Electric Dipole and Derivation of Electric field intensity at different points of an electric dipole

Electric Dipole:

When two equal (i.e., equal in magnitude) and opposite-charge particles are placed at a very small distance apart, the system is known as an electric dipole.

Electric Dipole Moment:

The product of magnitude of one point charged particle and the distance between the charges is called the 'electric dipole moment'. It is vector quantity and the direction of electric dipole moment is along the axis of the dipole pointing from negative charge to positive charge.

Let us consider the two charged particles, which have equal magnitude $+q$ coulomb and $-q$ coulomb, placed at a very small distance of $l$ in a dipole, so the electric dipole moment is →

$\overrightarrow{p}=q\times \overrightarrow{l}$

Unit: $C-m$ Or $Ampere-metre-sec$

Dimension: $[ALT]$

Electric field intensity due to an Electric Dipole:

The electric field intensity due to an electric dipole can be measured at three different points:

1. 1.) Electric field intensity at any point on the axis of an electric dipole


2. 2.) Electric field intensity at any point on the equatorial line of an electric dipole


3. 3.) Electric field intensity at any point on an electric dipole

1. Electric field intensity at any point on the axis of an electric dipole:

Let us consider an electric dipole $AB$ made up of two charges of $-q$ and $+q$ coulomb placed in a vacuum or air at a very small distance of $l$. Let take a point $P$ on the axis of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put a test charged particle $q_{0}$ at point $P$ for the measurement of the electric field intensity due to the dipole's charge.

So, the electric field intensity (magnitude only) at point $P$ due to a $+q$ charge of an electric dipole

$ E_{+q}=\frac{1}{4\pi\epsilon}\frac{q}{(r-\frac{l}{2})^{2}} \qquad(1)$

Electric field intensity (magnitude only) at point $P$ due $-q$ charge of electric dipole

$ E_{-q}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{(r+\frac{l}{2})^{2}} \qquad(2)$

The net electric field at point $P$ due to an electric dipole

$ E=E_{+q}-E_{-q}\qquad(3)$

Subtitute the value of $E_{+q}$ and $E_{-q}$ in equation $(3)$, Then the above equation $(3)$ can also be written as follows

$E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{1}{(r-\frac{l}{2})^{2}}-\frac{1}{(r+\frac{l}{2})^{2}} \right ]$

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r+\frac{l}{2})^{2}-(r-\frac{l}{2})^{2}}{(r+\frac{l}{2})^{2}(r-\frac{l}{2})^{2}} \right ]$

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r^{2}+\left(\frac{l}{2}\right)^{2}+2\frac{l}{2}r-r^{2}-\left(\frac{l}{2}\right)^{2}+2\frac{l}{2}r)}{(r+\frac{l}{2})^{2}(r-\frac{l}{2})^{2}} \right ]$

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{2rl}{(r^{2}-\frac{l^{2}}{4})^{2}} \right ]\qquad(4)$

Here $l \lt r $ so $\frac{l^{2}}{4} \lt \lt r^{2}$ therefore neglect the term $\frac{l^{2}}{4}$ in above equation $(4)$ so we can write above equation

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{2rl}{(r^{2})^{2}} \right ]$

$E=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2p}{r^{3}} \right ]\qquad (5)$

This is the equation of electric field intensity at a point on the axis of an electric dipole.

The vector form of the above equation $(5)$ is

$\overrightarrow{E}=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2\overrightarrow{p}}{r^{3}} \right ]$

2. Electric field intensity at any point on the equatorial line of an electric dipole:

Let us consider an electric dipole $AB$ made up of two charges of $+q$ and $-q$ coulomb placed in vacuum or air at a very small distance $l$. Let a point $P$ be on the equatorial line of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of the electric field intensity due to the dipole's charge.

