Chapter 15 Gravitation Part 1 – Physics free study material by TEACHING CARE online tuition and coaching classes
Introduction.
Newton’s law of Gravitation .
N ewton at the age of twentythree is said to have seen an apple falling down from tree in his orchid. This was the year 1665. He started thinking about the role of earth’s attraction in the motion of moon and other heavenly bodies.
By comparing the acceleration due to gravity due to earth with the acceleration required to keep the moon in its orbit around the earth, he was able to arrive the Basic Law of Gravitation.
Newton’s law of gravitation states that every body in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. The direction of the force is along the line joining the particles.
Thus the magnitude of the gravitational force F that two particles of masses m_{1}
and m2
separated by a
distance r exert on each other is given by
m m
F µ m1 m2
r 2
or F = G ^{1} ^{2}
r 2
Vector form : According to Newton’s law of gravitation
® – Gm1m2 ˆ = – Gm1m2 ® – Gm1m2 ®
F 12 =

r 2
r21
r 3 r21 =
®
 r21 
r21
ˆr12 = unit vector from A to B
®
Here negative sign indicates that the direction of F 12
is opposite to that of rˆ21 .
ˆr21 = unit vector from B to A,
®
®
Similarly
= – Gm1m2
rˆ = – Gm1m2 ®
= – Gm1m2 ®

F_{12} = gravitational force exerted
F 21 r 2 12
r 3 r12
®
 r12 
r12
on body A by body B
®
= Gm1m2 r 2
®
rˆ21
®
[Qrˆ12
= –rˆ21 ]
F 21 = gravitational force exerted
on body B by body A
\ It is clear that
F 12 = –
F 21 . Which is Newton’s third law of motion.
Here G is constant of proportionality which is called ‘Universal gravitational constant’.
If m1 = m2
and r = 1 then G = F
i.e. universal gravitational constant is equal to the force of attraction between two bodies each of unit mass whose centres are placed unit distance apart.
 The value of G in the laboratory was first determined by Cavendish using the torsional
 The value of G is 67×10^{–11} N–m^{2} kg^{–2} in S.I. and 6.67×10^{–8} dyne cm^{2}–g^{–2} in C.G.S. system.
 Dimensional formula [M 1 L3 T 2 ].
 The value of G does not depend upon the nature and size of the bodies.
 It also does not depend upon the nature of the medium between the two bodies.
 As G is very small hence gravitational forces are very small, unless one (or both) of the masses is
Properties of Gravitational Force.
 It is always attractive in nature while electric and magnetic force can be attractive or
 It is independent of the medium between the particles while electric and magnetic force depend on the nature of the medium between the
 It holds good over a wide range of It is found true for interplanetary to inter atomic distances.
 It is a central force e. acts along the line joining the centres of two interacting bodies.
 It is a twobody interaction e. gravitational force between two particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles
® ® ® ®
is the resultant of forces due to individual particles i.e.
While nuclear force is many body interaction
F = F 1 + F 2 + F 3 + ……..
 It is the weakest force in nature : As F_{nuclear} > F _{electromagnetic} > F _{gravitational} .
 The ratio of gravitational force to electrostatic force between two electrons is of the order of 10 ^{43} .
 It is a conservative force e. work done by it is path independent or work done in moving a particle round a closed path under the action of gravitational force is zero.
 It is an action reaction pair e. the force with which one body (say earth) attracts the second body (say moon) is equal to the force with which moon attracts the earth. This is in accordance with Newton’s third law of motion.
Note : @ The law of gravitation is stated for two point masses, therefore for any two arbitrary finite size bodies, as shown in the figure, It can not be applied as there is not unique value for the separation.
But if the two bodies are uniform spheres then the separation r may be taken as the distance between their centres because a sphere of uniform mass behave as a point mass for any point lying outside it.
Problem 1.  The gravitational force between two objects does not  depend on  [RPET 2003] 
(a) Sum of the masses
(c) Gravitational constant 
(b) Product of the masses
(d) Distance between the masses 
Solution : (a)
F = Gravitational constant ´ product of the masses .
(Distance between the masses)^{2}
Problem 2. Mass M is divided into two parts xM and (1 – x)M. For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is [EAMCET 2001]
 1 2
3 (c) 1 (d) 2
5
Solution : (a) Gravitational force F = Gm1m2
r 2
= GxM(1 – x)M r 2
= GM 2 x(1 – x) r 2
For maximum value of force
dF = 0 \
d é GM ^{2} x (1 –
ù = 0

