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
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)
• 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
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
• 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:
• 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
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
• 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
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
• 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
• Next we will see another notable discovery made by scientists of the seventeenth century. We will write it in 5 steps:
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)
• 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
• 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
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
• To fully understand this law, we must know some of the 'salient features of ellipse'
• 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 |
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
Let us see the important features of an ellipse:
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
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
• To fully understand this law, we can make use of fig.8.3 below:
• 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
• 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
• 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
♦ 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:
• 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
1. In fig.8.5(a), the planet is at Q
• 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 90o)
$\mathbf\small{\Rightarrow |\vec{L}|=m\;\times\;|\vec{r}|\;\times\;|\vec{v}_{\bot}|}$ (∵ 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
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 180o
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 180o
5. θ is zero or 180o 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'
 |
| Fig.8.3 |
• 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
• 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
• The second law is based on 'analysis of observations'
♦ Observations were made by Tycho Brahe
♦ Analysis of those observations were done by Kepler
♦ Observations were made by Tycho Brahe
♦ Analysis of those observations were done by Kepler
• 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
But this leads to more questions:
♦ 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 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
Let us see the implication of 'constant angular momentum'. It can be written in 6 steps:
 |
| 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 90o)
$\mathbf\small{\Rightarrow |\vec{L}|=m\;\times\;|\vec{r}|\;\times\;|\vec{v}_{\bot}|}$ (∵ 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
• 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:
• 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 180o
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 180o
5. θ is zero or 180o 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'
• 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
• 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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