' = ¹⁄₂ times the magnitude of ($\mathbf\small{\vec{r}\times[\vec{v}(\Delta t)]}$)
See the solved example 7.11 at the beginning of section 7.14
6. We have: Linear momentum (p) = mass × Linear velocity
• So we get: $\mathbf\small{\vec{p}=\vec{v}(m)}$
$\mathbf\small{\Rightarrow \vec{v}=\frac{\vec{p}}{m}}$
7. So the result in (5) becomes:
Area of △SQQ' = $\mathbf\small{\frac{1}{2}\times\vec{r}\times \left [\frac{\vec{p}}{m}(\Delta t)\right ]}$
8. Dividing both sides by Δt, we get:
$\mathbf\small{\frac{\text{Area of △SQQ'}}{\Delta t}=\frac{1}{2}\times\vec{r}\times \left [\frac{\vec{p}}{m}\right ]}$
$\mathbf\small{\Rightarrow \frac{\text{Area of △SQQ'}}{\Delta t}=\frac{1}{2m}\times \left [\vec{r}\times{\vec{p}}\right ]}$
9. But $\mathbf\small{\left [\vec{r}\times{\vec{p}}\right ]=\vec{L}}$
Where $\mathbf\small{\vec{L}}$ is the angular momentum of the planet
• So the result in (8) becomes:
$\mathbf\small{\frac{\text{Area of △SQQ'}}{\Delta t}=\frac{\vec{L}}{2m}}$
10. If Δt is very small, the distance traveled by the planet in that time duration will be very small
• Then Q' will be very close to Q
• In such a situation, the chord QQ' will nearly coincide with the arc length from Q to Q'
• So the area of △SQQ' will be same as the area of the sector SQQ'
11. So, if Δt is very small, we can modify the result in (9):
$\mathbf\small{\frac{\text{Area of sector SQQ'}}{\Delta t}=\frac{\vec{L}}{2m}}$
• Let us denote the area of sector SQQ' by ΔA
• Then the result becomes:
$\mathbf\small{\frac{\Delta A}{\Delta t}=\frac{\vec{L}}{2m}}$
12. But the observations made by Kepler gives us the following information:
• If we consider the same 'intervals of time' the 'areas swept' will also be the same
    ♦ Same 'intervals of time' means: Δt is a constant
    ♦ 'Areas swept' are same means: ΔA is a constant
13. So in the result in (11), the left side is a constant
• Then right side must also be a constant
• On the right side, '2' and 'm' are already constants
• So $\mathbf\small{\vec{L}}$ must be a constant
14. Thus from the Kepler's second law, we get an important information:
■ The angular momentum of a planet always remains the same

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Let us see the implication of 'constant angular momentum'. It can be written in 6 steps:

1. In fig.8.5(a), the planet is at Q

Fig.8.5

• At that instant, it's position vector is $\mathbf\small{\vec{r}}$
• At that instant, it's velocity vector is $\mathbf\small{\vec{v}}$  
2. Resolve $\mathbf\small{\vec{v}}$ into two components as shown in fig.b
• One component is parallel to $\mathbf\small{\vec{r}}$. It is denoted as $\mathbf\small{\vec{v}_{\shortparallel}}$
• The other component is perpendicular to $\mathbf\small{\vec{r}}$. It is denoted as $\mathbf\small{\vec{v}_{\bot }}$
3. Let the angular momentum of the planet be $\mathbf\small{\vec{L}}$
We have: $\mathbf\small{\vec{L}=\vec{r}\;\times\;\vec{p}}$
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;m\vec{v}}$
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;m(\vec{v}_{\shortparallel}+\vec{v}_{\bot })}$ (∵ $\mathbf\small{\vec{v}}$ will be the vector sum of it's components)
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;(m\vec{v}_{\shortparallel}+m\vec{v}_{\bot })}$
$\mathbf\small{\Rightarrow \vec{L}=(\vec{r}\;\times\;m\vec{v}_{\shortparallel})+(\vec{r}\;\times\;m\vec{v}_{\bot })}$
$\mathbf\small{\Rightarrow \vec{L}=(\vec{0})+(\vec{r}\;\times\;m\vec{v}_{\bot })}$ (∵ cross product of two parallel vectors is a null vector)
$\mathbf\small{\Rightarrow \vec{L}=\vec{r}\;\times\;m\vec{v}_{\bot }}$
$\mathbf\small{\Rightarrow |\vec{L}|=m\;\times\;|\vec{r}|\;\times\;|\vec{v}_{\bot}|\;\times\;\sin 90}$ (∵ angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{v}_{\bot }}$ is 90^o)
$\mathbf\small{\Rightarrow |\vec{L}|=m\;\times\;|\vec{r}|\;\times\;|\vec{v}_{\bot}|}$ (∵ sin 90 is 1)
• Thus we get the magnitude of the angular momentum of the planet
4. We see that, the magnitude depends upon 3 items:
(i) The mass m 
(ii) The magnitude of the position vector
(iii) The magnitude of the velocity
5. Mass m is a constant
• Magnitude of the position vector is same as the 'distance of the planet from the sun'
• So we can write:
If the angular momentum is to remain constant, either one of the two changes must take place:
(i) When the 'distance from sun' decrease, the velocity must increase
(ii) When the 'distance from sun' increase, the velocity must decrease
6. Thus we find that:
• When the planet is near the perihelion, it's velocity increases
• When the planet is near the aphelion, it's velocity decreases

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• But this is not all. Kepler's second law, and it's explanation, leads to more questions and discoveries. This can be elaborated in 7 steps as follows:

1. Consider a particle in rotational motion
• We know that, the angular momentum of that particle remains constant if there is no net external torque acting on it
• That is: Angular momentum of that particle remains constant if $\mathbf\small{\vec{\tau}}$ = 0
2. The external torque is given by: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• We have:$\mathbf\small{|\vec{\tau}|=|\vec{r}|\times |\vec{F}|\times \sin \theta}$
Where θ is the angle between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{F}}$
3. So, if $\mathbf\small{\vec{\tau}}$ is to become zero, one of the two conditions must be satisfied:
(i) $\mathbf\small{|\vec{F}|}$ must be zero OR
(ii) θ must be zero or 180^o
4. We cannot consider (i) because, a force must be present. Otherwise the planet cannot move
• So we conclude that: The external torque is zero because, θ is zero or 180^o
5. θ is zero or 180^o implies that: The $\mathbf\small{\vec{F}}$ is acting along the same direction as $\mathbf\small{\vec{r}}$
• So it was concluded that, the 'force in action' is a central force
6. What is a central force ?
• The answer can be written in 3 steps:
(i) Suppose a force is acting at a point
(ii) Consider the line joining that point and the 'origin of the system'
(iii) If the force in (i) is acting along the line in (ii), then that force is a central force
We will learn more about central force in higher classes
7. In our present case, sun is the origin (That is why we draw the position vector $\mathbf\small{\vec{r}}$ from the sun) 
• So we can conclude that, a force is acting along the line joining the sun and the planet
■ Let us write a summary of the above 7 steps:
• An external force must act on a planet in order to keep that planet in it's orbit
• But this external force must not cause any change to the angular momentum of that planet
• This can be achieved only if the external force is a 'central force'

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• Thus Kepler's second law enabled scientists to reach an important milestone
• An 'important milestone' because, the presence of a central force was detected for the first time
• It was Sir Isaac Newton who finally discovered what this central force really is
• In the next section, we will see Kepler's third law

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