Þ d (x – x ^{2} ) = 0
dx
dx
Þ 1 – 2x = 0
dx ê r ^{2}
Þ x = 1 / 2
 ú
û
Problem 3. The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth [SCRA 1998]
 Is the same (b) Is smaller (c) Is greater (d) Varies with its phase
Solution : (a) Earth and moon both exerts same force on each other.
Problem 4. Three identical point masses, each of mass 1kg lie in the x–y plane at points (0, 0), (0, 0.2m) and (0.2m, 0).
The net gravitational force on the mass at the origin is
(a)
(c)
1.67 ´ 10 ^{–}^{9} (ˆj + ˆj)N
1.67 ´ 10^{–}^{9}(ˆi – ˆj)N
(d)
3.34 ´ 10^{–}^{10}(ˆi + ˆj)N
3.34 ´ 10^{–}^{10}(ˆi + ˆj)N
Solution : (a) Let particle A lies at origin, particle B and C on y and x axis respectively
GmAmB ˆ
6.67 ´ 10 ^{–}^{11} ´ 1 ´ 1 ˆ
9 ˆ

FAC =
2 i =
AB
(0.2)^{2}
i = 1.67 ´ 10 i N
Similarly
FAB = 1.67 ´ 10 ^{–}^{9} ˆj N
\ Net force on particle A
F = F AC + F AB = 1.67 ´ 10 ^{–}^{9} (ˆi + ˆj) N
Problem 5. Four particles of masses m, 2m, 3m and 4m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be
(a)
24m^{2}G
a2
(b)
6m^{2}G
a2
a2
 Zero
Solution : (c) If two particles of mass m are placed x distance apart then force of
attraction
Gmm = F x 2
(Let)
Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experience four forces
FPA = force at point P due to particle
A = Gmm = F x 2
Similarly FPB = G2mm = 2F , FPC = G3mm = 3F and FPD = G4mm = 4 F
x ^{2} x ^{2} x ^{2}
Hence the net force on P
F net
= F PA + F PB + F PC + F PD = 2 F
\ F net = 2
Gmm = 2
x 2
[ x = a =
half of the diagonal of the square]
= .
a 2
Acceleration Due to Gravity.
The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.
The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g.
Consider a body of mass m is lying on the surface of earth then gravitational force on the body is given by
F = GMm
R 2
…..(i)
Where M = mass of the earth and R = radius of the earth.
If g is the acceleration due to gravity, then the force on the body due to earth is given by Force = mass ´ acceleration
or F = mg …..(ii)
From (i) and (ii) we have mg
= GMm R 2
\ g = GM R 2
Þ g = G æ 4 pR ^{3} r ö
…..(iii)
[As mass (M) = volume ( 4 pR ^{3} ) × density (r)]


R 2 ç 3 ÷
\ g = 4 prGR
3
3
…..(iv)
 From the expression g = GM = 4 prGR
it is clear that its value depends upon the mass radius and density
R ^{2} 3
of planet and it is independent of mass, shape and density of the body placed on the surface of the planet. i.e. a given planet (reference body) produces same acceleration in a light as well as heavy body.
 The greater the value of (M / R ^{2} ) or rR, greater will be value of g for that
 Acceleration due to gravity is a vector quantity and its direction is always towards the centre of the
 Dimension [g] = [LT^{–2}]
 it’s average value is taken to be 9.8 m/s^{2} or 981 cm/sec^{2} or 32 feet/sec^{2}, on the surface of the earth at mean sea
 The value of acceleration due to gravity vary due to the following factors : (a) Shape of the earth,
 Height above the earth surface, (c) Depth below the earth surface and (d) Axial rotation of the
Problem 6. Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the ratio of densities

