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1800-102-2727We have seen in our school physics practicals that when we bring one metallic bar near a magnet, it will be attracted towards the magnet. But this attraction will be applied till a certain distance. Therefore the magnet creates a region of attraction for a metallic body and when the metal comes in the region, the metal gets attracted to the magnet. This region of attraction for a magnet is known as the magnetic field of the magnet. But the magnet also has a repulsive magnetic field depending on the nature of the poles. Now we have studied Newton’s law of gravitation which basically tells us that all the objects get attracted to each other with a force of attraction known as the gravitational force. Like the magnet, the objects also have a field of attraction in which the gravitational force can be sensed. This region is conceptually known as the gravitational field of that object. The below figure shows the gravitational field of a mass M and a test mass is placed in it.
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The region or space where a force of attraction due to a mass is observed is known as the gravitational field. The force of attraction is nothing but the gravitational force. Like the electric field, the gravitational field can not be seen by our eyes, we can only feel it. Visually, the gravitational field is represented by lines directed towards the mass which is creating the gravitational field. This direction shows that, when an object A is placed in the gravitational field of another object B, object A will get attracted toward object B.
The direction of the gravitational field and the direction of gravitational force is shown in the above figure. It can be seen that the gravitational force will always be in the direction of the gravitational field.
Gravitational field intensity: It is the mathematical entity which shows the strength of the gravitational field is known as the gravitational field intensity and it is represented by a vector quantity ${\overrightarrow{E}}_{g}$. Mathematically it is defined as the gravitational force (${\overrightarrow{F}}_{g}$) acting on a unit mass when it is kept in the gravitational field.
${\overrightarrow{E}}_{g}=\frac{{\overrightarrow{F}}_{g}}{m}$
In the above figure, the gravitational field lines and the gravitational field is depicted where M is the mass of the object which is creating the gravitational field. The directions of gravitational force and the gravitational field.
The gravitational force acting on the mass in the gravitational field is given by,
${\overrightarrow{F}}_{g}=\frac{GMm}{{r}^{2}}\left(-\widehat{r}\right)$
The negative sign is showing that the direction of distance measurement is taken in the opposite direction of the gravitational force.
Therefore the gravitational field intensity of the gravitational field is given by,
${\overrightarrow{E}}_{g}=\frac{{\overrightarrow{F}}_{g}}{m}=\frac{GMm}{{r}^{2}m}\left(-\widehat{r}\right)$
${\overrightarrow{E}}_{g}=-\frac{GM}{{r}^{2}}\left(\widehat{r}\right)$
Where, $\widehat{r}=\frac{\overrightarrow{r}}{\left|\overrightarrow{r}\right|}=\frac{\overrightarrow{r}}{r}$
${\overrightarrow{E}}_{g}=-\frac{GM}{{r}^{3}}\left(\overrightarrow{r}\right)$
Since gravitational field intensity is a vector quantity, it will have a magnitude as well as the direction. Suppose there are n numbers of objects having their masses m_{1}, m_{2}, m_{3}….m_{n} as shown in the figure and we have to find out the field intensity at point P placed in between them. The gravitational field intensity of these objects is denoted as ${\left({\overrightarrow{E}}_{g}\right)}_{1}$, ${\left({\overrightarrow{E}}_{g}\right)}_{2}$, ${\left({\overrightarrow{E}}_{g}\right)}_{3}$…..${\left({\overrightarrow{E}}_{g}\right)}_{n}$. Then the resultant gravitational field intensity ${\left({\overrightarrow{E}}_{g}\right)}_{res}$ will be the vector sum of the field intensities due to all the objects at point P. This concept is nothing but the superposition principle of gravitational field intensities due to different masses.
${\left({\overrightarrow{E}}_{g}\right)}_{P}={\left({\overrightarrow{E}}_{g}\right)}_{1}+{\left({\overrightarrow{E}}_{g}\right)}_{2}+{\left({\overrightarrow{E}}_{g}\right)}_{3}+\dots \dots ..+{\left({\overrightarrow{E}}_{g}\right)}_{n}$
When a test mass is placed at a point near two objects of different masses having their own gravitational field then the test mass will be attracted by both the objects. Let a point P is present on the line passing through two objects of mass m_{1} and m_{2}.
Then there are mainly three possibilities for the position of the point P as given,
As seen from the figure above, if the point P is located on the left side of mass m1 then the field intensities due to both masses m_{1} and m_{2} will be in the same direction i.e. from left to right. Therefore the field intensities will be added vectorially.
${\left({\overrightarrow{E}}_{g}\right)}_{P}={\left({\overrightarrow{E}}_{g}\right)}_{1}+{\left({\overrightarrow{E}}_{g}\right)}_{2}$
As seen from the figure above, if the point P is located on the right side of mass m2 then the field intensities due to both masses m_{1} and m_{2} will be in the same direction again but this time the direction will be from right to left as shown. Therefore the field intensities will be added vectorially.
