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 = $m$
An object thrown vertically upward with Escape Velocity
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} $

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} $
$a=3.37 \times 10^{-2} ms^{-2} N$


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= $h$
Effect of Altitude on Gravitational Acceleration

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 = $R_{e}$
Effect of Depth on Gravitational Acceleration

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.
Effect of Shape of Earth on Gravitational Acceleration
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 $\lambda$.
Latitude at a Plane

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

Effect of Rotation of Earth about its own axis on Gravitational Acceleration
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 $T=24 \times 60 \times 100$

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

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:
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
Orbital Velocity of Satellite moving around the Earth

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

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

$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 $

Derivation of Gravitational Potential Energy due to Point mass and on the Earth

Definition of Gravitational Potential Energy:
When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is stored in the form of potential energy which is called gravitational potential energy.

Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential energy

$U=W_{\infty \rightarrow r}$

Derivation of Gravitational Potential energy due to a Point mass:
Gravitational Potential Energy due to a point mass

Let us consider,
The mass of the point object (i.e point mass)=$m$
The mass of the object that produces the gravitational field = $M$

If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is

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

If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

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

Therefore, the work done to bring the point mass from infinity to point $P$ that is at distance $r$ from point $O$ then work done to move the point object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M\: m }{r} $

This work done by the force is stored in the form of potential energy i.e

$U=W$

$U=-\frac{G\: M\: m }{r}$

Thus the above equation represents the gravitational potential energy of an object at point $P$

Gravitational Potential Energy on Earth:
Gravitational Potential Energy On Earth

Let us consider, The mass of Earth = $M_{e}$
The radius of earth = $R_{e}$
The mass of the object = $m$
The distance from centre $O$ of the earth to point $P$ = $r$
The distance from the surface of the earth to point $P$ = $h$

If the object is at a distance $x$ then the gravitational force is

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

If the object moves a very small distance element $dx$ that is at distance $x$ from centre point $O$ of the earth then the work done to move an object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M_{e} \: m}{x^{2}}.dx$

Therefore, the work done to bring the object from infinity to point $P$ that is at distance $r$ from centre point $O$ of the earth then the work done to move an object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M_{e} \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M_{e}\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M_{e}\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M_{e}\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M_{e}\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M_{e}\: m }{r} $

The above equaton shows that the work done by force is stored in the form of gravitational potential energy i.e.

$U=W$

$ U = -\frac{G\: M_{e}\: m }{r}$

Where $r=R_{e}+h$, then above equation can be written as

$U = -\frac{G\: M_{e}\: m }{R_{e}+h} \qquad(2)$

This is the equation of the gravitational potential energy at point $P$. The other form of the above equation i.e

$U = -\frac{g R_{e}^{2} \: m }{R_{e}+h} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

If the object is placed on the surface of the earth then $h=0$. So gravitational potential energy on the surface of the earth
$U=-\frac{G \: M_{e} \: m}{R_{e}}$

This is the equation of the gravitational potential energy of an object placed on the surface of the earth.

$U=-\frac{g R_{e}^{2} m}{R_{e}} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

$U=-g R_{e} m$ This is another form of the gravitational potential energy of an object placed on the surface of the earth.

Note:

We know that the gravitational potential energy at any point from above the surface of the earth

$U = -\frac{G\: M_{e}\: m }{R_{e}+h} $

$U = V m \qquad \left( \because V= -\frac{G\: M_{e}\: }{R_{e}+h} \right)$

$U = Gravitational \: Potential \times \: mass \: of \: an \: object$

Derivation of Gravitational Potential due to Point mass and on the Earth

Definition of Gravitational Potential:
When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is called gravitational potential.

Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential

$V=-\frac{W}{m}$

Derivation of Gravitational Potential due to a Point mass:
Gravitational Potential due to a point mass

Let us consider,
The mass of the point object (i.e point mass)=$m$
The mass of the object that produces the gravitational field = $M$

If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is

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

If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

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

Therefore, the work done to bring the point mass from infinity to point $P$ that is at distance $r$ from point $O$ then work done to move the point object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M\: m }{r} $

$\frac{W}{m} = -\frac{G\: M }{r} $

$ V = -\frac{G\: M}{r} \qquad \left( \because V=\frac{W}{m} \right)$

$ V = -\frac{G\: M}{r}$

Thus the above equation represents the gravitational potential at point $P$

Gravitational Potential on Earth:
Gravitational Potential On Earth

Let us consider, The mass of Earth = $M_{e}$
The radius of earth = $R_{e}$
The mass of the object = $m$
The distance from centre $O$ of the earth to point $P$ = $r$
The distance from the surface of the earth to point $P$ = $h$

If the object is at a distance $x$ then the gravitational force is

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

If the object moves a very small distance element $dx$ that is at distance $x$ from centre point $O$ of the earth then the work done to move an object from point $B$ to $A$

$dw=F.dx$

Now substitute the value of $F$ from equation $(1)$ in above equation

$dw=G \frac{M_{e} \: m}{x^{2}}.dx$

Therefore, the work done to bring the object from infinity to point $P$ that is at distance $r$ from centre point $O$ of the earth then the work done to move an object from infinity ($\infty$) to point $P$

$\int_{0}^{W} dw = \int_{\infty}^{r} G \frac{M_{e} \: m}{x^{2}}.dx $

$ [w]_{0}^{W} = G\: M_{e}\: m \int_{\infty}^{r} \frac{1}{x^{2}}.dx $

$ [W-0] = G\: M_{e}\: m [-\frac{1}{x} ]_{\infty}^{r} $

$ W = G\: M_{e}\: m [\left(-\frac{1}{r}\right) - \left(-\frac{1}{\infty}\right)]$

$ W = -\frac{G\: M_{e}\: m }{r} \qquad \left( \because \frac{1}{\infty} =0 \right)$

$ W = -\frac{G\: M_{e}\: m }{r} $

$\frac{W}{m} = -\frac{G\: M_{e} }{r} $

$ V = -\frac{G\: M_{e}}{r} \qquad \left( \because V=\frac{W}{m} \right)$

$ V = -\frac{G\: M_{e}}{r}$

Where $r=R_{e}+h$, then above equation can be written as

$V=-\frac{G \: M_{e}}{R_{e}+h} \qquad(2)$

This is the equation of gravitational potential at point $P$. The other form of the above equation i.e

$V=-\frac{g R_{e}^{2}}{R_{e}+h} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

If the object is placed on the surface of the earth then $h=0$. So gravitational potential on the surface of the earth

$V=-\frac{G \: M_{e}}{R_{e}}$

This is the equation of gravitational potential on the surface of the earth. The other form of the above equation is

$V=-\frac{g R_{e}^{2}}{R_{e}} \qquad \left( \because GM_{e}= g R_{e}^{2} \right)$

$V=-g R_{e}$

Gravitational field, Intensity of Gravitational field and its expression

Definition of Gravitational Field:

The region around an object in which another object experiences a gravitational force then the region of that object is called the gravitational field.

Gravitational Field Intensity:

The force applied per unit mass of an object that is placed in the gravitational field is called the intensity of the gravitational field.

$\overrightarrow{E}=- \frac{G \: M}{r^{2}} \hat{r}$

The Expression for gravitational field intensity:

Let us consider
The mass of a lighter object that experience the force = $m$
The mass of a heavy object that produces the gravitational field= $M$
The distance between the objects = $r$

The gravitational force between the objects is

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

Now the force per unit mass i.e Gravitational field intensity

$E=-\frac{F}{m} \qquad (2)$

Here the negative indicates that the direction of force is opposite to $\hat{r}$

The vector form of the gravitational field intensity

$\overrightarrow{E}=-\frac{F}{m} \hat{r}$

Where $\hat{r} \left (=\frac{\overrightarrow{r}}{r} \right)$ is the unit vector along the $\overrightarrow{r}$