æ r_{e} ö 5
of earth (r_{m}) and moon (r_{e}) is ç
è
÷ =
m ø 3
then radius of moon Rm in terms of Re will be [MP PMT 2003]
(a)
5 Re
(b)
1 Re
(c)
3 Re
(d)
1 Re
18 6 18 2 3
Solution : (a) Acceleration due to gravity g = 4 prGR \ g µ r R
or gm = rm . Rm
[As
g_{m} = 1
and r e = 5 (given)]
3
R_{m} æ g_{m} öæ r _{e} ö 1 5 5
ge r_{e} Re
ge 6
r _{m} 3
\ = ç
Re è g
÷ç
e øè r
÷ = ´
_{m} ø 6 3
\ Rm = 18 Re
Problem 7. A spherical planet far out in space has a mass
M_{0} and diameter
D0.
A particle of mass m falling freely near
the surface of this planet will experience an acceleration due to gravity which is equal to



[MP PMT 1987; DPMT 2002]
(a)
GM 0
/ D ^{2}
(b)
4mGM0 / D ^{2}
(c)
4GM 0
/ D ^{2}
(d)
GmM0
/ D ^{2}

Solution : (c) We know g = GM =
R 2
GM
(D / 2)^{2}
= 4GM D 2
If mass of the planet = M 0
and diameter of the planet = D0
. Then g = 4GM0 .


0
Problem 8. The moon’s radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is
[MP PMT 1997; RPET 2000; MP PET 2000, 2001]
 g
4
g (c)
5
g (d) g
6 8

GM gmoon
Mmoon R2
æ 1 öæ 4 ö^{2}
Solution : (b) Acceleration due to gravity g = \
= . earth = ç
÷ç ÷
R 2
gmoon = gearth ´ 16 = g .
gearth
Mearth
2
moon
è 80 øè 1 ø
80 5
Problem 9. If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth’s surface would [IITJEE 1981; CPMT 1981; MP PMT 1996, 97; Roorkee 1992; MP PET 1999]
 Decrease by 2% (b) Remain unchanged (c) Increase by 2% (d) Increase by 1%
Solution : (c) We know g µ 1
R 2
[As R decreases, g increases]
So % change in g = 2 (% change in R ) = 2 ´ 1% = 2%
\ acceleration due to gravity increases by 2%.
Problem 10. Mass of moon is
7.34 ´ 10^{22}kg.
If the acceleration due to gravity on the moon is 1.4m/s^{2}, the radius of the
moon is (G = 6.667 ´ 10 ^{–}^{11} Nm^{2} / kg ^{2} )
[AFMC 1998]
(a)
0.56 ´ 10^{4} m
1.87 ´ 10^{6} m
1.92 ´ 10^{6} m
1.01 ´ 10^{8} m
Solution : (b) We know g = GM
R 2
\ R = =
= 1.87 ´ 10^{6} m .
Problem 11. A planet has mass 1/10 of that of earth, while radius is 1/3 that of earth. If a person can throw a stone on earth surface to a height of 90m, then he will be able to throw the stone on that planet to a height [RPMT 1994] (a) 90m (b) 40m (c) 100m (d) 45m
GM g
M æ R
ö 2 1
æ 3 ö ^{2} 9
Solution : (c) Acceleration due to gravity g =
\ planet =
planet ç
earth ÷ =
´ ç ÷ =
R 2 g earth
M earth
ç Rplanet ÷
10 è 1 ø 10
è ø
If a stone is thrown with velocity u from the surface of the planet then maximum height H = u 2
2g
Hplanet = gearth Þ H
= 10 ´ H
= 10 ´ 90 = 100 metre.
Hearth gplanet
planet 9
earth 9
Problem 12. The radii of two planets are respectively R_{1}
and
R_{2} and their densities are respectively
r_{1} and
r_{2} . The ratio
of the accelerations due to gravity at their surfaces is [MP PET 1994]
(a)
g : g
= r_{1} : r _{2}
(b)
g : g
= R R
: r r


1 2 2 2
1 2
1 2 1 2 1 2
(c)
g1 : g 2 = R1 r _{2} : R2 r_{1}
(d)
g1 : g 2 = R1 r_{1} : R2 r _{2}
Solution : (d) Acceleration due to gravity g = 4 prGR
3
\ g1 : g 2 = R1 r_{1} : R2 r _{2} .
Variation in g Due to Shape of Earth.
Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The equatorial radius is
about 21 km longer than polar radius, from
g = GM
R 2



At equator GM
e



At poles GM
p
……(i)
…..(ii)



