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An in-depth exploration of newton's law of gravitation in vector form, discussing various scenarios such as gravitational forces between particles, gravitational fields, energy of satellites, and more. It also covers topics like uniform rings, discs, and spheres, and their gravitational effects. Particularly useful for understanding the principles of gravitation and their applications in physics.
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SYLLABUS
8. Gravitation
Physics Smart Booklet
Theory + NCERT MCQs + Topic Wise
Practice MCQs + NEET PYQs
r
Newton’s law of universal gravitation
Gravitation
Kepler’s laws of planetary motion
Based on the regularities in the motion of the planets, Kepler formulated a set of three laws known as Kepler’s laws
of planetary motion.
I Law (Law of orbits)
All planets move round the Sun in elliptical orbits with Sun at one of the foci.
II Law (Law of areas)
A line joining any planet and the Sun sweeps out equal areas in equal intervals of time.
Areal velocity: The area swept by the radius vector of a planet around the sun, per unit time is called areal velocity
of the planet. Areal velocity of a planet remains constant.
III Law (Law of periods)
The square of the period of any planet about the Sun is proportional to the cube of the semi-major axis of its
orbit. T
2 a
3
where T is the period and a is the semi major axis.
If T 1 and T 2 are the periods of any two planets and r 1 and r 2 are their mean distances from the Sun, then
3 2
3 1 2 2
2 1
r
r
Nearer planets move faster. For example the orbital speed of Earth is about 30 km s
− 1
. The speed of Jupiter is about
13.2 km s
− 1 with a period of 11.86 years and that of Saturn is 9.7 km s
− 1 with a period of 29.46 years.
th century, Saturn is the slowest. Infact, the Sanskrit name ‘shani’ refers to slowly
moving object. Saturn is seen for about 2½ years in each constellation
T 29. 2.5 years 12 12
=^
and passes
through 3 constellations in 7½ years, commonly known as ‘saade-sath’.
Newton’s Law of Gravitation: Every particle attracts every other particle with force that is proportional to the product of
the masses and inversely proportional to the square of their separation and acts along the straight line joining them.
2
1 2
r
mm F G
G is a universal constant, called the constant of gravitation.
11 2 2 G 6. 67 10 NmKg
− − =
The dimensional formula for G is[M LT ]
− 1 3 − 2
its centre.
particles within it.
superposition.
to zero at the centre of the earth as the particle is lowered down the centre. The reason for the initial increase is the
predominance of the effect of decrease in r over that of the shell of the earth’s crust that lies outside the radial
position of the particle. As the centre is approached, the effect of the outer shell predominates.
centre of the earth.
Newton’s law in vector form
2 12
1 2 3 12
1 2 21 rˆ r
Gmm r r
Gmm F (^)
→ →
where →
→ F 12 force exerted on particle of mass m 2 by particle of mass m 1
→ r 12 position vector of m 2 relative to m 1
rˆ 12 →^ unit vector in the direction of r 12
→
Gravity: It is the term used to describe the force on a body near the surface of a celestial body. The earth’s gravity is given
by
2 (R h )
GMm F
= where M → mass of the earth, R → average radius of the earth
h → height of a body of mass m above the surface of the earth.
Acceleration due to gravity (g): It is the acceleration of a body due to gravity. On the surface of the earth 2 R
g =
simultaneously.
3g
4 GR
=
.
Variation of g
(i) Due to altitude: Acceleration due to gravity at a height ‘h’ above surface of earth is
h (^2)
g (R h)
h
2h g g 1 R
(for h << R) R → radius of earth
Thus g decreases with altitude.
(ii) Due to depth: Acceleration due to gravity at a depth ‘d’ below the surface of earth is
d
d g g 1 R
for (d << R)
at d = R, i.e., centre of earth
gd = 0
g decreases with depth
(iii) Due to rotation of earth: Acceleration due to gravity at a latitude is given by
g = g − R
2 cos
2 → angular velocity of earth
(a) At poles: 2
2 g (^) p g R cos 2
= − = gmax
Uf −Ui = −
f
i
F. dr
2
1
r
r
2
1 2
r
Gmm dr = Gm 1 m (^2)
1 r 2
r
We choose gravitational potential energy of this system equal to zero when the separation between the particles is
infinity. Then potential energy of the system for any separation r is U(r) = − r
Gm 1 m 2
For a system of n particles, the potential energy is the sum of the potential energy of every pair of particles in the
system. In other words, the potential energy is additive.
