# Chapter Notes: Gravitation Physics Class 11

Notes for Gravitation chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

UNIVERSAL LAW OF GRAVITATION

Forces of mutual attraction acting between two point particles are directly proportional to the masses of these particles and inversely proportional to the square of the distance between them. The magnitude of the gravitational force is determined by the expression

$F = G{{{m_1}{m_2}} \over {{r^2}}}$

where m1 and m2 are masses of the interacting particles, r = distance between them.
The proportionality constant G is defined as the universal gravitational constant and its value is G = 6.6732*10-11 N m2/kg2.
Forces of gravity are directed along the line joining the interacting particles and are, therefore, called central forces, which is conservative.
The law of universal gravitation in the above form holds not only for two particles but also for
(a) bodies of an arbitrary shape whose dimensions are only a small fraction of the distance between the centers of mass of the bodies.
(b) bodies having a spherically symmetrical distribution of their mass.
The gravitational force is a real force and it is always of attractive nature.

Application 1

Three particles each of mass m are placed at the three corners of an equilateral triangle of side a. Find the force exerted by this system on another particle of mass m placed at (a) the centre of the triangle and (b) mid point of a side.

Solution

To solve the above problem we apply the gravitational interaction which follow the principle of superposition.
(a) When another mass m is placed at O, it experiences three forces ,  and . Since AO, BO and CO are equal hence . Angle between any two forces is same i.e. 120o. Therefore the resultant force exerted by the system on particle at O is zero.
(b)In this case the particle is placed at point D, which is equidistant from B and C.
$|{{\rm{\vec F}}_{\rm{B}}}| = |{{\rm{\vec F}}_{\rm{C}}}|$
But they are opposite in direction. Therefore the effective force at D will be due to mass m at A.
By geometry of the figure AO=a sin 60 = ${{\sqrt 3 \,a} \over 2}$.
Therefore,   ${F_A} = {{4G{m^2}} \over {3{a^2}}}$ along DA

ACCELERATION DUE TO GRAVITY

Earth attracts all bodies towards its centre. This property of the earth is called ‘gravity’ and the force with which it attracts a body is called the ‘force of gravity’ acting on that body. Thus when a body falls freely towards the earth’s surface, the force of gravity  produces an acceleration ${\rm{\vec g}}$  in it given by

${\rm{\vec g}} = {{{\rm{\vec F}}} \over m}$

This acceleration is called acceleration due to gravity. Its magnitude g is independent of  the mass, size, shape and composition of the body. It is directed radially inward  to the centre of the earth.

### Variation of ‘g’

(i)   ‘g’ above the earth surface at height h (h<< R).
$g' = g\left( {1 - {{2h} \over R}} \right)$; where R is radius of the earth and g is acceleration due to gravity on the surface of earth.
This shows that the acceleration due to gravity decreases in moving upward from the earth’s surface.
(ii)  ‘g’ below the earth surface at depth d
$g' = g\left( {1 - {d \over R}} \right)$
The acceleration due to gravity decreases in moving downward below the earth’s surface.
iii)  variation due to earth’s rotation $g' = g - R{\omega ^2}{\cos ^2}\lambda$
where g' is the acceleration due to gravity at latitude l and earth is rotating about its own axis with uniform angular velocity w. Here earth is assumed as solid sphere of radius R and mass M.
iv) Due to the shape of earth
From the geoide shape of earth we know that it is bulging at the equator and flattened at the poles. Hence g is maximum at pole and minimum at the equators.

Application 2

What is the acceleration due to gravity of earth at the surface of moon if the distance between earth and moon is 3.8 $\times$105 km and radius of earth is 6.4 $\times$ 103 km?

Solution

(i) If M and R be the mass and radius of the earth then the acceleration due to gravity due to earth on the surface of earth i.e.
$g = {{GM} \over {{R^2}}}$          ....(i)
Similarly, acceleration due to gravity at a distance r (>R) of the earth i.e.
${g'} = {{GM} \over {{r^2}}}$
If r be the distance between earth and moon then g' will give you the value of acceleration due to gravity on the moon due to earth. Therefore, from equation (i) and (ii)
$g' = {{{{\left( {6.4 \times {{10}^3}} \right)}^2}} \over {{{\left( {3.8 \times {{10}^5}} \right)}^2}}}g$
$\Rightarrow$     g = 0.00275 m/s2        [ g = 9.8 m/s2]

