In the previous section, we completed a discussion on the basics of gravitational force, gravity and acceleration due to gravity. We saw some solved examples also. In this section, we will see gravitational field
Gravitational field can be defined in 4 steps:
1. Consider a body A
2. There will be a 'particular space' around A
3. If any other body B enters this 'particular space', then B will experience a gravitational force of attraction towards A
4. This 'particular space' is called 'gravitational field of A'
5. So 'gravitational field' is a space
Next we have to define Intensity of gravitational field. It can be defined in 10 steps:
1. Consider a body A
2. There will be a gravitational field around A
3. If any other body B enters this gravitational field of A, then B will experience a gravitational force of attraction towards A
4. Suppose that B has a mass of 1 kg
5. Then the force experienced by B is called 'Intensity of gravitational field of A'
This is shown in fig.2.28 below:
6. So 'intensity of gravitational field' is a force
It's magnitude is given by: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A\times 1}{r^2}}$
• 'E' denotes the field
• The subscript G indicates that, the field is due to gravitational force
• The subscript A denotes the body which creates the field
• There is no need to indicate the second body B because, it will always be a unit (1 kg) mass
The Intensity of gravitational field of a body A is given by:
Eq.8.12: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{r^2}}$
• Where
♦ $\mathbf\small{|\vec{E}_{G(A)}|}$ is the Intensity of gravitational field of a body A
♦ mA is the mass of A
♦ r is the distance between the center of A and the center of the 1 kg mass (unit mass) B
• Direction of this force is along the line joining the centers of the two masses
7. We have to take care of a few points:
(i) When we bring the unit mass B into the field of A, there will be attraction
• As a result, B will move towards A
(ii) When B moves towards A, the distance r continuously decreases
(iii) When r decreases, the field continuously increases
8. We can give a 'value of the intensity' only if the field remains unchanged
• So we must take precautions to avoid such a change in the field
9. For that, we must provide force in the opposite direction so that, B does not move towards A
• Also, we must bring the unit mass B very slowly into the field. Advancing only very small distance in each second
• That is., we must bring B without any acceleration
• If B experience any acceleration during this movement, the corresponding forces will also come in the calculations. So we will not get the accurate value of the intensity
10. In this way, taking all precautions, we will be able to measure $\mathbf\small{|\vec{E}_{G(A)}|}$
• So, if $\mathbf\small{|\vec{F}_G|}$ is the gravitational force experienced by a body of mass m, we can write:
Eq.8.13: $\mathbf\small{|\vec{E}_G|=\frac{|\vec{F}_G|}{m}}$
• We have already seen how to calculate $\mathbf\small{|\vec{F}_G|}$ for various positions. So we can easily calculate $\mathbf\small{|\vec{E}_G|}$
Solved example 8.23
Derive the unit of the Intensity of gravitational field
Solution:
1. We have:
The Intensity of gravitational field of a body A is given by:
Eq.8.12: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{r^2}}$
• Where
♦ $\mathbf\small{|\vec{E}_{G(A)}|}$ is the Intensity of gravitational field of a body A
♦ mA is the mass of A
♦ r is the distance between the center of A and the center of the 1 kg mass (unit mass) B
2. Substituting the units on the right side, we get:
$\mathbf\small{|\vec{E}_{G(A)}|=\frac{\rm{(N\,m^2\,kg^{-2})(kg)}}{\rm{(m)^2}}=\rm{N\,kg^{-1}}}$
3. So the unit if intensity of gravitational field is N kg-1
Another method:
• We have:
Eq.8.13: $\mathbf\small{|\vec{E}_G|=\frac{|\vec{F}_G|}{m}}$
• So obviously, the unit is N kg-1
Note:
• N kg-1 can be further split up as: [(kg m s-2) (kg-1)] = [m s-2]
• So we can write:
'Intensity of gravitational field' and 'acceleration' are dimensionally equal
Solved example 8.24
Two masses 1000 kg and 10 kg are at a distance of 1 m apart. At which point on the line joining their centers, will the gravitational field intensity be zero?
