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:
• 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 RM. 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 (RM) 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 = 2πRM = 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: $\mathbf\small{|\vec{F}_{c(M)}|=\frac{m_M|\vec{v}|^2}{R_M}}$
(mM 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 mO
• 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 mO, 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
♦ RE denote the radius of the Earth. It is equal to 6.378 × 106 m
♦ RM denote the distance from the center of the Earth to the center of the Moon
It is equal to 3.84 × 108 m
♦ mO denote the mass of the object
♦ mM 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 RM and RE, 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
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:
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.
1. Consider the rotation of the Moon around the Earth. It is shown in fig.8.8 below:
 |
| Fig.8.8 |
• The radius of this orbit is RM. 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 (RM) 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 = 2πRM = 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: $\mathbf\small{|\vec{F}_{c(M)}|=\frac{m_M|\vec{v}|^2}{R_M}}$
(mM 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 mO
• 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 mO, 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
♦ RE denote the radius of the Earth. It is equal to 6.378 × 106 m
♦ RM denote the distance from the center of the Earth to the center of the Moon
It is equal to 3.84 × 108 m
♦ mO denote the mass of the object
♦ mM 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 RM and RE, 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
■ Newton proposed the Universal Law of Gravitation. It states that:
• 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 |
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.
• 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
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
♦ mE and mS are the masses of the Earth and the Sun respectively
♦ aE 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 TE 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
♦ mJ and mS are the masses of the Jupiter and the Sun respectively
♦ aJ 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 TJ 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
♦ mI and mJ are the masses of Io and Jupiter respectively
♦ aI is the radius of Io's orbit
♦ TI 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
$\mathbf\small{\Rightarrow \frac{m_J}{m_S}=0.000957}$
$\mathbf\small{\Rightarrow m_S= \frac{m_J}{0.000957}}$
$\mathbf\small{\Rightarrow m_S=1045\;m_J}$
• 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
Solved example 8.5
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
♦ mE and mS are the masses of the Earth and the Sun respectively
♦ aE 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 TE 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
♦ mJ and mS are the masses of the Jupiter and the Sun respectively
♦ aJ 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 TJ 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
♦ mI and mJ are the masses of Io and Jupiter respectively
♦ aI is the radius of Io's orbit
♦ TI 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
$\mathbf\small{\Rightarrow \frac{m_J}{m_S}=0.000957}$
$\mathbf\small{\Rightarrow m_S= \frac{m_J}{0.000957}}$
$\mathbf\small{\Rightarrow m_S=1045\;m_J}$
In the next section, we will see direction