Gravitational potential
The gravitational potential at any point is defined the negative of the work done by the gravitational force to bring
unit mass from infinity to that point.
Or
The gravitational potential at any point is the negative of the work done by a force in displacing a unit mass from that
point to infinity.
If V is the gravitational potential at any point, the potential energy U of a mass m at that point is given by
U = mV.
Gravitational potential at a point due to a point mass
Let P be a point at a distance r, along the x-axis, from a point mass M kept at the origin O.
By definition of potential given earlier, the potential at the point P is
r
Vp =− … ( 3 )
The following table gives potential due to various regular bodies.
Body Position Potential
At a point on its axis at a
distance r from its
center
2 2 a r
Uniform thin spherical shell of
radius a
At a distance r from its
center such that r
a
− r
At a distance r from its
center such that r <
a
− a
At a distance r from its
center such that r
a
− r
At a distance r from its
center such that r <
a
− 3 2 a
(3a^2 − r^2 )
Negative potential energy indicates that the gravitational force is attractive. Kinetic energy is always positive.
But potential energy can either be positive or negative. As long as the total energy (KE + PE) is
negative, the object is bound within the system.
Earth’s gravitational field
The gravitational force on a particle of mass m held at a distance h from the surface of the earth is
(R h)
GMm F 2
= towards the centre of the earth.
Hence gravitational field produced at a height h is G (^2) (R h)
m
f
= = , towards the earth.
At the surface of the earth, G (^2) R
f = , … (5)
towards the centre of the earth. We see that the gravitational field due to earth
is numerically equal to the acceleration due to gravity of the earth. The
variation of gravitational field due to earth with distance x from its centre is shown
in the figure.
Gravitational potential due to earth at a height h is R h
At the surface
2 R
gr g R
For convenience in special cases potential at the surface is taken to be zero. Anyhow it is only the difference in
potential is that significant.
As the height above the surface increases potential increases and the gravitational field is in the direction of
decreasing potential, that is towards the surface.. dr
dV F =− … (7)
Gravitational field due to a spherical shell at any point inside it is zero. Potential is constant equal to R
at all points on it. Hence gravitational field due to a shell is zero
= 0 dr
dV
. However, even inside
the spherical shell there will be gravitational field due to other objects. There is no gravitational
shielding.
The following table gives intensity of field due to various objects.
Object Position Field intensity
2 r
center 2 2 3 / 2 (a r )
GMr
center
(^222) r x
r
a
2 GMr
of radius a
At a distance r outside the shell
2 r
At a distance r inside the shell Zero
radius a
At a distance r outside the sphere
2 r
At a distance r inside the sphere
3 a
GMr
We have, the centripetal force 2
2
r
GMm
r
Kinetic energy is 2 r
GMm mv 2
2 = = … ( 1 )
Potential energy r
GMm U =− … ( 2 )
Total energy of the satellite E = K + U = 2 r
GMm 1 2
1
r
GMm =−
− … (3)
Close to the earth, 2
mgR
GMm E =− =−
As height increases, the kinetic energy decreases, potential energy increases.
(f) Rocket launching: It is a multistage process. Initially at the lift off, the launching rocket rises vertically to pass through
denser atmospheric layer with least fuel consumption. The first stage rocket falls off at about 60 km height, and the
second stage rocket is fired. The second stage rocket is gradually fitted by the guidance system. When it reaches the
desired height, the tracking system guides the rocket to move horizontally. At this stage, small rockets are fired to
separate the capsule from the second stage rocket and project it into space with the speed required to follow a
predetermined orbit.
Communication and Indian remote sensing (IRS) satellites
The artificial satellites are broadly classified into two types.
(i) Communication satellites and
(ii) Remote sensing satellites.
Communication satellites
simultaneously.
Remote sensing satellites
water resources, mineral deposits, agricultural crops etc.
fiction writer Arthur C. Clarke who proposed the idea of a communication satellite in 1945) is in the equatorial
plane of the earth. Its direction of angular motion coincides with that of the earth about its polar axis.