${\left({\overrightarrow{E}}_{g}\right)}_{P}={\left({\overrightarrow{E}}_{g}\right)}_{1}+{\left({\overrightarrow{E}}_{g}\right)}_{2}$
If the point P is located in between of the mass m_{1} and mass m_{2} then the field intensities due to both masses m_{1} and m_{2} will be in the opposite direction as shown. Therefore the field intensities will be added vectorially.
${\left({\overrightarrow{E}}_{g}\right)}_{P}={-\left({\overrightarrow{E}}_{g}\right)}_{1}+{\left({\overrightarrow{E}}_{g}\right)}_{2}$
Now, when the point P is in between of mass m_{1} and mass m_{2}, then there will be one
location of point P where the resultant field intensity will be zero. This point is known as
the null point or the point of zero intensity or the point of zero gravitational field
Intensity.
Therefore at the point of zero intensity,
${\left({\overrightarrow{E}}_{g}\right)}_{P}={-\left({\overrightarrow{E}}_{g}\right)}_{1}+{\left({\overrightarrow{E}}_{g}\right)}_{2}=0$
${\left({\overrightarrow{E}}_{g}\right)}_{1}={\left({\overrightarrow{E}}_{g}\right)}_{2}$
$\frac{G{m}_{1}}{{x}^{2}}=\frac{G{m}_{2}}{{\left(l-x\right)}^{2}}$
After solving the above equation, we will get the location of the null point or the point of
Zero intensity,
$x=\frac{\sqrt{{m}_{1}}l}{\sqrt{{m}_{1}}+\sqrt{{m}_{2}}}$
$l-x=\frac{\sqrt{{m}_{2}}l}{\sqrt{{m}_{1}}+\sqrt{{m}_{2}}}$
Consider a uniform ring creating a gravitational field around it. The mass of the ring be M and the radius be R as shown.
Consider a small elemental mass dm on the ring then the field intensity due to this element at a point P on the axis of the ring is,
$d{E}_{x}=d{E}_{g}cos\theta $
Where x is the distance of point P from the centre point of the ring (O).
$d{E}_{x}=\frac{Gdm}{{r}^{2}}\frac{x}{r}$ $\left[since,cos\theta =\frac{x}{r}\right]$
$d{E}_{x}=\frac{Gxdm}{{r}^{3}}$
Then the field intensity due to the ring at point P will be
${E}_{x}={E}_{g}={\int}_{}^{}\frac{Gxdm}{{r}^{3}}=\frac{Gx}{{r}^{3}}{\int}_{}^{}dm$
In the above expression, dm=M and $r=\sqrt{{x}^{2}+{R}^{2}}$. Then
${E}_{g}=\frac{GMx}{{\left({x}^{2}+{R}^{2}\right)}^{\frac{3}{2}}}\dots \left(i\right)$
In the vector form, the above equation will be written as,
${\overrightarrow{E}}_{g}=-\frac{GMx}{{\left({x}^{2}+{R}^{2}\right)}^{\frac{3}{2}}}\widehat{x}=-\frac{GM}{{r}^{2}}cos\theta \widehat{x}$
Point of maximum gravitational field: The gravitational field will be maximum when.
$\frac{{dE}_{g}}{dx}=0$
With this, we can get the location of maximum gravitational field at the axis of the ring by differentiating equation (i) with respect to x and the location will be,
$(x{)}_{{E}_{max}}=\pm \frac{R}{\sqrt{2}}$
Variation of E_{g} with x for a uniform ring
A uniform thin spherical shell can be said to be a composition of infinite elementary rings with different radii. Consider a uniform thin spherical shell creating a gravitational field, The mass of the spherical shell is M and radius is R as shown below.
Then the expression for the gravitational field intensity due to the shell at a point P kept at a distance of r is,
${E}_{g}={\int}_{}^{}\frac{GM}{4R{r}^{2}}\left[1+\frac{{r}^{2}-{R}^{2}}{{z}^{2}}\right]dz\dots \left(i\right)$
Case 1: If the point P is outside the shell (r>R):
Then the limits of z in equation (i) will be from r-R to r+R. Then the gravitational field at at point P due to the ring will be,
${E}_{g}=\frac{GM}{{r}^{2}}$
${\overrightarrow{E}}_{g}=-\frac{GM}{{r}^{2}}\widehat{r}$
Case 2: If the point P is inside the shell (r<R):
Then the limits of z in equation (i) will be from R-r to R+r. Then the gravitational field at point P due to the ring will be,
${E}_{g}=0$
It means that the gravitational field inside the thin spherical shell is zero.