So, the electric field intensity (magnitude only) at point $P$ due to a $+q$ charge of an electric dipole

$E_{+q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}} \right ] \qquad(1)$

Electric field intensity (magnitude only) at point $P$ due to $-q$ charge of electric dipole

$E_{-q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}} \right ] \qquad(2)$

The net electric field at point $P$ due to an electric dipole

$E=E_{+q}\:cos\theta + E_{-q}\: cos\theta \qquad(3)$

Put the value of $E_{+q}$ and $E_{-q}$ in the above equation $(3)$, So equation $(3)$ can also be written as follows

$E=2\left [ \frac{q}{4\pi\epsilon_{0} }\frac{1}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}} \right ]cos\theta \qquad (4)$

From figure, In $\Delta \: POB$,

$cos\:\theta=\frac{l}{\sqrt{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}}}$

Put the value of $cos\theta$ in equation $(4)$, so equation $(4)$ can also be written as follows

$E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times l}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}^{3/2}} \right ] \qquad (5)$

Here $l \lt r$ so $\frac{l^{2}}{4} \lt \lt r^{2}$ so neglect the term $\frac{l^{2}}{4}$ in above equation $(5)$ so we can write above equation

$ E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times l}{r^{3}} \right ] \qquad (6)$

$E=\frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{3}} \qquad (7)$

This is the equation of electric field intensity at a point on the equatorial line of an electric dipole.

The vector form of the above equation $(7)$ is

$\overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}} \frac{\overrightarrow{p}}{r^{3}}$

3. Electric field intensity at any point of an electric dipole:

Let us consider an electric dipole $AB$ of length $l$ consisting of the charge $+q$ and $-q$. Let's take a point $P$ in general, and denote its position vector by $\overrightarrow{r}$ from the center point $O$ of the electric dipole AB to point $P$.

The electric dipole moment is a vector quantity that has a direction from $-q$ charge to $+q$ charge. So the electric dipole moment's direction is resolved into two components, one component $ pcosθ$ is along the vector position $\overrightarrow{r}$, and the other component $psinθ$ is normal to vector position $\overrightarrow{r}$. So

Electric field intensity due to dipole moment of component $p\:cos\theta$ {Electric field along the Axial Line }

$ \overrightarrow{E_{\parallel }}=\frac{1}{4\pi\epsilon_{0}} \frac{2pcos\theta}{r^{3}} \qquad(1)$

Electric field intensity due to dipole moment of component $p\:sin\theta$ {Electric field along the Equatorial Line}

$ \overrightarrow{E_{\perp}}=\frac{1}{4\pi\epsilon_{0}} \frac{psin\theta}{r^{3}} \qquad(2)$

The resultant electric field vector $E$ at point $P$

$ \overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2}+2 E_{\perp}E_{\parallel}cos90^{\circ}}$

From figure, The angle between $E_{\perp}$ and $E_{\parallel}$ is $90^{\circ}$. So

$\overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2} }\qquad (3)$

Now substitute the value of equation $(1)$ and equation $(2)$ in equation $(3)$. Then

$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(sin^{2}\theta+4cos^{2}\theta)}$

$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(1+3cos^{2}\theta)}$

This is the equation of electric field intensity at any point due to an electric dipole.

The direction of the resultant electric field intensity vector $\overrightarrow{E}$ from the axial line is

$tan\alpha =\frac{\overrightarrow{E}_{\perp }}{\overrightarrow{E_{\parallel }}} \qquad(4)$

Put the value of $\overrightarrow{E_{\perp}}$ and $\overrightarrow{E_{\parallel}}$ in equation (4). we get

$tan\alpha =\frac{sin\theta}{2cos\theta}$

$tan\alpha =\frac{1}{2}tan\theta$

Here $\alpha$ is the angle between the resultant electric field intensity $\overrightarrow{E}$ and the axial line.

Read More

Definition and Expression of escape velocity of an object on the planet

Definition of Escape Velocity:

The minimum velocity, by which an object is thrown vertically in an upward direction and that object goes out from the gravitation field of the planet and does not come back, is called the escape velocity.

Deduction Escape Velocity Expression:

Let us consider the following:

The mass of the planet =$M$
The radius of the planet = $R$
The mass of the object = $m$

The gravitational force on an object at position $P$ which is a distance $x$ from the surface of the planet

$F=G\frac{M m}{x^{2}} \qquad(1)$

The work done by the force to move the object a very small distance $dx$ from position $A$ to $B$

$W=F.dx$

$dw=G\frac{M m}{x^{2}}dx$

The total work done to move the object from the surface of the planet to infinity

$\int^{W}_{0}dw= \int^{\infty}_{R}G\frac{M m}{x^{2}}dx$

$\left[ w \right]^{W}_{0}=G M m \int^{\infty}_{R} \frac{dx}{x^{2}}$

On solving the above equation

$W=GM m \int^{\infty}_{R} \frac{dx}{x^{2}}$

$W=GM m \left[ -\frac{1}{x} \right]^{\infty}_{R}$

$W=GM m \left[ -\frac{1}{\infty} + \frac{1}{R} \right]$

$W=GM m \left[\frac{1}{R} \right]$

$W=\frac{G M m}{R}$

The above work done is given to the object in the form of kinetic energy to projectile from the surface of the planet to infinity. i.e.