Solution:
1. In fig.8.29 below, mA (=1000 kg) and mB (=10 kg) are placed at a distance of (r=1 m) apart
2. Assume that mC (=1 kg) is placed at a distance of x from A
• Then force experienced by C due to A = $\mathbf\small{|\vec{F}_{G(A,C)}|=\frac{G\,\,m_A\,m_C}{r^2}=\frac{G\times 1000 \times 1}{x^2}=\frac{1000G}{x^2}}$
3. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{1000G}{x^2}}$
• This intensity acts from C towards the left
4. Similarly, the force experienced by C due to B = $\mathbf\small{|\vec{F}_{G(B,C)}|=\frac{G\,\,m_B\,m_C}{r^2}=\frac{G\times 10 \times 1}{(1-x)^2}=\frac{10G}{(1-x)^2}}$
5. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(B)}|=\frac{10G}{(1-x)^2}}$
• This intensity acts from C towards the right
6. We want C to experience zero net intensity
• For that, intensity towards the left must be equal to the intensity towards the right
• So we can write: $\mathbf\small{|\vec{E}_{G(A)}|=|\vec{E}_{G(B)}|}$
$\mathbf\small{\Rightarrow \frac{1000G}{x^2}=\frac{10G}{(1-x)^2}}$
$\mathbf\small{\Rightarrow \frac{100}{x^2}=\frac{1}{(1-x)^2}}$
$\mathbf\small{\Rightarrow \frac{10}{x}=\frac{1}{(1-x)}}$
$\mathbf\small{\Rightarrow 10-10x=x}$
$\mathbf\small{\Rightarrow x=\frac{10}{11}\,\rm{m}}$
7. So we can write:
The net intensity will be zero at a distance of $\mathbf\small{\frac{10}{11}\,\rm{m}}$ from A
Solved example 8.25
Find the distance of the point where gravitational field due to earth and moon cancel each other. Given that:
(i) Mass of the moon is $\mathbf\small{\frac{1}{81}}$ times the mass of the earth
(ii) Distance between earth and moon is 60 times the radius of the earth
Solution:
1. In fig.8.30 below, ME and MM are at a distance of 60RE apart
2. Assume that mC (=1 kg) is placed at a distance of x from earth
• Then force experienced by C due to earth = $\mathbf\small{|\vec{F}_{G(E,C)}|=\frac{G\,\,M_E\,m_C}{r^2}=\frac{G\times M_E \times 1}{x^2}=\frac{G\,M_E}{x^2}}$
3. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(E)}|=\frac{G\,M_E}{x^2}}$
• This intensity acts from C towards the earth
4. Similarly, the force experienced by C due to moon = $\mathbf\small{|\vec{F}_{G(M,C)}|=\frac{G\,\,M_M\,m_C}{r^2}=\frac{G\times M_M \times 1}{(60\,R_E-x)^2}=\frac{G\,M_M}{(60\,R_E-x)^2}}$
5. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(M)}|=\frac{G\,M_M}{(60\,R_E-x)^2}}$
• This intensity acts from C towards the moon
6. We want C to experience zero net intensity
• For that, intensity towards the earth must be equal to the intensity towards the moon
• So we can write: $\mathbf\small{|\vec{E}_{G(E)}|=|\vec{E}_{G(M)}|}$
$\mathbf\small{\Rightarrow \frac{G\,M_E}{x^2}=\frac{G\,M_M}{(60\,R_E-x)^2}}$
$\mathbf\small{\Rightarrow \frac{M_E}{x^2}=\frac{G\,\frac{M_E}{81}}{(60\,R_E-x)^2}}$
$\mathbf\small{\Rightarrow \frac{1}{x^2}=\frac{1}{81(60\,R_E-x)^2}}$
$\mathbf\small{\Rightarrow \frac{1}{x}=\frac{1}{\pm 9(60\,R_E-x)}}$
• Taking '+9', we get:
$\mathbf\small{540R_E-9x =x}$
$\mathbf\small{\Rightarrow 540R_E =10x}$
$\mathbf\small{\Rightarrow x=54R_E}$
• Taking '-9', we get:
$\mathbf\small{-540R_E+9x =x}$
$\mathbf\small{\Rightarrow 540R_E =8x}$
$\mathbf\small{\Rightarrow x=67.5R_E}$
7. So we have two distances: 54RE and 67.5RE
• But the distance between earth and the moon is only 60RE
• So the acceptable answer is 54RE
Solved example 8.26