A uniform solid sphere can be considered to be the composition of an infinite number of thin spherical shells. Consider a gravitational field is created by a uniform solid sphere at a point P. The mass and the radius of the sphere is M and R respectively. Then,
Case 1: If the point P is inside the sphere (r<R)
In this case the gravitational field created at point P will be,
${\overrightarrow{E}}_{g}=-\frac{GMr}{{R}^{3}}\widehat{r}$
Case 2: If the point P is at the surface of the sphere (r=R)
In this case the gravitational field created at point P will be,
${\overrightarrow{E}}_{g}=-\frac{GM}{{R}^{2}}\widehat{r}$
Case 3: If the point P is outside the sphere (r>R)
In this case the gravitational field created at point P will be,
${\overrightarrow{E}}_{g}=-\frac{GM}{{r}^{2}}\widehat{r}$
Consider a uniform disc creating a gravitational field around it.
If the mass of the disc is M and the radius is R then the gravitational field at a point on the axis of the disc will be,
${\overrightarrow{E}}_{g}\left(x\right)=-\frac{2GM}{{R}^{2}}\left[1-\frac{x}{\sqrt{{x}^{2}+{R}^{2}}}\right]\widehat{x}$
Q1. A particle of mass 3 kg is slowly shifted from a point P(0 m,0 m) to a point Q(4 m, 5 m) in a gravitational field. If the gravitational field is expressed as ${\overrightarrow{E}}_{g}=10\widehat{i}+5\widehat{j}\frac{N}{kg}$ find the work done by the external agent to move the particle from point P to Q.
Answer. Given, the gravitational field is expressed as, ${\overrightarrow{E}}_{g}=\frac{\left(10\widehat{i}+5\widehat{j}\right)N}{kg}$
Since the particle is moving from point P(0 m,0 m) to point Q(4 m, 5 m) then the displacement vector will be,
$\overrightarrow{S}=\left(4-0\right)\widehat{i}+\left(5-0\right)\widehat{j}m$
$\overrightarrow{S}=4\widehat{i}+5\widehat{j}m$
Now the gravitational force acting on the particle is given by,
${\overrightarrow{F}}_{G}=m{\overrightarrow{E}}_{g}$
${\overrightarrow{F}}_{G}=3\left(10\widehat{i}+5\widehat{j}\right)N$
${\overrightarrow{F}}_{G}=\left(30\widehat{i}+15\widehat{j}\right)N$
Therefore the gravitational work done on the particle is,
${W}_{g}={\overrightarrow{F}}_{G}\cdot \overrightarrow{S}$
${W}_{g}=\left(30\widehat{i}+15\widehat{j}\right)\cdot \left(4\widehat{i}+5\widehat{j}\right)$
${W}_{g}=\left(120+75\right)J=195J$
Now, applying the work-energy theorem in this situation,
${W}_{g}+{W}_{ext}=\Delta KE$
Since, in the given problem, the particle is moved very slowly from point P to Q then the kinetic energy is considered to be zero.
${W}_{g}+{W}_{ext}=\Delta KE=0$
${W}_{ext}=-{W}_{g}=-195J$
Q2. Three uniform spherical shells are placed concentrically with radii r_{1}, r_{2} and r_{3} respectively. The mass of these concentric shells are M_{1}, M_{2} and M_{3}. Find the gravitational force acting on mass m at points P and Q as shown in the figure.
Answer. As seen from the given diagram,
The mass and radius of the innermost shell are M_{1} and r_{1}
The mass and radius of the middle shell are M_{2} and r_{2}
The mass and radius of the outermost shell are M_{3} and r_{3}
And the distance of the point P and Q from the centre of the concentric circle is x and y.
At point P: Since the point P is inside the outermost shell, therefore the gravitational field due to outer shell at point P is zero.
