In the previous section, we saw how the Cavendish's experiment is used to find the value of G. In this section, we will see the forces exerted by the earth on 'point masses at various positions'
1. In fig.8.26 (a) below, the earth is shown in blue color
• We want to find the force exerted by the earth on a point mass A which is shown in yellow color
• That is., we want $\mathbf\small{\vec{F}_{G(E,A)}}$
• This point mass A is situated at a distance of r from the center O of the earth
2. In fig.b, the earth is assumed to be made up of concentric spherical shells
• We will now calculate the 'force exerted on A' by each shell
• Note: In fig.b, only 4 shells are shown. Also, each shells have considerable thickness. This is for clarity only. In reality, we must assume a large number of shells. Each shell having a very small thickness
3. Force exerted by the first shell:
(i) Let the mass of the first shell be m1
• Based on what we have seen in section 8.6, we assume that, this mass is concentrated at the center O
• So we have a point mass m1 at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the first shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(1,A)}|=\frac{Gm_1m_A}{r^2}}$
4. Force exerted by the second shell:
(i) Let the mass of the second shell be m2
• We assume that this mass is concentrated at the center O
• So we have a point mass m2 at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the second shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(2,A)}|=\frac{Gm_2m_A}{r^2}}$
5. Force exerted by the third shell:
(i) Let the mass of the third shell be m3
• We assume that this mass is concentrated at the center O
• So we have a point mass m3 at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the third shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(3,A)}|=\frac{Gm_3m_A}{r^2}}$
6. Let there be a total of n shells
• Force exerted by the nth shell:
(i) Let the mass of the nth shell be mn
• We assume that this mass is concentrated at the center O
• So we have a point mass mn at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the nth shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(n,A)}|=\frac{Gm_nm_A}{r^2}}$
7. Total force magnitude acting on the point mass at A is given by the sum of all the n forces
• We can simply add all the magnitudes because, all those forces act in the same direction (along the line joining O and A)
• So the total force magnitude
= $\mathbf\small{|\vec{F}_{G(1,A)}|+|\vec{F}_{G(2,A)}|+|\vec{F}_{G(3,A)}|+\;.\;.\;.\;+|\vec{F}_{G(n,A)}|}$
= $\mathbf\small{\frac{Gm_1m_A}{r^2}+\frac{Gm_2m_A}{r^2}+\frac{Gm_3m_A}{r^2}+\;.\;.\;.\;+\frac{Gm_nm_A}{r^2}}$
8. $\mathbf\small{\frac{Gm_A}{r^2}}$ is common. So we can write:
• The total force magnitude
= $\mathbf\small{\frac{Gm_A}{r^2}\left(m_1+m_2+m_3+\;.\;.\;.\;+m_n\right)}$
• But $\mathbf\small{\left(m_1+m_2+m_3+\;.\;.\;.\;+m_n\right)}$ is the total mass ME of the earth
• Thus we get:
♦ The total force magnitude = $\mathbf\small{\frac{GM_E\,m_A}{r^2}}$
♦ The direction of this total magnitude will be along the 'line joining O and A'
9. So to find the force exerted by the Earth on a point mass situated at a distance r away from the center of the earth can be calculated using 4 steps:
(i) Assume that, the total mass ME of the Earth is concentrated at the center O of the earth
• So a point mass ME is present at O
(ii) Another point mass mA is situated at a distance of r from O
(iii) Then the magnitude of the force exerted by the earth on A is given by:
Eq.8.3: $\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
(iv) This force acts along the line joining O and A
(i) Gravitational force
(ii) Gravity
• We have already seen the 'gravitational force'. It is the force of attraction between any two bodies in the universe
• When this force is acting between earth and a point mass, we call that force: force of gravity or simply gravity
• To be more specific, we can write:
Gravity is the force with which the earth pulls a point mass towards it's center
10. We can now derive a very useful result
(i) In the fig.8.26, the force is acting between points O and A
• This is because all mass of earth is assumed to be concentrated at O
(ii) We can find the gravity on any external point mass A
• A is at a distance of r from O
(iii) What if A is exactly on the surface of the earth?