$\frac{1}{2} m v^{2}_{e}= \frac{G M m}{R} $

Where $v_{e}$ is the escape velocity of the object.

$v^{2}_{e} = \frac{2G M m}{R}$

$v_{e} = \sqrt{\frac{2G M}{R}} \qquad(2)$

$v_{e} = \sqrt{\frac{2gR^{2}}{R}} \qquad \left(\because GM=gR^{2} \right)$

$v_{e} = \sqrt{2gR} \qquad(3)$

For earth put mass of the planet $M=M_{e}$ and $R=R_{e}$ in the above equation$(2)$ and equation$(3)$ and we get

$v_{e} = \sqrt{\frac{2G M_{e}}{R_{e}}} $

$v_{e} = \sqrt{2gR_{e}} $

Now substitute the value of the $g=9.8 m/s^{2}$ and radius of earth $R_{e}= 6.4 \times 10^{6} m$ then the escape velocity of the object

$v_{e}=11.2 m/s$

This is the escape velocity of the earth.

Now put $M=\frac{4}{3} \pi R^{3}$ in the above equation $(2)$ then we get

$v_{e} = \sqrt{\frac{2G \frac{4}{3} \pi R^{3}}{R}} $

$v_{e} = \sqrt{\frac{8 \pi \rho G R^{2}}{3}} \qquad(4)$

From the equation $(2)$, $(3)$, and $(4)$ we can conclude that

1.) The escape velocity of the object does not depend on the mass of the object.

2.) The escape velocity of the object is depend upon the mass and radius of the planet.

3.) If the velocity of the object is less than the escape velocity, then the object will reach a certain height and may either move in an orbit around the earth or may fall back to the planet.

4.) If the velocity of projection $(v)$ of the body from the surface of a planet is greater than the escape velocity $v_{e}$ of the planet, the body will escape out from the gravitational field of that planet and will move the interstellar space with velocity $v'$ which can be obtained by using the conservation of energy.

$\frac{1}{2}mv^{2}+\left( -\frac{GMm}{R} \right)= \frac{1}{2}mv'^{2}+0$

$\frac{1}{2}mv^{2}+\left( -\frac{GMm}{R} \right)= \frac{1}{2}mv'^{2}$

$v'^{2}= v^{2} - \frac{2GM}{R}$

$v'^{2}= v^{2} - v^{2}_{e} \qquad \left( \because v^{2}_{e} = \frac{2GM}{R} \right)$

$v'= \sqrt{v^{2} - v^{2}_{e}}$

The relation between orbital velocity and escape velocity of an object:

We know that the orbital velocity of any object revolving near the planet is

$v_{\circ} = \sqrt{gR} \qquad(1)$

The escape velocity of an object which is placed on the planet is

$v_{e} = \sqrt{2gR} \qquad(2)$

From the above equation $(1)$ and equation $(2)$, we get

$v_{e} = \sqrt{2} v_{\circ}$

Alternative Method to Derive Expression for Escape Velocity:

The potential energy of an object on the surface of the planet

$U=-\frac{GMm}{R}$

If the object is thrown vertically in the upward direction with escape velocity $v_{e}$, then the kinetic energy of the object:

$K=\frac{1}{2}mv_{e}^{2}$

We know that the total energy of the object is zero at infinity and then

$K+U=0$

Now substitute the value of $K$ and $U$ in the above equation

$\left( \frac{1}{2}mv_{e}^{2} \right)+\left( -\frac{GMm}{R} \right)=0$

$ \frac{1}{2}mv_{e}^{2} -\frac{GMm}{R} =0$

$ \frac{1}{2}mv_{e}^{2} = \frac{GMm}{R} $

$ \frac{1}{2}v_{e}^{2} = \frac{GM}{R} $

$v_{e}^{2} = \frac{2GM}{R} $

$v_{e} = \sqrt {\frac{2GM}{R}} $

$v_{e} = \sqrt{\frac{2gR^{2}}{R}} \qquad \left(\because GM=gR^{2} \right)$

$v_{e} = \sqrt{2gR} $

Read More

Numerical Problems and Solutions of Centripetal Force and Centripetal Acceleration