In fig.8.31(a) below, two solid spheres A and B of masses 800 kg and 600 kg respectively are placed at a center to center distance of 0.25 m. Calculate the intensity of the gravitational field at a point C. Given that, C is 0.2 m from the center of A and 0.15 m from the center of B
Solution:
1. Assume that mC (=1 kg) is placed at C
• Then force experienced by C due to A = $\mathbf\small{|\vec{F}_{G(A,C)}|=\frac{G\,\,m_A\,m_C}{r^2}=\frac{G\times 800 \times 1}{0.2^2}=20000G}$
2. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(A)}|=20000G}$
• This intensity acts from C towards A
3. Similarly, the force experienced by C due to B = $\mathbf\small{|\vec{F}_{G(B,C)}|=\frac{G\,\,m_B\,m_C}{r^2}=\frac{G\times 600 \times 1}{0.15^2}=\frac{8000G}{3}}$
4. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(B)}|=\frac{8000G}{3}}$
• This intensity acts from C towards B
5. So the point C is acted upon by two intensities: $\mathbf\small{|\vec{E}_{G(A)}|}$ and $\mathbf\small{|\vec{E}_{G(B)}|}$
• We must calculate the resultant of those two
• Considering the squares of the sides, we see that: 0.22 + 0.152 = 0.252
• So triangle ABC is a right angled triangle. AB is the hypotenuse
• The point C, which is opposite to the hypotenuse, is the 90o vertex
6. Thus we can write:
• The two intensities at point C, are perpendicular to each other. This is shown in fig.8.31(b)
• Since they are perpendicular to each other, we can easily calculate the magnitude of the resultant intensity:
$\mathbf\small{|\vec{E}_{G_{resultant}}|=\sqrt{|\vec{E}_{G(A)}|^2+|\vec{E}_{G(B)}|^2}}$
= $\mathbf\small{\sqrt{(20000G)^2+\left(\frac{8000G}{3}\right)^2}}$ = 20177G N kg
7. The direction can be specified as follows:
• The two intensities $\mathbf\small{|\vec{E}_{G(A)}|}$ and $\mathbf\small{|\vec{E}_{G(B)}|}$ form the adjacent sides of a rectangle. The resultant $\mathbf\small{|\vec{E}_{G_{resultant}}|}$ will be the diagonal of that rectangle
• Let this diagonal make an angle 𝛼 with $\mathbf\small{|\vec{E}_{G(B)}|}$. This is shown in fig.b
• We get: ${}\mathbf\small{\tan \alpha=\frac{|\vec{E}_{G(A)}|}{|\vec{E}_{G(B)}|}=\frac{20000G}{\frac{8000G}{3}}=7.5}$
⇒ 𝛼 = 82.4o
Solved example 8.27
Derive an expression for the intensity of gravitational field at a point in the interior of a uniform solid sphere
Solution:
1. Earlier, we derived Eq.8.5:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ ME = Mass of the earth
2. The earth is assumed to be a uniform solid sphere. So we can use this equation
3. In our present case, we have a sphere A of mass mA and radius R
• Also we have a point mass mB at a depth d below the surface of the sphere
• Then the magnitude of the force exerted by the sphere A on mB is given by:
$\mathbf\small{|\vec{F}_{G(A,B_{depth})}|=\frac{G\,m_A\,m_B}{R^3}r}$
• Where:
♦ r = R - d
♦ R = Radius of the sphere A
♦ mA = Mass of the sphere A
♦ mB = Point mass B
4. So we obtained the force experienced by B
• If the mass mB is 1 kg, that force can be considered as the intensity
• So we get:
Intensity of gravitational field at a depth d below the surface of a sphere A:
$\mathbf\small{|\vec{E}_{G(A_{depth})}|=\frac{G\,m_A}{R^3}r}$
• Where:
♦ r = R - d
♦ R = Radius of the sphere A
♦ mA = Mass of the sphere A
♦ mB = Point mass B
Solved example 8.28
Gravitational field at the surface of a solid sphere A is 1.5 × 10-4 N kg-1. Radius of this sphere is R. A point P is marked in the interior of the sphere, at a distance of 0.5R from the center. What is the intensity of gravitational field at P due to the sphere?