• This is shown in fig.c
(iv) We see that, r will become approximately equal to RE, the radius of the earth
(v) We have ME at O
• Also we have mA at a distance of RE from O
(vi) So the force exerted by ME on mA will be equal to $\mathbf\small{\frac{GM_E\,m_A}{R_E^2}}$
(vii) Thus we can write:
Eq.8.4:
Gravity on a point mass situated on the earth's surface:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
1. In fig.8.26 (a) below, the earth is shown in blue color
 |
| Fig.8.26 |
• That is., we want $\mathbf\small{\vec{F}_{G(E,A)}}$
• This point mass A is situated at a distance of r from the center O of the earth
2. In fig.b, the earth is assumed to be made up of concentric spherical shells
• We will now calculate the 'force exerted on A' by each shell
• Note: In fig.b, only 4 shells are shown. Also, each shells have considerable thickness. This is for clarity only. In reality, we must assume a large number of shells. Each shell having a very small thickness
3. Force exerted by the first shell:
(i) Let the mass of the first shell be m1
• Based on what we have seen in section 8.6, we assume that, this mass is concentrated at the center O
• So we have a point mass m1 at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the first shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(1,A)}|=\frac{Gm_1m_A}{r^2}}$
4. Force exerted by the second shell:
(i) Let the mass of the second shell be m2
• We assume that this mass is concentrated at the center O
• So we have a point mass m2 at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the second shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(2,A)}|=\frac{Gm_2m_A}{r^2}}$
5. Force exerted by the third shell:
(i) Let the mass of the third shell be m3
• We assume that this mass is concentrated at the center O
• So we have a point mass m3 at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the third shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(3,A)}|=\frac{Gm_3m_A}{r^2}}$
6. Let there be a total of n shells
• Force exerted by the nth shell:
(i) Let the mass of the nth shell be mn
• We assume that this mass is concentrated at the center O
• So we have a point mass mn at O and another point mass mA at a distance r from O
(ii) Then the magnitude of the 'force exerted by the nth shell' on the point mass A will be given by:
$\mathbf\small{|\vec{F}_{G(n,A)}|=\frac{Gm_nm_A}{r^2}}$
7. Total force magnitude acting on the point mass at A is given by the sum of all the n forces
• We can simply add all the magnitudes because, all those forces act in the same direction (along the line joining O and A)
• So the total force magnitude
= $\mathbf\small{|\vec{F}_{G(1,A)}|+|\vec{F}_{G(2,A)}|+|\vec{F}_{G(3,A)}|+\;.\;.\;.\;+|\vec{F}_{G(n,A)}|}$
= $\mathbf\small{\frac{Gm_1m_A}{r^2}+\frac{Gm_2m_A}{r^2}+\frac{Gm_3m_A}{r^2}+\;.\;.\;.\;+\frac{Gm_nm_A}{r^2}}$
8. $\mathbf\small{\frac{Gm_A}{r^2}}$ is common. So we can write:
• The total force magnitude
= $\mathbf\small{\frac{Gm_A}{r^2}\left(m_1+m_2+m_3+\;.\;.\;.\;+m_n\right)}$
• But $\mathbf\small{\left(m_1+m_2+m_3+\;.\;.\;.\;+m_n\right)}$ is the total mass ME of the earth
• Thus we get:
♦ The total force magnitude = $\mathbf\small{\frac{GM_E\,m_A}{r^2}}$
♦ The direction of this total magnitude will be along the 'line joining O and A'
9. So to find the force exerted by the Earth on a point mass situated at a distance r away from the center of the earth can be calculated using 4 steps:
(i) Assume that, the total mass ME of the Earth is concentrated at the center O of the earth
• So a point mass ME is present at O
(ii) Another point mass mA is situated at a distance of r from O
(iii) Then the magnitude of the force exerted by the earth on A is given by:
Eq.8.3: $\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
(iv) This force acts along the line joining O and A
■ At this stage, we can write the difference between two commonly used terms:
(ii) Gravity
• We have already seen the 'gravitational force'. It is the force of attraction between any two bodies in the universe
• When this force is acting between earth and a point mass, we call that force: force of gravity or simply gravity
• To be more specific, we can write:
Gravity is the force with which the earth pulls a point mass towards it's center
(i) In the fig.8.26, the force is acting between points O and A
• This is because all mass of earth is assumed to be concentrated at O
(ii) We can find the gravity on any external point mass A
• A is at a distance of r from O
(iii) What if A is exactly on the surface of the earth?