From (i) and (ii) ge = ^{p}

g p 2
Since
Requator
 Rpole
\ g pole > gequator
and
g = g + 0.018 ms 2

Therefore the weight of body increases as it is taken from equator to the pole.
Problem 13. Where will it be profitable to purchase 1 kg sugar (by spring balance) [RPET 1996]
 At poles (b) At equator (c) At 45° latitude (d) At 40°latitude
Solution : (b) At equator the value of g is minimum so it is profitable to purchase sugar at this position.
Problem 14. Force of gravity is least at [CPMT 1992]
(a) The equator (b) The poles
(c) A point in between equator and any pole (d) None of these
Solution : (a)
Variation in g With Height.
Acceleration due to gravity at the surface of the earth
g = GM
R 2
…..(i)
Acceleration due to gravity at height h from the surface of the earth
g‘ =
GM
(R + h)^{2}
…..(ii)
æ R ö ^{2}
From (i) and (ii)
g‘ = gç R + h ÷
…..(iii)
è ø
= g R 2
r 2
…..(iv) [As r = R + h]
 As we go above the surface of the earth, the value of g decreases because g¢µ 1 .
r 2
 If r = ¥ then g¢= 0 , e., at infinite distance from the earth, the value of g becomes zero.
 If h << R e., height is negligible in comparison to the radius then from equation (iii) we get
æ R ö 2
æ h ö^{–}^{2}
é 2hù
g¢ = gç R + h ÷ = gç1 + R ÷ = g ê1 – R ú
[As h << R ]
è ø è ø ë û
 If h << R then decrease in the value of g with height :
Absolute decrease Dg = g – g¢ = 2hg
R
Fractional decrease
Dg = g – g¢ = 2h
g g R
Percentage decrease
Dg ´ 100% = 2h ´ 100%
g R
Problem 15. The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth) [MP PET 2003]
 g
9
g¢ æ
R ö^{2} æ
R ö^{2}
1
g (c)
3
g
g (d) g
4
Solution : (a)
g = ç R + h ÷ = ç R + 2R ÷ = 9
\ g ¢ = .
9
è ø è ø
Problem 16. The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R) [EAMCET (Engg.) 2000]
(a) 8R (b) 9R (c) 10 R (d) 20R

Solution : (b) Acceleration due to gravity at height h is given by g¢ = gæ
R ö^{2}
÷
g æ R ö^{2} R 1
è R + h ø
Þ 100 = gç R + h ÷ Þ R + h = 10
Þ h = 9R .
è ø
Problem 17. At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of earth is (R = radius of earth) [CBSE PMT 2000; AIIMS 2000]
(a) 28N (b) 16N (c) 32N (d) 72N
æ R ö^{2}
æ ö ^{2}

ç R ÷
æ 2 ö^{2} 4 4
Solution : (c) Weight of the body at height R,
W¢ = Wç
÷ = W ç
R ÷ = Wç ÷ = W =
´ 72 = 32N.
è R + h ø
ç R + ÷


è ø
è 3 ø 9 9
Problem 18. If the distance between centres of earth and moon is D and the mass of earth is 81 times the mass of moon, then at what distance from centre of earth the gravitational force will be zero [RPET 1996]
(a) D/2 (b) 2D/3 (c) 4D/3 (d) 9D/10
Solution : (d) If P is the point where net gravitational force is zero then FPA = FPB
Gm1m =
x2
Gm2m
(d – x)^{2}
By solving x =
For the given problem d = D , m
= earth, m
= moon and m = 81m
\ m = m1
So x =
1
= = D
2
= 9D
1 2 2 81
1 + 1 10
9
Variation in g With Depth.
Acceleration due to gravity at the surface of the earth
g = GM
R 2
= 4 prGR
3
…..(i)
Acceleration due to gravity at depth d from the surface of the earth
g¢ = 4 prG(R – d) 3
…..(ii)
From (i) and (ii)
g¢ = g é1 – d ù
ëê R úû
 The value of g decreases on going below the surface of the From equation (ii) we get So it is clear that if d increase, the value of g decreases.
g¢ µ (R – d).
 At the centre of earth d = R \ g¢= 0 , e., the acceleration due to gravity at the centre of earth becomes zero.
 Decrease in the value of g with depth
Absolute decrease Dg = g – g¢ = dg
R
Fractional decrease
Dg = g – g¢ = d
g g R
Percentage decrease
Dg ´ 100% = d ´ 100%
g R
 The rate of decrease of gravity outside the earth ( if h << R ) is double to that of inside the
Problem 19. Weight of a body of mass m decreases by 1% when it is raised to height h above the earth’s surface. If the body is taken to a depth h in a mine, change in its weight is [KCET 2003; MP PMT 2003]
 2% decrease (b) 5% decrease (c) 1% increase (d) 0.5% increase
Solution : (b) Percentage change in g when the body is raised to height h ,
Dg ´ 100% = 2h ´ 100 = 1%
g R
Percentage change in g when the body is taken into depth d, Dg ´ 100% = d ´ 100% = h ´ 100% [As d = h ]
g R R
\ Percentage decrease in weight = 1 æ 2h ´ 100ö = 1 (1%) = 0.5% .