Solution:
1. From the previous solved example 8.26, we have:
Intensity of gravitational field at a depth d below the surface of a sphere A:
$\mathbf\small{|\vec{E}_{G(A_{depth})}|=\frac{G\,m_A}{R^3}r}$
• Where:
♦ r = R - d
♦ R = Radius of the sphere A
♦ mA = Mass of the sphere A
♦ mB = Point mass B
2. If we assume d to be zero, we will get the intensity at the surface
• For that, we have to put R in place of r
• So we get: $\mathbf\small{|\vec{E}_{G(A_{surface})}|=\frac{G\,m_A}{R^3}R=\frac{G\,m_A}{R^2}}$
• So we can write: $\mathbf\small{\frac{G\,m_A}{R^2}=1.5\times 10^{-4}\,\rm{N\,m^{-1}}}$
3. Consider the expression in (1) again. We want the intensity at 0.5R
• We get: $\mathbf\small{|\vec{E}_{G(A_{depth=0.5R})}|=\frac{G\,m_A}{R^3}0.5R=\frac{0.5G\,m_A}{R^2}}$
4. But from (2), we know the value of $\mathbf\small{\frac{G\,m_A}{R^2}}$
• So the result in (3) becomes:
$\mathbf\small{|\vec{E}_{G(A_{depth=0.5R})}|=0.5 \times 1.5 \times 10^{-4}=0.75\times 10^{-4}\,\rm{N\,m^{-2}}}$
Gravitational field can be defined in 4 steps:
1. Consider a body A
2. There will be a 'particular space' around A
3. If any other body B enters this 'particular space', then B will experience a gravitational force of attraction towards A
4. This 'particular space' is called 'gravitational field of A'
5. So 'gravitational field' is a space
Next we have to define Intensity of gravitational field. It can be defined in 10 steps:
1. Consider a body A
2. There will be a gravitational field around A
3. If any other body B enters this gravitational field of A, then B will experience a gravitational force of attraction towards A
4. Suppose that B has a mass of 1 kg
5. Then the force experienced by B is called 'Intensity of gravitational field of A'
This is shown in fig.2.28 below:
 |
| Fig.8.28 |
It's magnitude is given by: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A\times 1}{r^2}}$
■ Note the convention for writing the intensity vector:
• The subscript G indicates that, the field is due to gravitational force
• The subscript A denotes the body which creates the field
• There is no need to indicate the second body B because, it will always be a unit (1 kg) mass
So we can write:
Eq.8.12: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{r^2}}$
• Where
♦ $\mathbf\small{|\vec{E}_{G(A)}|}$ is the Intensity of gravitational field of a body A
♦ mA is the mass of A
♦ r is the distance between the center of A and the center of the 1 kg mass (unit mass) B
• Direction of this force is along the line joining the centers of the two masses
7. We have to take care of a few points:
(i) When we bring the unit mass B into the field of A, there will be attraction
• As a result, B will move towards A
(ii) When B moves towards A, the distance r continuously decreases
(iii) When r decreases, the field continuously increases
8. We can give a 'value of the intensity' only if the field remains unchanged
• So we must take precautions to avoid such a change in the field
9. For that, we must provide force in the opposite direction so that, B does not move towards A
• Also, we must bring the unit mass B very slowly into the field. Advancing only very small distance in each second
• That is., we must bring B without any acceleration