= R v R 3
8 G v (^) e e
r
GMm U =− ; 2 r
GMm K =− K 2 r
GMm E =W+K=− =−
GMm E =− holds for an elliptical orbit of semi major axis ‘a’ if r is replaced by a. So, total
mechanical energy of a satellite remains the same if it is put in different elliptical orbits with the same a, but
different values of eccentricity e (including 2000)
burn, are less than their corresponding values in the circular orbit.
W = −2K and E = −K for a circular orbit do not apply for the new elliptical orbit.
it in the orbit since gravity takes care of that
temporarily (iii) Cardiovascular system of an astronaut does not need to work hard to pump blood around the body
(iv) There is no preferred direction, no upside down or right side up. All orientations of an astronaut are equally
comfortable.
no sedimentation. (ii) There is no convection. (iii) A column of air expands on heating but stays where it is. (iii)
Surface tension is much more evident. It is because of surface tension that near normal dinning is possible in space.
Space shuttle straws come with small clamps to pinch them closed and keep the drinks from climbing.
Illustrations
1. A uniform solid sphere of mass M and radius ‘a’ is surrounded symmetrically by a uniform thin spherical shell of
equal mass and radius 2a. Find the gravitational field at a distance
3
2
(^) a from the centre.
4
9
GM
a
4
16
GM
a
2
3
GM
a
21
25
GM
a
Ans (A)
The point P is at a distance
3
2
(^) a from the centre.
As P is inside the cavity of the thin spherical shell,
the field here due to the shell is zero.
The field due to the solid sphere is 2 2
4
3 9
2
= =
GM GM g a a
2. A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown.
A particle of mass m 1 is placed on the line joining the two centres at a distance x from the point of contact of the
sphere and the shell [if r < x < 2 r ]. The magnitude of gravitational force due to sphere and shell on this particle is
1 3
Gmm x
r
1 3
Gmm ( x − r )
r
1 2 ( − )
Gmm
x r
1 2 ( − )
GMm
x r
Ans (B)
P (^) M
circular
Changing the orbit of a satellite
elliptical M
Three elliptical orbits with the same semimajor axis a, but e values (e = 0 for circular orbit of r = a)
2a 2a 2a
2 3 2 3
1 1 1
2 2 2
or 8
4 R 2 (^) = 4 R 1 = 4 10 km
1 4 1 1 1
2 4 1 2 2
2 2 10 km h
2 and 10 km h
R
T
R
T
−
−
= =
= =
v
v
In the given situation | v 1 − v 2 | = 10
4 km h
− 1
7. A body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity.
If the radius of the earth is R, the velocity of the body at a height h is
(A) gR (B)
2 2
3( + )
gR
R h
2
gR
R h
2 2
gR
R h
Ans (D): From conservation of energy
2 (^2 )
GMm m GM GMm m R R R h
v
2 1 2 2 2
2
GM GM GM GM gR
R h R R R h R h
v v =
8. A solid sphere of uniform density and radius 4 m is located with its centre at the origin ‘O’ of
coordinates. Two spheres of equal radius 1 m with their cavities at A (−2, 0, 0) and B (2, 0, 0)
respectively are taken out, leaving behind spherical cavities. The mass of each sphere taken
out is M. The gravitational field at B is
21
5
GM
31
16
31
8
GM
Ans (C)
Mass of whole sphere of radius 4 M (without cavities) is
3
0 3
4 (4) 3 64 4 (1) 3
= =
M
M M
The gravitational field at B = field due to whole sphere − field due to sphere A
0 = 3 − 2
GM r GM
R AB
3 2
(64 ) 2 31
4 4 16
G M GM GM
= − =
9. Three uniform spheres each having a mass m and radius R are kept in such a way that each touches the other two.
Find the magnitude of the gravitational force on any of the spheres due to the other two.
2
2
Gm
2
2
3
2 2
Gm
R
2
2
3
4 2
Gm
R
2
2 2 2
Gm
Ans (A)
= 60 , PQS is an equilateral triangle of side 2 R.