• This is shown in fig.c
(iv) We see that, r will become approximately equal to RE, the radius of the earth
(v) We have ME at O
• Also we have mA at a distance of RE from O
(vi) So the force exerted by ME on mA will be equal to $\mathbf\small{\frac{GM_E\,m_A}{R_E^2}}$
(vii) Thus we can write:
Eq.8.4:
Gravity on a point mass situated on the earth's surface:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
Point mass below the surface of the earth
1. In fig.8.27 (a) below, the earth is shown in blue color
• We want to find the gravity on a point mass A which is shown in yellow color
• That is., we want $\mathbf\small{\vec{F}_{G(E,A)}}$
2. A is situated below the surface of the earth
• The depth below the surface is d
• This d should be measured along the radius. This is indicated by the white dotted line from the center O
3. In fig.b, the earth is assumed to be made up of concentric spherical shells
• We will now calculate the 'force exerted on A' by each shell
• Note: In fig.b, only 5 shells are shown. Also, each shells have considerable thickness. This is for clarity only. In reality, we must assume a large number of shells. Each shell having a very small thickness
4. Force exerted by the first shell:
• A is inside the first shell. So the force exerted by the first shell is zero. We can ignore the first shell
5. Force exerted by the second shell:
• A is inside the second shell. So the force exerted by the second shell is zero. We can ignore the second shell
6. Force exerted by the third shell:
• A is inside the third shell. So the force exerted by the third shell is zero. We can ignore the third shell
7. Force exerted by the fourth shell
• A is outside the fourth shell. So this shell will exert a force
• We assume that, the mass m4 of this shell is concentrated at O
• Also note that, the distance OA is r
• So the force exerted by the fourth shell is $\mathbf\small{|\vec{F}_{G(4,A)}|=\frac{Gm_4m_A}{r^2}}$
8. In this way, we have to calculate the forces exerted by all the shells whose radii are less than r
• But 'all the shells whose radii are less than r' together constitute a single solid sphere of radius r
• Let Mr be the mass of that sphere
• Mr can be assumed to be concentrated at O
• So the force exerted by that sphere on A will be equal to $\mathbf\small{\frac{GM_r\,m_A}{r^2}}$
9. This can be used as a general result:
Gravity on a point mass mA situated at a depth d below the surface of the earth = $\mathbf\small{\frac{GM_r\,m_A}{r^2}}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ Mr = Mass of a 'concentric sphere of radius r' in the interior of the earth
10. So our next aim is to find Mr. This can be done in 2 steps:
(i) Let us assume that the density of the earth is uniform
• Let us denote this density as $\mathbf\small{\rho}$
(ii) Then we have:
• Mass of the earth
= Mass of the sphere of radius r
= Volume of the sphere of radius r × density
$\mathbf\small{\Rightarrow M_r=\frac{4 \pi r^3}{3}\times \rho}$
$\mathbf\small{\Rightarrow M_r=\frac{4 \pi \rho r^3}{3}}$
11. Now the result in (9) becomes:
• Gravity on a point mass mA situated at a depth d below the surface of the earth
= $\mathbf\small{G \left( \frac{4 \pi \rho r^3}{3}\right) m_A\times\frac{1}{r^2}}$
• Rearranging this, we get:
Eq.8.5:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ $\mathbf\small{\rho}$ = density of the earth
12. In the above result, consider the ratio $\mathbf\small{\left( \frac{4 \pi \rho }{3}\right)}$
• We have: Radius of the earth = $\mathbf\small{R_E}$
• Multiplying both numerator and denominator by $\mathbf\small{R_E^3}$, we get:
$\mathbf\small{\left( \frac{4 \pi \rho }{3}\right)=\left( \frac{4 \pi \rho }{3}\right)\frac{R_E^3}{R_E^3}=\left( \frac{4 \pi R_E^3\rho }{3}\right)\frac{1}{R_E^3}}$
• But $\mathbf\small{\left( \frac{4 \pi R_E^3\rho }{3}\right)}$ is mass of the earth ME
• So we get: $\mathbf\small{\left( \frac{4 \pi \rho }{3}\right)=\frac{M_E}{R_E^3}}$
13. So the result in (11) becomes:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{M_E}{R_E^3}\right)r}$
14. This can be rearranged and we get:
Eq.8.6:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ ME = Mass of the earth
15. We can now derive a very useful result. It can be written in just 2 steps:
(i) In fig.c, the point mass A is on the surface of the earth
• So r will become RE
(ii) Then using Eq.8.4, we get:
Gravity on a point mass mA situated on the surface of the earth
= $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^3}R_E=\frac{G\,M_E\,m_A}{R_E^2}}$
• This is the same result that we wrote as Eq.8.4 earlier in this section
Eq.8.4:
Magnitude of the gravity on a point mass mA situated on the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
Eq.8.5:
Magnitude of the gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
OR
Eq.8.6:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
1. In fig.8.27 (a) below, the earth is shown in blue color