Solutions to Numerical Problems of the Centripetal Force and Centripetal Acceleration

Numerical Problems and Solutions

Q.1 An object of mass $0.20 \: Kg$ is being revolved in a circular path of diameter $2.0 \: m$ on a frictionless horizontal plane by means of a string. It performs $10$ revolutions in $3.14 \: s$. Find the centripetal force acting on an object.

Solution:
Given that:
Mass of body $(m)=0.20 \: Kg$
Diameter of Circular path $(d)=2.0: m$
Number of revolutions $(n)=10$
Time taken to complete $(10)$ revolution$(t)=3.14 s$
The centripetal force acting on a body $(F)=?$
Now the centripetal force:
$F=m r \omega^2$
$F=m r \left( \frac{2 \pi n}{t} \right)^2$
Now Substitute the given values in the centripetal force formula:
$F=0.20 \times 1 \left( \frac{2 \times 3.14 \times 10}{3.14} \right)^2$
$F=0.8 N$

Q.2 A car of mass $1200 Kg$ is moving with a speed of $10.5 ms^{-1}$ on a circular path of radius $20 m$ on a level road. What will be the frictional force between the car and the road so that the car does not slip?

Solution:
Given that:
Mass of car $(m) = 1200 \: Kg$
Speed of car $(v) = 10.5: ms^{-1}$
The radius of circular path $(r)= 20 m$
The frictional force $(F)=?$
The frictional force $(F)$ should be equal to the required centripetal force i.e
$F=\frac{mv^{2}}{r}$
Now Substitute the given values in the centripetal force formula:
$F=\frac{1200 \times (10.5)^{2}}{20}$
$F=6.615 \times 10^{3} N$

Q.3 A string can bear a maximum tension of $50 N$ without breaking. A body of mass $1 kg$ is tied to one end of $2 m$ long piece of the string and rotated in a horizontal plane. Find the maximum linear velocity with which the string would not break.

Solution:
Given that:
Maximum tension on the string without breaking $(F)=50 N$
mass of the a body $(m)=1 Kg$
length of the string $(l)=2 m$
The maximum velocity with which the string would not break $(v)=?$
We know that the centripetal force:
$F=\frac{mv^{2}}{r}$
$v=\sqrt{\frac{F \: r}{m}}$
Now Substitute the given values in the centripetal force formula:
$v=\sqrt{\frac{50 \times 2 2}{1}}$
$v=10 ms^{-1}$

Q.4 The moon revolves around the earth in $2.36 \times 10^{6} s$ in a circular orbit of radius $3.85 \times 10^{5} Km$. Calculate the centripetal acceleration produced in the motion of the moon.

Solution:
Given that:
The radius of circular orbit of the moon $(r)=3.85 \times 10^{5} Km = 3.85 \times 10^{8} m$
Time taken to complete one revolution $(T)=2.36 \times 10^{6} s$
Centripetal acceleration produced in the motion of the moon $=?$
The centripetal acceleration:
$a= \omega^{2} r $
$a= \left( \frac{2 pi}{T} \right) r \qquad \left( \because \omega= \frac{2 pi}{T} \right) $
Now Substitute the given values in the centripetal acceleration formula:
$a= \left( \frac{2 \times 3.14}{2.36 \times 10^{6}} \right) \times 3.85 \times 10^{8} $
$a=2.73 \times 10^{-3} ms^{-2} N$

Q.5 Calculate the centripetal acceleration at a point on the equator of the earth. The radius of the earth is $6.4 \times 10^{6} m$ and it completes one rotation per day about its axis.