Forces on S:
2 2
2 2 (2 ) 4
Gm Gm F R R
= = and F = F
2 2 2 Fnet = F + F + 2 F cos60 = 3 F
2
2
net
Gm F R
10. The distance between two bodies A and B is r. Taking the gravitational force according to the law of inverse square
of r , the acceleration of body A is a. If the gravitatinal force follows an inverse fourth power law, then what will be
the acceleration of the body A?
a
r
a
r
a
r
a
r
Ans (D)
2 2
A B B A A
Gm m F Gm F a a r m r
= = = = ...(i)
4 4
A B B A A
Gm m F Gm F a a r m r
(^) = = = = ...(ii)
From (i) and (ii) 2 2
a a a a r r
11. Value of g on the surface of earth is 9.8 ms
− 2
. Find its value on the surface of a planet whose mass and radius both
are two times that of earth.
(A) 9.8 ms
− 2 (B) 19.6 ms
− 2 (C) 4.9 ms
− 2 (D) 10.32 ms
− 2
Ans (C)
2 2
9.8 ms ;
g R
− = =
2 2 2
. 4.9 ms (2 ) 2
g R R
− = = =
12. Calculate the change in the value of g at latitude 45 (when compared with g at equator). The radius of earth = 6.
10
3 km.
(A) 0.021 ms
− 2 (B) 0.0013 ms
− 2 (C) 0.043 ms
− 2 (D) 0.017 ms
− 2
Ans (D)
2 2 2 g (^) equator = g − R cos 0= g − R
2 2 2 45 cos 45^ 2
R g (^) g R g
= − = −
2 2 2 ( ) 2 2
R R g g g R
= (^) − (^) − − =
6 2 6.37 10 2
2 24 60 60
(^) = ^
= 0.017 ms
− 2
13. A body is weighed by a spring balance, 1000 N at the north pole. If only the rotation of earth is accounted for, how
much will it weigh at the equator?
Ans (B)
2 2 g (^) pole = g − R cos 90 = g mg = 1000
2 2 2 g = gequator = g − R cos 0 = g − R
2 (^1000 ) ( ) 1000 1 997 N
R mg g R g g
= − = (^) − (^) =
14. A particle of mass 20 g experiences a gravitational force of 4N along + ve X-direction. Find the gravitational field at
that point (magnitude)
(A) 50 N kg
− 1 (B) 100 N kg
− 1 (C) 200 N kg
− 1 (D) 150 N kg
− 1
Ans (C) 3 m 20 g 20 10 kg; F 4 N
− = = =
3 1 3
0.2 10 200 N kg 20 10
m
− −
15. Two spheres one of mass m has radius r. Another sphere has mass 4m and radius 2 r. The centre to centre distnace
betwen them is 12 r. Find the distance from the centre of smaller sphere where net graviational field is zero.
(A) 2 r (B) 4 r (C) 5 r (D) 3.5 r
We have F = 2
1 2
r
Gmm
2
1
F
2
1 2
2
1 2
( 3 r )
Gmm
r
Gmm
% decrease in force = 100 F
1
1 2
20. The ratio between masses of two planets is 2 : 3 and the ratio between their radii is 3 : 2. The ratio between
acceleration due to gravity on these two planets is
(A) 4 : 9 (B) 8 : 27 (C) 9 : 4 (D) 27 : 8
Ans (B)
We know that 2 R
g =
2
1
g
2 2
2
2 1
1
2
1
M
2 1
2 2
R
2
21. If the change in the value of g at a height h above the surface of the earth is same as at a depth d below it, then (both
d and h are much smaller than the radius of the earth)
(A) d = 2
h (B) d = h (C) d = 2h (D) d = h^2
Ans (C)
The acceleration due to gravity at height h is g
/ = g
2 h 1
The acceleration due to gravity at a depth d is g
// = g
d 1
Given that, g
/ = g
//
g
/
2 h 1 = g
//
d 1 2h = d
22. If R is the radius of the earth and g is acceleration due to gravity on the earth’s surface, the mean density of earth is
3 gR
4 gR
3 g
Rg
Ans (C)
We have, g = 2 R
2
3 = 4 RG
3 g
3
3 2
2 3
NCERT LINE BY LINE QUESTIONS
3
−
9
−
E
2
2
2
E
o
NCERT BASED PRACTICE QUESTIONS
sum
1 2
2
1 2
2
2
(c) (^2 2 )
2
2
E