ç ÷
2 è R ø
Problem 20. The depth at which the effective value of acceleration due to gravity is g
4
is (R = radius of the earth)
[MP PET 2003]
(a) R (b)
3R (c)
4
R (d) R
2 4
Solution : (b)
g¢ = gæ1 – d ö Þ g = gæ1 – d ö Þ d = 3R



ç ÷ ç ÷
è ø è ø 4
Problem 21. Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine 100 km below the earth’s surface (Given R = 6400km) [AFMC 2000; Pb. PMT 2000]
(a)
9.66m / s^{2}
(b)
7.64m / s ^{2}
(c)
5.06m / s ^{2}
(d)
3.10m / s^{2}
Solution : (a) Acceleration due to gravity at depth d,
g ¢ = gé1 – d ù = gé1 –
100 ù = 9.8é1 – 1 ù
= 9.8 ´ 63 = 9.66m / s ^{2} .
ëê R úû
ëê 6400 úû
ëê 64 úû 64
Problem 22. The depth d at which the value of acceleration due to gravity becomes
1 times the value at the surface, is
n
[R = radius of the earth] [MP PMT 1999]
(a)
R (b)
R æ n – 1 ö
(c)
R (d)
R æ n ö
n ç n ÷ n^{2} ç n + 1 ÷
è ø è ø
Solution : (b)
g¢ = gæ1 – d ö Þ
g = gæ1 – d ö
Þ d = 1 – 1
Þ d = æ n – 1 ö R





ç ÷ ç ÷ ç ÷
è ø è ø n è ø
Variation in g Due to Rotation of Earth.
As the earth rotates, a body placed on its surface moves along the circular path and hence experiences centrifugal force, due to it, the apparent weight of the body decreases.
Since the magnitude of centrifugal force varies with the latitude of the place, therefore the apparent weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body.
If the body of mass m lying at point P, whose latitude is l, then due to rotation of earth its apparent weight can be given by mg¢ = mg + Fc
or mg¢ =
Þ mg¢ =
[As
F = mw 2 r = mw 2 R cos l ]

By solving we get g¢ = g – w 2 R cos 2 l
Note : @The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of earth makes with equatorial plane. It is denoted by l .
@ For the poles l = 90o and for equator l = 0o
 Substituting l = 90o
in the above expression we get g pole
= g – w 2 R cos 2 90o
\ g pole = g
…..(i)
i.e., there is no effect of rotational motion of the earth on the value of g at the poles.
 Substituting l = 0o
in the above expression we get
geqator
= g – w 2 R cos 2 0o
\ g equator
= g – w ^{2} R
…..(ii)
i.e., the effect of rotation of earth on the value of g at the equator is maximum.
From equation (i) and (ii)
g pole – gequator
= Rw 2 = 0.034m / s 2
 When a body of mass m is moved from the equator to the poles, its weight increases by an amount