• If B experience any acceleration during this movement, the corresponding forces will also come in the calculations. So we will not get the accurate value of the intensity
10. In this way, taking all precautions, we will be able to measure $\mathbf\small{|\vec{E}_{G(A)}|}$
• We see that, this intensity is defined as the force per unit mass
Eq.8.13: $\mathbf\small{|\vec{E}_G|=\frac{|\vec{F}_G|}{m}}$
• We have already seen how to calculate $\mathbf\small{|\vec{F}_G|}$ for various positions. So we can easily calculate $\mathbf\small{|\vec{E}_G|}$
Now we will see some solved examples:
Solved example 8.23
Derive the unit of the Intensity of gravitational field
Solution:
1. We have:
The Intensity of gravitational field of a body A is given by:
Eq.8.12: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{G\,\,m_A}{r^2}}$
• Where
♦ $\mathbf\small{|\vec{E}_{G(A)}|}$ is the Intensity of gravitational field of a body A
♦ mA is the mass of A
♦ r is the distance between the center of A and the center of the 1 kg mass (unit mass) B
2. Substituting the units on the right side, we get:
$\mathbf\small{|\vec{E}_{G(A)}|=\frac{\rm{(N\,m^2\,kg^{-2})(kg)}}{\rm{(m)^2}}=\rm{N\,kg^{-1}}}$
3. So the unit if intensity of gravitational field is N kg-1
Another method:
• We have:
Eq.8.13: $\mathbf\small{|\vec{E}_G|=\frac{|\vec{F}_G|}{m}}$
• So obviously, the unit is N kg-1
Note:
• N kg-1 can be further split up as: [(kg m s-2) (kg-1)] = [m s-2]
• So we can write:
'Intensity of gravitational field' and 'acceleration' are dimensionally equal
Solved example 8.24
Two masses 1000 kg and 10 kg are at a distance of 1 m apart. At which point on the line joining their centers, will the gravitational field intensity be zero?
Solution:
1. In fig.8.29 below, mA (=1000 kg) and mB (=10 kg) are placed at a distance of (r=1 m) apart
 |
| Fig.8.29 |
• Then force experienced by C due to A = $\mathbf\small{|\vec{F}_{G(A,C)}|=\frac{G\,\,m_A\,m_C}{r^2}=\frac{G\times 1000 \times 1}{x^2}=\frac{1000G}{x^2}}$
3. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(A)}|=\frac{1000G}{x^2}}$
• This intensity acts from C towards the left
4. Similarly, the force experienced by C due to B = $\mathbf\small{|\vec{F}_{G(B,C)}|=\frac{G\,\,m_B\,m_C}{r^2}=\frac{G\times 10 \times 1}{(1-x)^2}=\frac{10G}{(1-x)^2}}$
5. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(B)}|=\frac{10G}{(1-x)^2}}$
• This intensity acts from C towards the right
6. We want C to experience zero net intensity
• For that, intensity towards the left must be equal to the intensity towards the right
• So we can write: $\mathbf\small{|\vec{E}_{G(A)}|=|\vec{E}_{G(B)}|}$
$\mathbf\small{\Rightarrow \frac{1000G}{x^2}=\frac{10G}{(1-x)^2}}$
$\mathbf\small{\Rightarrow \frac{100}{x^2}=\frac{1}{(1-x)^2}}$
$\mathbf\small{\Rightarrow \frac{10}{x}=\frac{1}{(1-x)}}$
$\mathbf\small{\Rightarrow 10-10x=x}$
$\mathbf\small{\Rightarrow x=\frac{10}{11}\,\rm{m}}$
7. So we can write:
The net intensity will be zero at a distance of $\mathbf\small{\frac{10}{11}\,\rm{m}}$ from A
Solved example 8.25
Find the distance of the point where gravitational field due to earth and moon cancel each other. Given that:
(i) Mass of the moon is $\mathbf\small{\frac{1}{81}}$ times the mass of the earth
(ii) Distance between earth and moon is 60 times the radius of the earth