Then the resultant gravitational field at point P due to these shells are,
${\left({E}_{g}\right)}_{P}={\left({E}_{g}\right)}_{{M}_{1}}+{\left({E}_{g}\right)}_{{M}_{2}}+{\left({E}_{g}\right)}_{{M}_{3}}$
${\left({E}_{g}\right)}_{P}=\frac{G{M}_{1}}{{x}^{2}}+\frac{G{M}_{2}}{{x}^{2}}+0$
${\left({E}_{g}\right)}_{P}=\frac{G\left({M}_{1}+{M}_{2}\right)}{{x}^{2}}\dots \left(i\right)$
The gravitational force acting at point P will be
${\left({F}_{G}\right)}_{P}=m{\left({E}_{g}\right)}_{P}$
${\left({F}_{G}\right)}_{P}=\frac{G\left({M}_{1}+{M}_{2}\right)m}{{x}^{2}}$
At point Q: Since the point Q is outside of all the shells, therefore the gravitational field due to outer shell at point Q is,
${\left({E}_{g}\right)}_{Q}={\left({E}_{g}\right)}_{{M}_{1}}+{\left({E}_{g}\right)}_{{M}_{2}}+{\left({E}_{g}\right)}_{{M}_{3}}$
${\left({E}_{g}\right)}_{Q}=\frac{G{M}_{1}}{{y}^{2}}+\frac{G{M}_{2}}{{y}^{2}}+\frac{G{M}_{3}}{{y}^{2}}$
${\left({E}_{g}\right)}_{Q}=\frac{G\left({M}_{1}+{M}_{2}+{M}_{3}\right)}{{y}^{2}}\dots \left(ii\right)$
And the gravitational force acting at point Q will be
${\left({F}_{G}\right)}_{Q}=m{\left({E}_{g}\right)}_{Q}$
${\left({F}_{G}\right)}_{Q}=\frac{G\left({M}_{1}+{M}_{2}+{M}_{3}\right)m}{{y}^{2}}$
Q3. Two objects are placed 5 m apart and the mass of these objects is 130 kg and 60 kg. A point P is placed in the gravitational field of both the objects such that the point P is placed at a distance of 3 m from 130 kg object and at a distance of 4 m from 60 kg object. Then the magnitude of the gravitational field at point P due to both the objects is (in $\frac{N}{kg}$)?
Answer. The pictorial diagram for the given situation is given below,
As seen from the above image, M_{1}=130 kg and M_{2}=60 kg
Also from the geometry of the triangle in the diagram, the angle between ${\overrightarrow{E}}_{1}$ and ${\overrightarrow{E}}_{2}$ is 90°.
Therefore the resultant of the gravitational field due to mass M_{1} and M_{2} will be,
${\overrightarrow{E}}_{net}={\overrightarrow{E}}_{1}+{\overrightarrow{E}}_{2}$
${E}_{net}=\sqrt{{{E}_{1}}^{2}+{{E}_{2}}^{2}}\dots .\left(i\right)$
Now the gravitational field intensity due to object 1 is,
${E}_{1}=\frac{G{M}_{1}}{{a}^{2}}$
${E}_{1}=\frac{G\times 130}{{3}^{2}}=\frac{130G}{9}\dots \left(ii\right)$
And the gravitational field intensity due to object 2 is,
${E}_{2}=\frac{G{M}_{2}}{{b}^{2}}$
${E}_{1}=\frac{G\times 60}{{4}^{2}}=\frac{60G}{16}\dots \left(iii\right)$
From equations (i), (ii) and (iii), the net gravitational field at point P is
${E}_{net}=\sqrt{{\left(\frac{130G}{9}\right)}^{2}+{\left(\frac{60G}{16}\right)}^{2}}=\sqrt{222.7}G\frac{N}{kg}$
Q4. Find the null point between two objects A and B 7 m apart. The mass of object A is 20 kg and the mass of the object B is 15 kg.
A. Given, the distance between the objects A and B, l=7 m
Mass of the object A, m_{A}=20 kg
Mass of the object A, m_{B}=15 kg
Then the null point is the point where the gravitational fields of these two objects will cancel each other.
Let the null point is x m from the object A then
${E}_{A}={E}_{B}$
$\frac{G{m}_{A}}{{x}^{2}}=\frac{G{m}_{B}}{{\left(l-x\right)}^{2}}$
$20{\left(l-x\right)}^{2}=15{x}^{2}$
$20\left({l}^{2}+{x}^{2}-2lx\right)=15{x}^{2}$
$20{x}^{2}+20{l}^{2}-40lx=15{x}^{2}$
$5{x}^{2}+20\times {7}^{2}-40\times 7\times x=0$
$5{x}^{2}-280x+980=0$
x=52.25 m and x=3.75 m
The value 3.75 m is acceptable therefore the null point will be at a distance of 3.75 m from the object A and 3.25 m from object B.
Q1. What is a point of zero intensity?
Answer. A point of zero intensity is the null point in the gravitational field of a multiple mass system where the resultant gravitational field becomes zero. If we take two masses at some distance apart then the point of zero intensity will always lie between the masses.
Q2. What will be the direction of the gravitational field of a mass?
Answer. The direction of the gravitational field of a mass will always be towards the mass. This direction of the gravitational field indicates that the gravitational force is always attractive in nature.
Q3. What is the gravitational field intensity due to a uniform spherical shell inside it?
Answer. The gravitational field intensity due to a uniform spherical shell itself will be zero. Or we can say that the gravitational field intensity inside a spherical shell is constant at all the points and that constant value is zero.
Q4. Is gravitational field intensity a scalar quantity?
Answer. No, the gravitational field intensity is a vector quantity. The gravitational field intensity has a magnitude and direction too.