m(g – g ) = mw 2 R
 Weightlessness due to rotation of earth : As we know that apparent weight of the body decreases due to rotation of If w is the angular velocity of rotation of earth for which a body at the equator will become weightless
g¢ = g – w 2 R cos 2 l
Þ 0 = g – w ^{2} R cos ^{2} 0^{o}
Þ g – w ^{2} R
\ w =
[As l = 0o
for equator]
or time period of rotation of earth T = 2p
w
= 2p
Substituting the value of
R = 6400 ´ 103 m
and
g = 10m / s 2
we get
w = 1 = 1.25 ´ 10 ^{3} rad
and
T = 5026.5 sec = 1.40 hr.
800
Note : @ This time is about
sec
1 times the present time period of earth. Therefore if earth starts rotating 17
17
times faster then all objects on equator will become weightless.
@ If earth stops rotation about its own axis then at the equator the value of g increases by
w 2 R and consequently the weight of body lying there increases by mw 2 R .
@ After considering the effect of rotation and elliptical shape of the earth, acceleration due to gravity at the poles and equator are related as


g = g + 0.034 + 0.018m / s 2 \ g = g + 0.052m / s 2
Problem 23. The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms^{–2}) [EAMCET 2000]
(a) 2.5×10^{–3}rad/sec (b) 5.0×10^{–1}rad/sec (c)
10 ´ 10^{1} rad / sec
(d)
7.8 ´ 10 ^{–}^{2} rad / sec
Solution : (a) Effective acceleration due to gravity due to rotation of earth g¢ = g – w ^{2}R cos^{2} l
Þ 0 = g – w ^{2}R cos^{2} 60^{o} Þ w 2R = g Þ w =
4
= 2 = 2 rad 800 sec
[As g¢ = 0
and l = 60^{o} ]
Þ w =
1
400
= 2.5 ´ 10 ^{–}^{3} rad .
sec
Problem 24. If earth stands still what will be its effect on man’s weight [AFMC 1994]
 Increases (b) Decreases (c) Remains same (d) None of these
Solution : (a) When earth stops suddenly, centrifugal force on the man becomes zero so its effective weight increases.
Problem 25. If the angular speed of earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly
(a)
1.5 hours
8 hours
18 hours
24 hours
Solution : (a) Time period for the given condition T = 2p
= 1.40 hr » 1.5 hr
nearly.
Mass and Density of Earth.
Newton’s law of gravitation can be used to estimate the mass and density of the earth.
As we know
g = GM , so we have
R 2
M = gR 2
G
\ M = 9.8 ´ (6.4 ´ 10^{6} )^{2}
6.67 ´ 10 ^{11}
= 5.98 ´ 10^{24} kg » 10^{25} kg
and as we know
g = 4 prGR , so we have
3
r = 3g 4pGR
\ r = 3 ´ 9.8 = 5478.4 kg / m^{3}
4 ´ 3.14 ´ 6.67 ´ 10 ^{11} ´ 6.4 ´ 10^{6}
Inertial and Gravitational Masses.
 Inertial mass : It is the mass of the material body, which measures its