Now substitute the value of $F$ in the above equation $(2)$. Therefore we get,

$E=-G \frac{M\:m}{m r^{2}}$

$E=- \frac{G \: M}{r^{2}}$

The vector form of the above equation

$\overrightarrow{E}=- \frac{G \: M}{r^{2}} \hat{r}$

Deduction of Newton's Law of gravitation from Kepler's Law

Deduction of Newton's Law of gravitational force from Kepler's Law:

Let us consider:
The mass of a planet = $m$
The radius of the circular path of a planet=$r$
The mass of the sun = $M$
The velocity of the planet = $v$
The time of the revolution=$T$

The attraction force between the planet and the sun is achieved by the centripetal force i.e

$F=\frac{m v^{2}}{r} \qquad(1)$

The orbital velocity of the planet:

$v=\frac{Circumference \: of \: the \: circular \: path}{Time \: period}$

$v=\frac{2\pi r}{T} \qquad(2)$

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

$F=\frac{m}{r} \left( \frac{2\pi r}{T} \right)^{2}$

$F=\frac{4 \pi^{2} mr}{T^{2}} \qquad(3)$

According to Kepler's third law i.e

$T^{2}=Kr^{3} \qquad(4)$

From equation$(3)$ and equation$(4)$, we get

$F=\frac{4 \pi^{2}mr}{Kr^{3}}$

$F=\frac{4 \pi^{2}}{K}\frac{mr}{r^{3}}$

$F=\frac{4 \pi^{2}}{K}\frac{m}{r^{2}} \qquad(5)$

The source of this force is the sun. So R.H.S of equation $(5)$ should be related to the sun. Since $m$ and $r$ are related to the planet. So the quantity $\frac{4 \pi^{2}}{K}$ should be related to some constant of the sun. Let $\frac{4 \pi^{2}}{K}$ be proportional to the mass of the sun. i.e

$\frac{4 \pi^{2}}{K m} \propto M$

$\frac{4 \pi^{2}}{K m} = G M $

Here $G$ is the proportionality constant and it is called the universal gravitation constant. Now substitute the $\frac{4 \pi^{2}}{K m} = G M$ in equation $(5)$. then we get

$F=G \frac{M m}{r^{2}}$

The above equation represents the force of attraction between the sun and the planet. It is also known as Newton's law of gravitational force.

Force of Friction | Expression for the acceleration and the work done on inclined rough surface| Advantage and Disadvantage

Definition of Friction:
When an object slides over the surface of another object then each surface applies a parallel force to the other. This force is called friction force. The direction of the force of friction on every object with respect to the other is always opposite to the motion of the second object.
$F=\mu R$
Properties of Friction Force:

  • The force of friction always acts in the opposite direction of the motion of the object.
  • The force of friction always opposes the motion of the object. So it does not help in motion.
  • Friction force always helps an object to be in the rest of the position. i.e If an object is in the rest of the position that is due to only frictional force.
  • The frictional force is act on both states of the object i.e Rest and motion.

  • Modern Concept of Friction:

    It is now believed that frictional forces arise because of a fundamental force occurring in nature commonly known as an electric force. Every object has surface irregularities at the atomic level.

    When two surfaces come in contact then the real area in contact is much smaller than the actual area of the surface. At the contact points, the distance between the particles becomes very small and as such the molecular electrical forces begin to act between the surfaces and molecular bonds are formed.

    When one of the surfaces is pulled over the other, the molecular bonds are broken. As a result, the material of the objects is deformed and new bonds are formed. The successive loading and deformation processes result in loss of energy which appears in the form of heat. To compensate for the energy loss, a force is required to be applied to start the motion.

    Types of forces of friction: There are two types of forces of friction

    1. Force of Static Friction
    2. Force of Dynamic Friction

    1. Force of Static Friction:
    When the friction forces acting between two surfaces at rest with respect to each other are called the force of static friction.

    Note: The maximum value of the force of static friction is always equal to the minimum value of force required to start the motion. Once the motion starts, the force of friction after reducing becomes such that it is sufficient to maintain uniform motion.