Solution:
1. In fig.8.30 below, ME and MM are at a distance of 60RE apart
 |
| Fig.8.30 |
• Then force experienced by C due to earth = $\mathbf\small{|\vec{F}_{G(E,C)}|=\frac{G\,\,M_E\,m_C}{r^2}=\frac{G\times M_E \times 1}{x^2}=\frac{G\,M_E}{x^2}}$
3. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(E)}|=\frac{G\,M_E}{x^2}}$
• This intensity acts from C towards the earth
4. Similarly, the force experienced by C due to moon = $\mathbf\small{|\vec{F}_{G(M,C)}|=\frac{G\,\,M_M\,m_C}{r^2}=\frac{G\times M_M \times 1}{(60\,R_E-x)^2}=\frac{G\,M_M}{(60\,R_E-x)^2}}$
5. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(M)}|=\frac{G\,M_M}{(60\,R_E-x)^2}}$
• This intensity acts from C towards the moon
6. We want C to experience zero net intensity
• For that, intensity towards the earth must be equal to the intensity towards the moon
• So we can write: $\mathbf\small{|\vec{E}_{G(E)}|=|\vec{E}_{G(M)}|}$
$\mathbf\small{\Rightarrow \frac{G\,M_E}{x^2}=\frac{G\,M_M}{(60\,R_E-x)^2}}$
$\mathbf\small{\Rightarrow \frac{M_E}{x^2}=\frac{G\,\frac{M_E}{81}}{(60\,R_E-x)^2}}$
$\mathbf\small{\Rightarrow \frac{1}{x^2}=\frac{1}{81(60\,R_E-x)^2}}$
$\mathbf\small{\Rightarrow \frac{1}{x}=\frac{1}{\pm 9(60\,R_E-x)}}$
• Taking '+9', we get:
$\mathbf\small{540R_E-9x =x}$
$\mathbf\small{\Rightarrow 540R_E =10x}$
$\mathbf\small{\Rightarrow x=54R_E}$
• Taking '-9', we get:
$\mathbf\small{-540R_E+9x =x}$
$\mathbf\small{\Rightarrow 540R_E =8x}$
$\mathbf\small{\Rightarrow x=67.5R_E}$
7. So we have two distances: 54RE and 67.5RE
• But the distance between earth and the moon is only 60RE
• So the acceptable answer is 54RE
Solved example 8.26
In fig.8.31(a) below, two solid spheres A and B of masses 800 kg and 600 kg respectively are placed at a center to center distance of 0.25 m. Calculate the intensity of the gravitational field at a point C. Given that, C is 0.2 m from the center of A and 0.15 m from the center of B
 |
| Fig.8.31 |
1. Assume that mC (=1 kg) is placed at C
• Then force experienced by C due to A = $\mathbf\small{|\vec{F}_{G(A,C)}|=\frac{G\,\,m_A\,m_C}{r^2}=\frac{G\times 800 \times 1}{0.2^2}=20000G}$
2. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(A)}|=20000G}$
• This intensity acts from C towards A
3. Similarly, the force experienced by C due to B = $\mathbf\small{|\vec{F}_{G(B,C)}|=\frac{G\,\,m_B\,m_C}{r^2}=\frac{G\times 600 \times 1}{0.15^2}=\frac{8000G}{3}}$
4. But mass of C is 1 kg. So the above force is actually the intensity
• We can write: $\mathbf\small{|\vec{E}_{G(B)}|=\frac{8000G}{3}}$
• This intensity acts from C towards B
5. So the point C is acted upon by two intensities: $\mathbf\small{|\vec{E}_{G(A)}|}$ and $\mathbf\small{|\vec{E}_{G(B)}|}$
• We must calculate the resultant of those two
• Considering the squares of the sides, we see that: 0.22 + 0.152 = 0.252
• So triangle ABC is a right angled triangle. AB is the hypotenuse
• The point C, which is opposite to the hypotenuse, is the 90o vertex
6. Thus we can write:
• The two intensities at point C, are perpendicular to each other. This is shown in fig.8.31(b)
• Since they are perpendicular to each other, we can easily calculate the magnitude of the resultant intensity:
$\mathbf\small{|\vec{E}_{G_{resultant}}|=\sqrt{|\vec{E}_{G(A)}|^2+|\vec{E}_{G(B)}|^2}}$