If an external force F acts on a body of mass m_{i}, then according to Newton’s second law of motion
F = mi a
or mi = a
Hence inertial mass of a body may be measured as the ratio of the magnitude of the external force applied on it to the magnitude of acceleration produced in its motion.
 It is the measure of ability of the body to oppose the production of acceleration in its motion by an external
 Gravity has no effect on inertial mass of the
 It is proportional to the quantity of matter contained in the
 It is independent of size, shape and state of
 It does not depend on the temperature of body.
 It is conserved when two bodies combine physically or
 When a body moves with velocity v , its inertial mass is given by
m = , where m_{0} = rest mass of body, c = velocity of light in vacuum,
 Gravitational Mass : It is the mass of the material body, which determines the gravitational pull acting upon
If M is the mass of the earth and R is the radius, then gravitational pull on a body of mass mg
is given by
F = GMmg
or m
= F = F
R 2 g
(GM / R ^{2} ) E
Here mg
is the gravitational mass of the body, if
E = 1 then mg = F
Thus the gravitational mass of a body is defined as the gravitational pull experienced by the body in a gravitational field of unit intensity,
(3) Comparison between inertial and gravitational mass
 Both are measured in the same
 Both are scalars
 Both do not depends on the shape and state of the body
 Inertial mass is measured by applying Newton’s second law of motion where as gravitational mass is measured by applying Newton’s law of
 Spring balance measure gravitational mass and inertial balance measure inertial
(4) Comparison between mass and weight of the body
Mass (m)  Weight (W) 
It is a quantity of matter contained in a body.  It is the attractive force exerted by earth on any body. 
Its value does not change with g  Its value changes with g. 
Its value can never be zero for any material particle.  At infinity and at the centre of earth its value is zero. 
Its unit is kilogram and its dimension is [M].  Its unit is Newton or kg–wt and dimension are [ MLT ^{–}^{2} ] 
It is determined by a physical balance.  It is determined by a spring balance. 
It is a scalar quantity.  It is a vector quantity. 
Problem 26. Gravitational mass is proportional to gravitational [AIIMS 1998]
(a) Field (b) Force (c) Intensity (d) All of these
Solution : (d)
Problem 27. The ratio of the inertial mass to gravitational mass is equal to [CPMT 1978]
(a) 1/2 (b) 1 (c) 2 (d) No fixed number
Solution : (b)
Gravitational Field.
The space surrounding a material body in which gravitational force of attraction can be experienced is called its gravitational field.
Gravitational field intensity : The intensity of the gravitational field of a material body at any point in its field is defined as the force experienced by a unit mass (test mass) placed at that point, provided the unit mass (test mass) itself does not produce any change in the field of the body.
So if a test mass m at a point in a gravitational field experiences a force F then
I = F m
 It is a vector quantity and is always directed towards the centre of gravity of body whose gravitational field is
 Units : Newton/kg or m/s^{2}
 Dimension : [M^{0}LT^{–2}]
 If the field is produced by a point mass M and the test mass m is at a distance r from it then by
Newton’s law of gravitation
F = GMm
r 2
then intensity of gravitational field
I = F
m
= GMm / r 2
m
\ I = GM
r 2
 As the distance (r) of test mass from the point mass (M), increases, intensity of gravitational field decreases
I = GM ; \
r 2
I µ 1
r 2
 Intensity of gravitational field I = 0 , when r = ¥ .
 Intensity at a given point (P) due to the combined effect of different point masses can be calculated by vector sum of different intensities
Inet
= I1 + I 2 + I 3 + ……..
 Point of zero intensity : If two bodies A and B of different masses m_{1} and m_{2}
Let P be the point of zero intensity i.e., the intensity at this point is equal and apposite due to two bodies A and B and if any test mass placed at this point it will not experience any force.
are d distance apart.
For point P
I1 + I 2 = 0
Þ – Gm1 +
x 2
Gm2 = 0 (d – x)^{2}
By solving x = and (d – x) =
 Gravitational field line is a line, straight or curved such that a unit mass placed in the field of another mass would always move along this Field
lines for an isolated mass m are radially inwards.
 As
I = GM
r 2
and also
g = GM \
R 2
I = g
Thus the intensity of gravitational field at a point in the field is equal to acceleration of test mass placed at that point.
Gravitational Field Intensity for Different Bodies.
(1) Intensity due to uniform solid sphere
(2)
Intensity due to spherical shell
 Intensity due to uniform circular ring
At a point on its axis  At the centre of the ring 
I = GMr
(a 2 + r 2 )3 / 2 
I = 0 
(4)
Intensity due to uniform disc
Problem 28. Knowing that mass of Moon is
M where M is the mass of Earth, find the distance of the point where
81
gravitational field due to Earth and Moon cancel each other, from the Moon. Given that distance between Earth and Moon is 60 R. Where R is the radius of Earth [AIIMS 2000]
(a) 2 R (b) 4 R (c) 6 R (d) 8 R Solution : (c) Point of zero intensity x =
mass of the earth m_{1} = M , Mass of the moon m_{2} = M
81
and distance between earth & moon d = 60R
Point of zero intensity from the Earth x =
So distance from the moon = 60R – 54 R = 6R .
Problem 29. The gravitational potential in a region is given by
= 9 ´ 60R = 54 R
10
V = (3x + 4y + 12z) J/kg. The modulus of the gravitational
field at (x = 1, y = 0, z = 3) is [BHU 1997]
(a)
20 N kg ^{–}^{1}
(b)
13 N kg ^{–}^{1}
(c)
12 N kg ^{–}^{1}
(d)
5 N kg ^{–}^{1}
Solution : (b)
I = ç i + j + k ÷ = (3i + 4 j + 12k)
[As V = (3x + 4y + 12z) (given)]