    Due to the force of static friction between two dry and unlubricator surfaces, the maximum force is according to the following two laws:

    i.) The maximum force does not normally depend upon the area of contact.
    ii.) It is directly proportional to the normal reaction.

    Coefficient of Static Friction:
    The ratio of the maximum value of force in static friction and normal reaction is called the coefficient of static friction.
    $\mu_{s}=\frac{F}{R}$
    2. Force of Dynamic Friction:
    When the force of friction acting between the two surfaces in relative motion is called the force of dynamic friction.

    Note: Due to dynamic friction between the dry and unlubricated surfaces the value of dynamic friction is according to the two laws:

    i.) It does not depend upon the surfaces of contact.
    ii.) It is equal to the maximum normal reaction force.

    Coefficient of Dynamic Friction:
    The ratio of force in dynamic friction and normal reaction is called the coefficient of dynamic friction.
    $\mu_{k}=\frac{F}{R}$
    Limiting friction:

    It is the maximum value of friction that acts when the object just begins to move is called Limiting Friction. The limiting friction is directly proportional to the normal reaction in the object. i.e

    $F \propto R$

    $F = \mu R$

    Angle of friction:

    The angle between the effective resistance $S$ (i.e resultant of frictional force and normal reaction) and normal reaction $R$ is called the angle of friction.

    Let us consider, A object of mass $m$ placed on a horizontal surface. If $F$ force is applied on the object then the value of limiting friction

    $F \propto R$

    $F = \mu R$

    $\mu=\frac{F}{R} \qquad(1.1)$
    Angle of Friction
    From figure in $\Delta ROS$

    $tan \theta =\frac{RS}{RO}$

    $tan \theta =\frac{RS}{RO}$

    $tan \theta =\frac{F}{R} \qquad \left( \because RS=F \: and \: RO=R \right)$

    From the equation $(1.1)$ and the above equation, We can conclude that

    $\mu=tan\theta$

    Angle of Repose and Angle of Sliding:

    The angle of repose or angle of sliding is defined as the minimum angle of inclination of a surface with the horizontal, such that an object placed on the surface just begins to slide down.

    The various forces acting on the object are shown in the figure below:
    Angle of Repose Or Angle of Sliding

    i.) Weight $mg$ of the object acting vertically downwards.
    ii.) Normal Reaction $R$ acting perpendicular to inclined surface $AB$
    iii.) Force of friction $f$ acting up in the inclined surface $AB$

    Now from the figure apply the equilibrium condition:

    $R=mg \: cos\theta \qquad(1.2)$

    $f=mg \: sin\theta \qquad(1.3)$

    Now divide by equation $(1.3)$ to equation $(1.2)$

    $\frac{f}{R}=\frac{mg \: sin\theta}{mg \: cos\theta}$

    $\frac{f}{R}=\frac{sin\theta}{cos\theta}$

    $\frac{f}{R}=tan \theta$

    $\frac{\mu R}{R}=tan \theta \qquad \left( \because f=\mu R \right)$

    $\mu =tan \theta $

    Thus the angle of friction is equal to the angle of repose.

    1.) Expression for the acceleration and work done on the object moves downward on a rough inclined surface without applied force:

    Let us consider, A object of mass $m$ moving sliding downward on a rough surface that is inclined at an angle $\theta$ from horizontal. If the angle $\theta$ of the inclined surface is greater than the angle of repose, the object slides down with an acceleration $a$ without any applied force.