Solution:
Given that:
Radius of earth $(r)=6.4 \times 10 ^{6} m$
Time taken to complete one rotation per day about its axis$(T)=24 \times 60 \times 60 s$
centripetal acceleration at a point on equator $=?$
The centripetal acceleration:
$a= \omega^{2} r $
$a= \left( \frac{2 pi}{T} \right) r \qquad \left( \because \omega= \frac{2 pi}{T} \right) $
Now Substitute the given values in the centripetal acceleration formula:
$a= \left( \frac{2 \times 3.14}{24 \times 60 \times 60} \right) \times 6.4 \times 10^{6} $
$a=3.37 \times 10^{-2} ms^{-2} N$

Read More

Variation of Acceleration Due to Gravity

Variation in Gravitational Acceleration:

The value of $g$ varies above and below the surface of the earth. It also changes with the latitude at different places and due to the rotation of the earth about its axis. The earth is not a perfect sphere but is approximately an ellipsoid. The radius of the earth is more at the equator and lesser at the poles. The acceleration due to gravity also changes due to the shape of the earth.

1.) Effect of Altitude on Acceleration due to Gravity
2.) Effect of depth below the earth's surface on Acceleration due to Gravity
3.) Effect of the shape of the earth 4.) Effect of rotation about its own axis

1.) Effect of Altitude on Acceleration due to Gravity:

Let us consider: The mass of the earth = $M_{e}$
The radius of the earth = $R_{e}$
A satellite is moving around the earth from the surface of the earth at height= $h$

The gravitational acceleration on the surface of the earth at point $A$

$g=\frac{GM_{e}}{R_{e}^{2}} \qquad (1.1)$

The gravitational acceleration on the altitude $h$ from the surface of the earth at point $B$

$g'=\frac{GM_{e}}{\left( R_{e} + h \right)^{2}} \qquad(1.2)$

Now divide equation $(1.2)$ by equation $(1.1)$

$\frac{g'}{g}= \frac{\frac{GM_{e}}{\left( R_{e} + h \right)^{2}}}{\frac{GM_{e}}{R_{e}^{2}}}$

$\frac{g'}{g}= \frac{R_{e}^{2}}{\left( R_{e} + h \right)^{2}}$

$\frac{g'}{g}= \frac{R_{e}^{2}}{R_{e}^{2} \left( 1 + \frac{h}{R_{e}} \right)^{2}}$

$\frac{g'}{g}= \frac{1}{ \left( 1 + \frac{h}{R_{e}} \right)^{2}}$

$\frac{g'}{g}= \left( 1 + \frac{h}{R_{e}} \right)^{-2}$

Now apply the binomial theorem

$\frac{g'}{g}= 1 - \frac{2h}{R_{e}} + \frac{3h^{2}}{R_{e}^{2}}$

Here $h \lt R_{e}$ then $h^{2} \lt \lt R_{e}^{2}$ so neglect the term of higher power of $\frac{3h^{2}}{R_{e}^{2}}$. Therefore we get

$\frac{g'}{g}= 1 - \frac{2h}{R_{e}} $

$g'= \left( 1 - \frac{2h}{R_{e}} \right) g $

From the above equation it is clear that the value $\left( 1 - \frac{2h}{R_{e}} \right) \lt 1$ then

$g' \lt g$

So, the gravitational acceleration $g$ decreases above the earth's surface.

2.) Effect of depth below the earth's surface on Acceleration due to Gravity:

Let us consider:
The mass of the earth = $M_{e}$
The mass of the earth at depth $h$ from the surface of the earth = $M'_{e}$
The radius of the earth = $R_{e}$

The gravitational acceleration on the surface of the earth at point $A$

$g=\frac{GM_{e}}{R_{e}^{2}} \qquad (2.1)$

The gravitational acceleration at point $B$ on the depth $h$ from the surface of the earth

$g'=\frac{GM'_{e}}{\left( R_{e} - h \right)^{2}} \qquad(2.2)$

Now divide equation $(2.2)$ by equation $(2.1)$

$\frac{g'}{g}= \frac{\frac{GM'_{e}}{\left( R_{e} - h \right)^{2}}}{\frac{GM_{e}}{R_{e}^{2}}}$

$\frac{g'}{g}= \frac{R_{e}^{2}}{\left( R_{e} - h \right)^{2}} \left( \frac{M'_{e}}{M_{e}} \right) \qquad(2.3)$

If $\rho$ is the density of the earth the total mass of the earth and the mass of the earth at depth $h$

$M_{e}=\frac{4}{3}\pi R_{e}^{3} \rho$

$M'_{e}=\frac{4}{3}\pi \left ( R_{e}-h\right)^{3} \rho$

Now subtitute the value of $M_{e}$ and $M'_{e}$ in equaion $(2.3)$

$\frac{g'}{g}= \frac{R_{e}^{2}}{\left( R_{e} - h \right)^{2}} \left( \frac{\frac{4}{3}\pi \left ( R_{e}-h\right)^{3} \rho}{\frac{4}{3}\pi R_{e}^{3} \rho} \right) $