= $\mathbf\small{\sqrt{(20000G)^2+\left(\frac{8000G}{3}\right)^2}}$ = 20177G N kg
7. The direction can be specified as follows:
• The two intensities $\mathbf\small{|\vec{E}_{G(A)}|}$ and $\mathbf\small{|\vec{E}_{G(B)}|}$ form the adjacent sides of a rectangle. The resultant $\mathbf\small{|\vec{E}_{G_{resultant}}|}$ will be the diagonal of that rectangle
• Let this diagonal make an angle 𝛼 with $\mathbf\small{|\vec{E}_{G(B)}|}$. This is shown in fig.b
• We get: ${}\mathbf\small{\tan \alpha=\frac{|\vec{E}_{G(A)}|}{|\vec{E}_{G(B)}|}=\frac{20000G}{\frac{8000G}{3}}=7.5}$
⇒ 𝛼 = 82.4o
Solved example 8.27
Derive an expression for the intensity of gravitational field at a point in the interior of a uniform solid sphere
Solution:
1. Earlier, we derived Eq.8.5:
Gravity on a point mass mA situated at a depth d below the surface of the earth:
$\mathbf\small{|\vec{F}_{G(E,A_{depth})}|=\frac{G\,M_E\,m_A}{R_E^3}r}$
• Where:
♦ r = RE - d
♦ RE = Radius of the earth
♦ ME = Mass of the earth
2. The earth is assumed to be a uniform solid sphere. So we can use this equation
3. In our present case, we have a sphere A of mass mA and radius R
• Also we have a point mass mB at a depth d below the surface of the sphere
• Then the magnitude of the force exerted by the sphere A on mB is given by:
$\mathbf\small{|\vec{F}_{G(A,B_{depth})}|=\frac{G\,m_A\,m_B}{R^3}r}$
• Where:
♦ r = R - d
♦ R = Radius of the sphere A
♦ mA = Mass of the sphere A
♦ mB = Point mass B
4. So we obtained the force experienced by B
• If the mass mB is 1 kg, that force can be considered as the intensity
• So we get:
Intensity of gravitational field at a depth d below the surface of a sphere A:
$\mathbf\small{|\vec{E}_{G(A_{depth})}|=\frac{G\,m_A}{R^3}r}$
• Where:
♦ r = R - d
♦ R = Radius of the sphere A
♦ mA = Mass of the sphere A
♦ mB = Point mass B
Solved example 8.28
Gravitational field at the surface of a solid sphere A is 1.5 × 10-4 N kg-1. Radius of this sphere is R. A point P is marked in the interior of the sphere, at a distance of 0.5R from the center. What is the intensity of gravitational field at P due to the sphere?
Solution:
1. From the previous solved example 8.26, we have:
Intensity of gravitational field at a depth d below the surface of a sphere A:
$\mathbf\small{|\vec{E}_{G(A_{depth})}|=\frac{G\,m_A}{R^3}r}$
• Where:
♦ r = R - d
♦ R = Radius of the sphere A
♦ mA = Mass of the sphere A
♦ mB = Point mass B
2. If we assume d to be zero, we will get the intensity at the surface
• For that, we have to put R in place of r
• So we get: $\mathbf\small{|\vec{E}_{G(A_{surface})}|=\frac{G\,m_A}{R^3}R=\frac{G\,m_A}{R^2}}$
• So we can write: $\mathbf\small{\frac{G\,m_A}{R^2}=1.5\times 10^{-4}\,\rm{N\,m^{-1}}}$
3. Consider the expression in (1) again. We want the intensity at 0.5R
• We get: $\mathbf\small{|\vec{E}_{G(A_{depth=0.5R})}|=\frac{G\,m_A}{R^3}0.5R=\frac{0.5G\,m_A}{R^2}}$
4. But from (2), we know the value of $\mathbf\small{\frac{G\,m_A}{R^2}}$
• So the result in (3) becomes:
$\mathbf\small{|\vec{E}_{G(A_{depth=0.5R})}|=0.5 \times 1.5 \times 10^{-4}=0.75\times 10^{-4}\,\rm{N\,m^{-2}}}$
• In the next section, we will see the graphical representation of the variation of intensity in various cases
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