Gravitational Field

The region around a body within which its gravitational force of attraction is perceptible is called its gravitational field. The intensity of the field at a point is defined as the force experienced by a unit mass when placed at that point in the given field due to mass M.
i.e.  ${\rm{\vec I}} = {{{\rm{\vec F}}} \over m} = {{ - GM} \over {{r^2}}}\hat r$; where  is a unit vector directed from mass m to that point.
Intensity at a point due to a spherical shell and a solid sphere can be realized respectively as Application 3

There are two concentric shells of masses M1 and M2 and radii R1 and R2. Find the force on a particle of mass m when the particle is located at
(i) r1 > R2        (ii) R1 < r2 < R2           (iii) r3 < R1

Solution

(i) From the figure, it is clear that the point P1 lies outside to both the shell. Therefore, gravitational Intensity at ${p_1} = {{G\left( {{M_1} + {M_2}} \right)} \over {r_1^2}}$.
Therefore, force on the particle of mass $m = {{G\left( {{M_1} + {M_2}} \right)m} \over {r_1^2}}$ (ii) When R1 < r2 < R2, the point P2 lies outside the smaller shell but inside the larger shell.
Therefore, Intensity at P2 = Intensity due to smaller shell + Intensity due to larger shell.
=${{G{M_1}} \over {r_2^2}} + 0$
Therefore force on mass $m = {{G{M_1}m} \over {r_2^2}}$
iii) When r3 < R1,  Point P3 lies inside to both the shells. The intensity of the field at P3
= Intensity due to smaller shell + Intensity due to larger shell
= 0+0=0

Gravitational Potential

It is defined as negative of work done by gravitational force per unit mass in shifting a unit test mass from infinity to the given point.
i.e. $V =- {W \over m}$
i.e. $V =- {{GM} \over r}$ is the gravitational potential at a point which is at a distance r from M. Gravitational Potential Energy

The gravitational potential energy of a particle placed in a gravitational field is measured by the amount of work done in displacing the particle from a reference position to its present position. Generally, the reference position is chosen at infinity from the attracting mass where the potential energy of the particle is taken as zero.
$U(r) =- {W_\infty } + U(\infty ) =- {W_\infty }$  $[U(\infty ) = 0]$
$U(r) =- {W_\infty } - \int\limits_\infty ^r {F\left( r \right)} dr$
We have ,  $F\left( r \right) =- {{GMm} \over {{r^2}}}$
$U\left( r \right) =- \int\limits_\infty ^r { - {{GMm} \over {{r^2}}}} dr =- {{GMm} \over r}$
This gives the gravitational potential energy of the particle at the point. The negative sign indicates that the potential energy decreases from zero as the particle is brought (from infinity) towards the attracting mass.
The potential energy can also be written as
$U(r) =- mg{{{R^2}} \over r}$
$U\left( r \right) =- \int\limits_\infty ^r { - {{GMm} \over {{r^2}}}} dr =- {{GMm} \over r} =- {{GMm} \over r}$
This gives the gravitational potential energy of the particle at the point. The negative sign indicates that the potential energy decreases from zero as the particle is brought (from infinity) towards the attracting mass.
The potential energy can also be written as
$U(r) =- mgU(r) =- mg{{{R^2}} \over r}$
At surface it is –mgR;       R = radius of the earth.

Gravitational Self Energy

The gravitational self energy of a body (or a system of particles) is defined as the work done in assembling the body (or system of particles) from infinitesimal elements that are initially at infinite distance apart.
The gravitational potential energy of two particles of masses m1 and m2 which are r12 distance apart is given as $- {{G{m_1}{m_2}} \over {{r_{12}}}}$
Therefore, the potential energy of n particles due to their mutual gravitational attraction is equal to the sum of the potential energy of all particles
i.e. ${U_s} = - G\sum\limits_{ij \atop i \ne j } {{{{m_i}{m_j}} \over {{r_{ij}}}}}$ (Counting every pair once only)

Application 4

Calculate the self gravitational potential energy of matter forming (a) a thin uniform shell of mass m and radius R and (b) a uniform sphere of mass m and radius R.