$\frac{g'}{g}= \frac{\left( R_{e} - h \right)}{R_{e}}$

$\frac{g'}{g}= \left( 1 - \frac{h}{R_{e}} \right)$

$g'= \left( 1 - \frac{h}{R_{e}} \right)g$

From the above equation it is clear that the value $\left( 1 - \frac{h}{R_{e}} \right) \lt 1$ then

$g' \lt g$

So, the gravitational acceleration $g$ decreases below the earth's surface.

3.) Effect of the shape of the earth:

Earth is not a perfect sphere. It is flattened at the poles and bulges out at the equator. The equatorial radius $R_{E}$ of the earth is about $21 Km$ Then greater than the polar radius $R_{P}$. Now from the equation $g=\frac{GM}{R^{2}}$ $\left (i.e. g \propto \frac{1}{R^{2}} \right)$, we can conclude that the higher radius has less gravitational acceleration because of that the value of $g$ is least at equator and maximum at the pole.

4.) Effect of rotation about its own axis:

Latitude at a plane is defined as the angle at which the line joining the place to the centre of the earth, makes with the equatorial plane. It is generally denoted by the $\lambda$.

From the figure given below, the latitude at a place $P = \angle OPC = \lambda$

Consider earth is in a perfect sphere of mass $M_{e}$, radius $R_{e}$ with centre $O$. The whole mass of the earth is supposed to be concentrated at the centre $O$. As the earth rotates about its polar axis from west to east, every particle lying on its surface moves along a horizontal circle with the same angular velocity as that of the earth. The centre of each circle lies on the polar axis.

Let us consider, a particle of mass $m$ at a place $P$ of latitude $\lambda$. If the earth is rotating about its polar axis $NS$ with constant angular velocity $\omega$, then the particle also rotates and describes a horizontal circle of radius $r$, Where

$r=PC=OP \: cos\lambda = R \: cos\lambda \qquad (3.1)$

The centrifugal force acting on the particle at $P$ is $m\: r \: \omega^{2}$. It acts along $PA$ directed away from the centre $C$ of the circle of rotation.

The true weight of the particle = $mg$

The centrifugal force at P = $m\: r \: \omega^{2}$

So the resultant of true weight and centrifugal force is $mg'$ can be found by Using the parallelogram law of the forces i.e

$mg'=\sqrt{(mg)^{2}+(mr\omega^{2})^{2}+2(mg)(mr\omega^{2})cos (180^{\circ}- \lambda)}$

$g'=\sqrt{(g)^{2}+(r\omega^{2})^{2}+2(g)(r\omega^{2})cos\lambda)}$

$g'=\sqrt{(g)^{2}+(\omega^{2} R \: cos \lambda )^{2}+2(g)( \omega^{2} R \: cos \lambda )cos\lambda)} \qquad \left( \because r=R \: cos \lambda \right)$

$g'=g \left( 1 + \frac{R^{2}\omega^{4}}{g^{2}} cos^{2} \lambda - \frac{2R \omega^{2}}{g} cos^{2} \lambda)\right)^{\frac{1}{2}}$

Here the value of $\frac{R \omega^{2}}{g}$ is very small, therefore the terms with its squares and of higher power can be neglected. So the above equation can be written as:

$g'=g \left( 1 - \frac{2R \omega^{2}}{g} cos^{2} \lambda)\right)^{\frac{1}{2}}$

Now apply the binomial theorem, and we get

$g'=g \left( 1 - \frac{1}{2}\frac{2R \omega^{2}}{g} cos^{2} \lambda)\right)$

$g'=g \left( 1 - \frac{R \omega^{2}}{g} cos^{2} \lambda)\right)$

$g'= \left( g - R \omega^{2} cos^{2} \lambda\right) \qquad(3.2)$

As, $cos \lambda$ and $\omega$ are positive, therefore $g' \lt g$. Thus from the above equation, it is clear that acceleration due to gravity:

i.) Decreases on account of the rotation of the earth
ii.) Increases with the increase in the latitude of the place.