è ¶x ¶y ¶z ø
It is uniform field Hence its value is same every where  I =
= 13 Nkg ^{–}^{1} .
Problem 30. The magnitudes of the gravitational field at distance r_{1} and r_{2} from the centre of a uniform sphere of radius R
and mass M are F_{1} and F_{2}
respectively. Then [IITJEE 1994]


F r F_{1} r ^{2}
(a)
1 = 1
if r_{1} < R and r_{2} < R
(b)
= ^{2} if r1 > R
and r_{2} > R
F2 r2
F r
2
2 1
F1 r ^{2}
(c)
1 = 1
if r1 > R
and r_{2} > R
(d)
= ^{1} if r1 < R and r2 < R



F2 r2 2 2
Solution : (a, b) We know that gravitational force µ Intensity µ 1
r 2
when r > R
[As I = GM ]
r 2
\ F1
F2
2




2 if
1
r1 > R
and r_{2} > R
and gravitational force µ Intensity µ r
when r < R
[As I = 4 prGr ]
3
\ F1
= r1
if r < R and r < R .
F2 r2 1 2
Problem 31. Infinite bodies, each of mass 3kg are situated at distances 1m, 2m, 4m, 8m…………. respectively on xaxis. The
resultant intensity of gravitational field at the origin will be
(a) G (b) 2G (c) 3G (d) 4G Solution : (d) Intensity at the origin I = I1 + I 2 + I 3 + I 4 + …….
= GM + GM + GM + GM + ……….




2 2 2 2
1 2 3 4
= GM é 1 + 1 + 1 + 1 +……….. ù
êë1^{2} 2^{2} 4 ^{2} 8 ^{2} úû
= GM é1 + 1 + 1 + 1 +…………. ù
ëê 4 16 64 úû
æ ö
ç 1 ÷ a
= GM ç ÷



ç 1 – 1 ÷
è ø
[As sum of G.P. =
1 – r ]
= GM ´ 4
3
= G´ 3 ´ 4 = 4G
3
[As M = 3kg
given]
Problem 32. Two concentric shells of mass
M1 and
M_{2} are having radii
r1 and
r2.
Which of the following is the correct
expression for the gravitational field on a mass m.
(a)
(b)
I = G(M1 + M 2 )
r 2
I = G(M1 + M 2 )
r 2
for r < r_{1}
for r < r_{2}
(c)
(d)
I = G M 2
r 2
I = GM1
r 2
for r1
for r1
< r < r2
< r < r2
Solution : (d) Gravitational field on a mass m due to outer shell (radius r_{2} ) will be zero because the mass is placed inside this
shell. But the inner shell (radius r ) behaves like point mass placed at the centre so I = GM1 for r < r < r
1 r 2 1 2
Problem 33. A spherical shell is cut into two pieces along a chord as shown in the figure. P is a point on the plane of the
chord. The gravitational field at P due to the upper part is I_{1}
relation between them
and that due to the lower part is
I_{2}. What is the
 I1 > I2
 I1 < I2
 I1 = I2
 No definite relation
Solution : (c) Intensity at P due to upper part = I_{1} and Intensity at P due to lower part = I _{2}
Net Intensity at p due to spherical shell I_{1} + I _{2} = 0
\ I1 = –I 2
Problem 34. A uniform ring of mass m is lying at a distance 1.73 a from the centre of a sphere of mass M just over the sphere where a is the small radius of the ring as well as that of the sphere. Then gravitational force exerted is
(a)
(b)
(c)
(d)
GMm
8a ^{2}
GMm
(1.73 a)^{2}
GMm a 2
1.73 GMm
8a ^{2}
Solution : (d) Intensity due to uniform circular ring at a point on its axis I =
Gmr
(a 2 + r 2 )3 / 2
\ Force on sphere F =
GMmr =
(a 2 + r 2 )3 / 2
= =
(4a 2 )3 / 2
8a ^{2}
[As r =
3a ]
Gravitational Potential.
At a point in a gravitational field potential V is defined as negative of work done per unit mass in shifting a test mass from some reference point (usually at infinity) to the given point i.e.,
V = – W
m
\ I = – dV
dr

= – F.dr
m
= ò I .dr
[As
F = I ]
m
i.e., negative gradient of potential gives intensity of field or potential is a scalar function of position whose space derivative gives intensity. Negative sign indicates that the direction of intensity is in the direction where the potential decreases.