Solution

Here it is supposed that initially the particles of the body are scattered at infinite distance from each other. Therefore in the formation of a body some external agent has to do some work in assembling the body. This energy is stored in the body as gravitational potential energy and is known as self-gravitational energy or mutual gravitational interaction.
(a) Potential of the shell  = $- {{GM} \over R}$
Net work done by external agent = $- \int\limits_0^m {{{GMdM} \over R}} = - {{G{m^2}} \over {2R}}$
self energy = $- {1 \over 2}{{G{m^2}} \over R}$
(b)Now let us consider a sphere of radius x and density r then mass of the sphere =${4 \over 3}\pi {x^3}\rho$
Gravitational potential of the surface= ${{- 4} \over 3}\pi G\rho {x^2}$ The work done by the external agent increasing surface x to x + dx,
$= -{4 \over 3}\pi G\rho {x^2}\left( {4\pi {x^2}dx\rho } \right) =- {{16{\pi ^2}} \over 3}G{\rho ^2}{x^4}dx$
Therefore, total work done by the external agent
$=- {{16{\pi ^2}G{\rho ^2}} \over 3}\int\limits_0^R {{x^4}dx}=- {{16{\pi ^2}G{\rho ^2}{R^5}} \over {15}}$
$\rho= {{3M} \over {4\pi {R^3}}}$ Therefore, self energy = $- {3 \over 5}{{G{m^2}} \over R}$

Launching speed of a Projectile

Let us suppose we have to launch a projectile having mass m to reach a height h. The gravitational potential energy of the projectile on the surface of Earth is
$U\left( R \right) =- {{GMm} \over R}$      ....(i)
Its gravitational potential energy at height h from the surface of earth is
$U\left( {R + h} \right) =- {{GMm} \over {\left( {R + h} \right)}}$    ...(ii)
Therefore, the change in potential energy
i.e. $U\left( {R + h} \right) - U\left( R \right) =- GMm\left\{ {{1 \over {R + h}} - {1 \over R}} \right\} = {{GM} \over {{R^2}}}{{mh} \over {\left( {1 + {h \over R}} \right)}} = {{mgh} \over {\left( {1 + {h \over R}} \right)}}$
[If h << R, $\Delta U = mgh$, which we used earlier]
This difference of P.E. is fulfilled by providing initial kinetic energy. If v be the velocity then
${1 \over 2}m{v^2} = {{mgh} \over {\left( {1 + {h \over R}} \right)}} \Rightarrow v = \sqrt {{{2gh} \over {\left( {1 + {h \over R}} \right)}}}$

MOTION OF PLANETS AND SATELLITES

Kepler’s Law

(i) The law of Elliptical Orbits – Each planet moves in an elliptical orbit with sun at one of its foci. (ii) The law of area – The radius vector of the planet relative to the Sun sweeps out equal area in equal time (iii)The Harmonic Law – The square of the period of revolution of the planet around the sun is proportional to the cube of the semi-major axis of the elliptical orbit.
Mathematically,  T2 µ a3
or    T2 = ka     where k is a constant and same for all planets.

Application 5

A planet moves around sun in an elliptical orbit of semi-major axis a  and eccentricity e. If the mass of sun is M, find the velocity at the perigee and apogee.

Solution

Let m be the mass of the planet.
As it is clear from the figure,
rP = a –  c
ra = a + c
Applying the conservation of angular momentum at the perigee and apogee, we get

mvprp = mvara
o
r          ${{{v_p}} \over {{v_a}}} = {{{r_a}} \over {{r_p}}} = {{a + c} \over {a - c}}$
Using conservation of mechanical energy, we get
${1 \over 2}mv_p^2 - {{GMm} \over {{r_p}}} = {1 \over 2}mv_a^2 - {{GMm} \over {{r_a}}}$
On rearranging, we get
$v_p^2 - v_a^2 = 2GM\left[ {{1 \over {{r_p}}} - {1 \over {{r_a}}}} \right]$     ....(ii)
Substituting the value of ${v_p} = {v_0}{{{r_a}} \over {{r_p}}}$
$v_a^2\left[ {{{r_a^2 - r_p^2} \over {r_p^2}}} \right] = 2GM\left[ {{{{r_a} - {r_p}} \over {{r_a}{r_p}}}} \right]$
or $v_a^2 = {{2GM} \over {{r_a} + {r_p}}}\left( {{{{r_p}} \over {{r_a}}}} \right)$
Since   rp = a c          and      ra = a + c, therefore,

$v_a^2 = {{GM} \over a}\left( {{{a - c} \over {a + c}}} \right)$
since $e = {c \over a}$   then  ${v_a} = \sqrt {{{GM} \over a}\left( {{{1 - e} \over {1 + e}}} \right)}$
Putting the value in (i), we get
${v_p} = \sqrt {{{GM} \over a}\left( {{{1 + e} \over {1 - e}}} \right)}$