At equator,
$\lambda = 0^{\circ}$
$g'=p_{e}$
Then from above equation $(3.2)$,

$g_{e}=g-R\omega^{2}$

Clearly, $g_{e}$ is minimum.

At pole,
$\lambda = 90^{\circ}$
$g'=g_{p}$
Then from above equation $(3.2)$,

$g_{p}=g-R\omega^{2 (0)}$

$g_{P}=g$

Clearly, $g_{p}$ is maximum.

Hence, the value of acceleration due to gravity is minimum at the equator and maximum at the poles. The value of acceleration due to gravity at the poles will remain unchanged whether the earth is at rest or rotating.

The difference in the value of acceleration due to gravity at the pole and equator:

We know that:

Acceleration due to gravity at the equator is

$g_{e}=g-R\omega^{2}$

Acceleration due to gravity at the pole is

$g_{p}=g$

The difference in gravitational acceleration at the equator and pole

$g_{p} - g_{e}= g - \left(g-R\: \omega^{2} \right)$

$g_{p} - g_{e}= R\: \omega^{2}$

$g_{p} - g_{e}= R\: \left( \frac{2 \pi }{T}\right)^{2}$

Now put $T=24 \times 60 \times 100$

$g_{p} - g_{e} \approx 0.034 m/sec^{2}$

Read More

Total energy of an orbiting Satellite and its Binding Energy

Definition:

The total mechanical energy associated with an orbiting satellite around the planet is the sum of kinetic energy (i.e., due to orbital motion) and potential energy (i.e., the gravitational potential energy of the satellite).

Derivation of the total mechanical energy of the orbiting satellite around the planet:

Let us consider

The mass of the satellite = $m$

The mass of the planet = $M$

The distance between satellite to a planet from the center of the planet = $r$

The radius of planet =$R$

The potential energy of the satellite is

$U=-\frac{G \: M \: m}{r} \qquad(1)$

The kinetic energy of the satellite is

$K=\frac{1}{2}m v_{e}^{2}$

$K=\frac{1}{2} m \left[ \sqrt{\frac{G M}{r}} \right]^{2} \quad \left( \because v_{e}^{2}=\sqrt{\frac{G \: M}{r}} \right)$

$K=\frac{1}{2} \left( \frac{G M m}{r} \right) \qquad(2)$

The total energy (i.e. mechanical energy) of the satellite is

$E= K+U$

Now put the value of kinetic and potential energy from equation $(1)$ and equation $(2)$ in the above equation

$E= \frac{1}{2} \left[ \frac{G M m}{r} \right]+ \left[ -\frac{G M m}{r} \right]$

$E= -\frac{1}{2} \left[ \frac{G M m}{r} \right]$

$E= -\frac{1}{2} \left[ \frac{G M m}{R+h} \right] \left( \because r=R+h \right)$

The above equation shows that the total mechanical energy associated with orbiting satellite is negative.

The total mechanical energy associated with the orbiting satellite around the Earth:

Put $M=M_{e}$ and $R=R_{e}$ then

$E= -\frac{1}{2} \left[ \frac{G M_{e} m}{R_{e}+h} \right] \left( \because r=R_{e}+h \right)$

$E= -\frac{1}{2} \left[ \frac{g R_{e}^{2} m}{R_{e}+h} \right] \left( \because GM_{e}=g R_{e}^{2} \right)$

If the satellite revolves around near the Earth (i.e., $h=0$) then the total mechanical energy of the satellite

$E= -\frac{1}{2} \left[ \frac{g R_{e}^{2} m}{R_{e}+0} \right]$

$E= -\frac{1}{2} \left[ \frac{g R_{e}^{2} m}{R_{e}} \right]$

$E= -\frac{1}{2} \left[ g R_{e} m \right]$

Binding Energy of the Satellite:

The minimum amount of mechanical energy required to free the revolving satellite around the planet from its orbit is called the binding energy of the revolving satellite.

We know that, the revolving satellite's total mechanical energy in an orbit of the planet is

$E= -\frac{1}{2} \left[ \frac{G M m}{R+h} \right]$

At infinite distance between a satellite to a planet, the total mechanical energy of the satellite is zero. Therefore, if an orbiting satellite is provided postive energy that is equal to the total mechanical energy, its total mechanical energy becomes zero, and the satellite escapes from the planet's orbit. This total positive mechanical energy is called the gravitational binding energy of the satellite. i.e.