 |
| Fig.8.27 |
• That is., we want $\mathbf\small{\vec{F}_{G(E,A)}}$
2. A is situated below the surface of the earth
• The depth below the surface is d
• This d should be measured along the radius. This is indicated by the white dotted line from the center O
3. In fig.b, the earth is assumed to be made up of concentric spherical shells
• We will now calculate the 'force exerted on A' by each shell
• Note: In fig.b, only 5 shells are shown. Also, each shells have considerable thickness. This is for clarity only. In reality, we must assume a large number of shells. Each shell having a very small thickness
4. Force exerted by the first shell:
• A is inside the first shell. So the force exerted by the first shell is zero. We can ignore the first shell
5. Force exerted by the second shell:
• A is inside the second shell. So the force exerted by the second shell is zero. We can ignore the second shell
6. Force exerted by the third shell:
• A is inside the third shell. So the force exerted by the third shell is zero. We can ignore the third shell
7. Force exerted by the fourth shell
• A is outside the fourth shell. So this shell will exert a force
• We assume that, the mass m4 of this shell is concentrated at O
• Also note that, the distance OA is r
• So the force exerted by the fourth shell is $\mathbf\small{|\vec{F}_{G(4,A)}|=\frac{Gm_4m_A}{r^2}}$
8. In this way, we have to calculate the forces exerted by all the shells whose radii are less than r
• But 'all the shells whose radii are less than r' together constitute a single solid sphere of radius r
• Let Mr be the mass of that sphere
• Mr can be assumed to be concentrated at O
• So the force exerted by that sphere on A will be equal to $\mathbf\small{\frac{GM_r\,m_A}{r^2}}$
9. This can be used as a general result:
Gravity on a point mass mA situated at a depth d below the surface of the earth = $\mathbf\small{\frac{GM_r\,m_A}{r^2}}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ Mr = Mass of a 'concentric sphere of radius r' in the interior of the earth
10. So our next aim is to find Mr. This can be done in 2 steps:
(i) Let us assume that the density of the earth is uniform
• Let us denote this density as $\mathbf\small{\rho}$
(ii) Then we have:
• Mass of the earth
= Mass of the sphere of radius r
= Volume of the sphere of radius r × density
$\mathbf\small{\Rightarrow M_r=\frac{4 \pi r^3}{3}\times \rho}$
$\mathbf\small{\Rightarrow M_r=\frac{4 \pi \rho r^3}{3}}$
11. Now the result in (9) becomes:
• Gravity on a point mass mA situated at a depth d below the surface of the earth
= $\mathbf\small{G \left( \frac{4 \pi \rho r^3}{3}\right) m_A\times\frac{1}{r^2}}$
• Rearranging this, we get:
Eq.8.5:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ $\mathbf\small{\rho}$ = density of the earth
12. In the above result, consider the ratio $\mathbf\small{\left( \frac{4 \pi \rho }{3}\right)}$
• We have: Radius of the earth = $\mathbf\small{R_E}$
• Multiplying both numerator and denominator by $\mathbf\small{R_E^3}$, we get:
$\mathbf\small{\left( \frac{4 \pi \rho }{3}\right)=\left( \frac{4 \pi \rho }{3}\right)\frac{R_E^3}{R_E^3}=\left( \frac{4 \pi R_E^3\rho }{3}\right)\frac{1}{R_E^3}}$
• But $\mathbf\small{\left( \frac{4 \pi R_E^3\rho }{3}\right)}$ is mass of the earth ME
• So we get: $\mathbf\small{\left( \frac{4 \pi \rho }{3}\right)=\frac{M_E}{R_E^3}}$
13. So the result in (11) becomes:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{M_E}{R_E^3}\right)r}$
14. This can be rearranged and we get:
Eq.8.6:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ ME = Mass of the earth
15. We can now derive a very useful result. It can be written in just 2 steps:
(i) In fig.c, the point mass A is on the surface of the earth
• So r will become RE
(ii) Then using Eq.8.4, we get:
Gravity on a point mass mA situated on the surface of the earth
= $\mathbf\small{\frac{G\,M_E\,m_A}{R_E^3}R_E=\frac{G\,M_E\,m_A}{R_E^2}}$
• This is the same result that we wrote as Eq.8.4 earlier in this section
So we have successfully calculated the 'gravity experience by a point mass' when it is at various positions. There are three possible positions. We can write a summary:
Eq.8.3:
Magnitude of the gravity on a point mass mA at any distance r:
$\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
Eq.8.3:
Magnitude of the gravity on a point mass mA at any distance r:
$\mathbf\small{|\vec{F}_{G(E,A)}|=\frac{GM_E\,m_A}{r^2}}$
Magnitude of the gravity on a point mass mA situated on the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{surface})}|=\frac{GM_E\,m_A}{R_E^2}}$
Eq.8.5:
Magnitude of the gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=G\,m_A \left( \frac{4 \pi \rho }{3}\right)r}$
OR
Eq.8.6:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• In the next section, we will see the acceleration at the three positions
No comments:
Post a Comment