    The various forces acting on the object are shown in the figure below:
    When the object moves downward on a rough inclined surface without applied force
    i.) Weight $mg$ of the object acting vertically downwards.
    ii.) Normal Reaction $R$ acting perpendicular to inclined surface $AB$
    iii.) Force of friction $f$ acting up in the inclined surface $AB$

    Apply the equilibrium condition from the figure:

    $R=mg \: cos\theta \qquad(1.4)$

    The net force on the object moving down the inclined surface

    $F=mg \: sin\theta - f \qquad(1.5)$

    $F=mg \: sin\theta - \mu R $

    Now substitute the value of $R$ from equation $(1.4)$ in above equation

    $F=mg \: sin\theta - \mu mg \: sin \theta$

    $ma=mg \left( \: sin\theta - \mu \: sin \theta \right) \qquad( F=ma)$

    $a=g \left( \: sin\theta - \mu \: sin \theta \right) $

    If the object is displaced the distance $S$ by the acceleration $a$ then work done by force:

    $W=F.S$

    $W= mg \left(\: sin\theta - \mu \: sin \theta \right). S$

    2.) Expression for acceleration and work done on the object moving upward on a rough inclined surface by the applied force:

    Let us consider, A object of mass $m$ moving upward on a rough inclined surface by applying the force $F$ sliding downward on a rough surface so the various forces acting on the object are shown in the figure below:
    When the object moves upward on a rough inclined surface by the applied force
    i.) Weight $mg$ of the object acting vertically downwards.
    ii.) Normal Reaction $R$ acting perpendicular to inclined surface $AB$
    iii.) Force of friction $f$ acting up in the inclined surface $AB$

    In equilibrium condition from the figure:

    $R=mg \: cos\theta \qquad(1.6)$

    The net force on the object moving down the inclined surface

    $F=mg \: sin\theta + f \qquad(1.7)$

    $F=mg \: sin\theta + \mu R $

    Now substitute the value of $R$ from equation $(1.6)$ in above equation

    $F=mg \: sin\theta + \mu mg \: sin \theta$

    $ma=mg \left( \: sin\theta + \mu \: sin \theta \right) \qquad( \because F=ma)$

    $a=g \left( \: sin\theta + \mu \: sin \theta \right) $

    If the object is displaced the distance $S$ by the force $F$ then work done by force:

    $W=F.S$

    $W= mg \left(\: sin\theta + \mu \: sin \theta \right). S$

    3.) Expression for acceleration and work done on the object moves on a rough horizontal surface by the applied force:

    Let us consider, A object is moving on a rough horizontal surface by applying the force $F$ on the object of mass $m$ and then the net force on the object from the figure below
    When the object moves on a rough horizontal surface by the applied force
    $F-f=ma$

    $F-\mu R=ma \qquad \left(\because f=\mu R \right)$

    $F-\mu mg=ma \qquad \left(R=mg \right)$

    $F=m \left(\mu g+a \right)$

    $a=\frac{F-\mu mg}{m}$

    If the object is displaced the distance $S$ by the force $F$ then work done by force:

    $W=F.S$

    $W= m \left(\mu g+a \right). S$

    The Advantages of Friction:

  • The two objects will not stick to each other if there is no friction between the surface.
  • The parts of machinery are held together with the help of nuts and bolts but without friction, these can not be held.
  • A person can not walk or stand on the surface without friction.
  • The brakes of any vehicle will not work without friction.
  • It is not possible to transfer motion from one part of a machine to the other part without the help of friction.
  • Sandpaper is used in cleaning because this cleaning is only possible with the help of friction.
  • Adhesives will lose their purpose.

  • The Disadvantage of Friction:

  • Friction always opposes the relative motion between any two objects in contact. Therefore, extra energy is lost to overcome friction. Thus, friction involves the unnecessary loss of energy. This shows that the output is always less than the input.
  • Friction between the parts of machinery causes wears and tear. Therefore, the lifetime of the parts of the machinery reduces.
  • Frictional forces produce heat, which causes damage to the machinery.

  • Methods of Reducing friction:

  • By Polishing: Polishing makes the surface smoother. Therefore, friction reduces.
  • By Streamlining: Friction due t air is considerably reduced by streamlining the shape of the object (sharp in front) moving through the air.
  • By Lubrication: Oil, grease, and many other materials are used as lubricants. These lubricants fill up the irregularities of the surface making them smother. Hence, friction decreases.
  • By proper selection of materials.
  • By using ball bearings.
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