$E= +\frac{1}{2} \left[ \frac{G M m}{R+h} \right]$

The gravitational binding energy associated with orbiting satellite around the earth

$E= +\frac{1}{2} \left[ \frac{G M_{e} m}{R_{e}+h} \right]$

If a satellite revolves around near Earth ($h=0$), then the binding energy

$E= +\frac{1}{2} \left[ \frac{G M_{e} m}{R_{e}} \right]$

Read More

Expression for Orbital velocity of Satellite and Time Period

Orbital Velocity of Satellite:

When any satellite moves about the planet in a particular orbit then the velocity of the satellite is called the orbital velocity of the satellite.

Expression for Orbital Velocity of Satellite:

Let us consider:

The mass of the satellite = $m$

The mass of planet= $M$

The radius of Planet =$R$

The satellite is moving about the planet at height=$h$

The satellite is moving about the planet with orbital velocity=$v_{\circ}$

The distance from the center of the planet to satellite=$r$

The force of gravitation between the planet and the satellite

$F=G \frac{M \: m}{r^{2}} \qquad(1)$

This force work act as a centripetal force to revolve the satellite around the planet i.e.

$F=\frac{m v_{\circ}^{2}}{r} \qquad(2)$

From the above equation $(1)$ and equation $(2)$, we get

$\frac{m v_{\circ}^{2}}{r} = G \frac{M \: m}{r^{2}}$

$v_{\circ}= \sqrt{\frac{G \: M }{r}}$

Where $r=R+h$ then

$v_{\circ}= \sqrt{\frac{G \: M }{R+h}} \qquad(3)$

This is the equation of the orbital velocity of the satellite.

If any satellite revolves around the earth then the orbital velocity

$v_{\circ}= \sqrt{\frac{G \: M_{e} }{R_{e}+h}}$

$v_{\circ}= \sqrt{\frac{gR_{e}^{2}}{R_{e}+h}} \qquad \left( \because G \: M_{e} = gR_{e}^{2} \right)$

$v_{\circ}= R_{e} \sqrt{\frac{g}{R_{e}+h}} $

This is the equation of the orbital velocity of a satellite revolving around the earth.

If the satellite is orbiting very close to the surface of the earth ( i.e $h=0$) then the orbital velocity

$v_{\circ}= R_{e} \sqrt{\frac{g}{R_{e}+0}} $

$v_{\circ}= \sqrt{g R_{e}} $

Now subtitute the value of radius of earth (i.e $R_{e}=6.4 \times 10^{6} \: m$) and gravitational accelertaion ($g=9.8 \:m/sec^{2}$) then orbital velocity

$v_{\circ}=7.92 \: Km/sec$

The time period of Revolving Satellite:

The time taken by satellite to complete on revolution around the planet is called the time period of the satellite.

Let us consider the time period of the revolving satellite is $T$ Then

$T= \frac{Distance \: covered \: by \: Satellite \: in \: one \: revolution}{Orbital \: Velocity}$

$T= \frac{2 \pi r}{v_{\circ}}$

$T= \frac{2 \pi \left( R+h \right)}{v_{\circ}}$

$T= \frac{2 \pi \left( R+h \right)}{v_{\circ}}$

Now subtitute the value of orbital velocity $v_{\circ}$ from equation $(3)$ in above equation then

$T= \frac{2 \pi \left( R+h \right)}{\sqrt{\frac{G \: M }{R+h}}}$

$T= 2 \pi \sqrt{ \frac{ \left( R+h \right)^{3}}{G M}}$

This is the equation of the time period of the revolution of satellites.

If a satellite revolves around the earth then the time period

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+h \right)^{3}}{G M_{e}}}$

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+h \right)^{3}}{g R_{e}^{2}}} \qquad \left( \because G \: M_{e} = gR_{e}^{2} \right)$

This is the equation of the time period of revolution of satellite revolving around the earth.

If the satellite revolves very nearly around the earth (i.e $h=0$) then the time period of the satellite from the above equation

$T= 2 \pi \sqrt{ \frac{ \left( R_{e}+0 \right)^{3}}{g R_{e}^{2}}} $

$T= 2 \pi \sqrt{ \frac{ R_{e}}{g}} $

$T= 84.6 \: min $

Read More