Orbital Velocity

Orbital Velocity of a planet around the sun (or of a satellite around a planet)

Let m be the mass of the planet or satellite which revolves round the sun/planet of mass M in a orbit of radius r from the centre of the Sun/Planet with velocity vo. The required centripetal force is provided by the gravitational attraction between Sun and planet (or planet and satellite) i.e.
${{mv_o^2} \over r} = {{GMm} \over {{r^2}}}$
${v_o} = \sqrt {{{GM} \over r}}$
Now if planet be Earth and an artificial satellite is orbiting earth then
${v_o} = \sqrt {{{g{R^2}} \over r}}$
If r = R + h, where R is radius of earth and h is the height of the satellite from the surface of earth, then
${v_o} = \sqrt {{{g{R^2}} \over {\left( {R + h} \right)}}}$
If orbit is very close to surface of earth, then
${v_o} = \sqrt {gR}$

Energy of a Satellite
Energy of a satellite around a planet is of two types (i) P.E. and (ii) K.E.

$K = {1 \over 2}m{v^2}$   and   $U =- {{GMm} \over r}$
$E = K + U = {1 \over 2}m{v^2} - {{GMm} \over r}$            [since ${v^2} = {{GM} \over r}$]
$E =- {{GMm} \over {2r}}$ Velocity of satellite and nature of path.
(i)  V = Vo ® circular path around the planet.
(iii) V > Vo ® Elliptical path around the planet.
(iv) V < Ve ® Elliptical path around the planet.
(v) V = Ve ® Parabolic path escape from the planet
(vi) V > Ve ® Hyperbolic path escape from the planet.

Energy Graph

The variations of kinetic energy K have been shown by the graph as shown, potential energy U and total energy E with radius for a satellite in a circular orbit. From following graph it is clear that the value of U and E are negative and that of K is positive. As the r increases three curves have the tendency to approach the value of zero. Escape Velocity

It is defined as the minimum velocity needed for a particle projected upward so as to escape from the planet.
We know that the potential energy of a particle of mass m on the surface of the planet is given as
$U\left( {{R_P}} \right) =- {{GMm} \over {{R_P}}}$      ...(1)
From (1) it is clear that the amount of work required to move the particle from the surface of planet to infinity would be${{G{M_P}m} \over {{R_P}}}$.If this energy is converted into kinetic energy by any means then the corresponding acquired velocity by the particle will be the escape velocity (ve) i.e.
${1 \over 2}mv_e^2 = {{G{M_P}m} \over {{R_P}}}$
${v_e} = \sqrt {{{2G{M_P}} \over {{R_P}}}}$
If the planet is earth then
${v_e} = \sqrt {2gR}= 11.2$km/s

Application 6

A sky lab of mass 2$\times$103 kg is first launched from the surface of earth in a circular orbit of the radius 2R (from the centre of earth) and then it is shifted from this circular orbit to another circular orbit of radius 3R. Calculate the minimum energy required
(a)  to place the lab in the first orbit
(b)  to shift the lab from first orbit to the second orbit, Given R = 6400 km and
g = 10 m/s2.

Solution

According to the problem sky lab exists in three energy levels, our task is to calculate the total energy of the three level. i.e. on the surface, first orbit and second orbit.  Energy difference between first orbit and surface of the earth is the answer of (a) and that between first orbit and second orbit is the answer of (b).
Total mechanical energy of the sky lab on the surface of earth
${E_1} = KE + PE = 0 +- {{GMm} \over R} =- {{GMm} \over R}$
Total mechanical energy of the sky lab in first orbit i.e.
${E_2} =- {{GMm} \over {4R}}$
Total   mechanical energy of the skylab in the second orbit i.e.
${E_3} =- {{GMm} \over {6R}}$
(a)Required energy  $\Delta {E_1} =- {{GMm} \over {4R}} + {{GMm} \over R} = {3 \over 4}{{GMm} \over R}$
$\Rightarrow \Delta {E_1} = {3 \over 4}mgR = 9.6 \times {10^{10}}$
(b)In this case required energy
$\Delta {E_2} = - {{GMm} \over {6R}} + {{GMm} \over {4R}} = {1 \over {12}}{{GMm} \over R}$

$\Delta {E_2} = {1 \over {12}}mgR = 1.1 \times {10^{10}}J$

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