8 Gravitation
NEWTON'S UNIVERSAL LAW OF GRAVITATION
GM e G(pRe3r) 4
Gravitational force is an attractive force between any two point g= = = prGRe ……(iv)
2 2 3
Re Re
masses M1 and M2 separated by any distance r.
where r is the density of earth.
æM M ö This is the relation between universal gravitational constant
It is given by F = G ç 1 2 2 ÷ 'G' and acceleration due to gravity 'g'.
è r ø
Change in the Value of Acceleration due to Gravity (g)
where G is the universal gravitational constant. (i) Due to rotation or latitude of earth : Let us consider a
Its value, G = 6.67 × 10–11 Nm2 kg–2 particle P at rest on the surface of earth at a latitude f.
Dimensions of G are [M–1 L3 T–2]
g ( f ) = g - Re w 2 cos 2 f
(a) The gravitational force is the central force and follows
inverse square law. It acts along the line joining the particles. w
(b) Since the work done by the gravitational force is
independent of the path followed and hence it is a r
conservative force. f
(c) It is the weakest force in nature. It is 1038 times smaller R g¢
than nuclear force and 1036 times smaller than electric force.
Strongest force being nuclear force (for small range)
followed by electric force.
(d) Gravitation is independent of the presence of other bodies
At poles f = 90º, so g(f) = g ...(v)
around it.
At equator f = 0º so g(f) = g –w2Re ...(vi)
ACCELERATION DUE TO GRAVITY (g) (ii) Due to shape of earth : The shape of the earth is not
The force of attraction exerted by earth on a body of mass m is perfectly spherical. It is flattened at poles. The polar radius
the force of gravity. is 21 km less than the equatorial radius. Hence acceleration
due to gravity at poles is greater than at the equator.
So the force of gravity from Newton’s gravitational law is gp = 9.83 ms–2 = value of g at poles
GM e m ge = 9.78 ms–2 = values of g at equator
F= ……(i) i.e. gp > ge
r2 (iii) At a depth 'd' below the earth surface : The acceleration
where Me = mass of earth due to gravity at the surface of earth.
r = distance of the body from the centre of earth. GM e éæ 4 ö ù
g= = êç ÷pR e rú G (from eqn. iv) ...(vii)
The force of gravity can be written as F = mg ……(ii) Re 2
ë è 3 ø û
where g is called the acceleration due to gravity.
From the expression (i) and (ii), we get d
P
GM e R-d
g= ……(iii)
r2 O R
If body is located at the surface of earth i.e., r = Re (radius of
earth) then Earth
Surface of earth
, EBD_7751
212 PHYSICS
If a body is taken at a depth d below the earth surface, then
the body is attracted by inner core of mass M' of earth (Mi )1 a 2
=
radius (Re– d). Then acceleration due to gravity at point P (M i )2 a1
is
(ff applied force on (Mi)1 and (Mi)2 is same.)
GM ¢ G é () p(R e - d)3 r ù (b) Gravitational mass Mg defined by Newton's law of
g¢ = = ë û
gravitation
(R e - d) 2 (R e - d) 2
= G (()p(R e - d)r) ...(viii) Fg W Weight of body
Mg = g = g = Accelerati on due to gravity
From eqns. (vii) and (viii) we have
æ R -d ö æ d ö 8. If two spheres of same material, mass and radius are put in
g¢ = g ç e ÷ = g ç1 - ÷ ...(ix) contact, the gravitational attraction between them is directly
è Re ø è Re ø proportional to the fourth power of the radius.
The value of acceleration due to gravity at the centre of the
earth is zero.
(iv) At a height 'h' above the earth surface : The acceleration R
2 B A
due to gravity at the surface of earth from expression (iv) 1 æ4 ö 4 R
is defined as F= Gç ps ÷ R
4 è3 ø
GM e m ¬ 2R ® m
g= 2
Re
This system may be considered to be made up of two mass
At a height h above the earth surface, the acceleration due each separated by 2R distance.
to gravity is
9. If the earth stops rotating about its axis, the value of g at
GM e GM e the equator will increase by about 0.35%, but that at the
g¢ = = ...(x)
(R e + h ) 2 r2 poles will remain unchanged.
-2
If the earth starts rotating at the angular speed of about 17
æR +hö times the present value, there will be weightlessness on
Þ Þ g ¢ = gç e ÷ [from eqs. (ix) & (x)]
ç R ÷ equator, but g at the poles will remain unchanged. In such a
è e ø
case, the duration of the day will be about 84 min.
If h << Re
10. If the radius of planet decreases by n%, keeping the mass
æ 2h ö unchanged, the acceleration due to gravity on its surface
g ¢ » g ç1 -
è R ÷ø
...(xi)
e
Dg - 2 DR
increases by 2n%. i.e., =
Keep in Memory g R
11. If the mass of the planet increases by m% keeping the
1. The value of the acceleration due to gravity on the moon radius constant, the acceleration due to gravity on its
is about one sixth of that on the earth and on the sun is
Dg DM
about 27 times that on the earth. surface increases by m% i.e., = where R = constant.
g M
æ gE ö 12. If the density of planet decreases by r % keeping the radius
çè gs = 27g E and g M = ÷ø
6 constant, the acceleration due to gravity decreases by r%.
2. The value of g is minimum on the mercury, among all 13. If the radius of the planet decreases by r% keeping the
planets. density constant, the acceleration due to gravity decreases
3. For h<<R, the rate of decrease of the acceleration due to by r%.
gravity with height is twice as compared to that with depth.
Example 1.
4. The value of g increases with the increase in latitude. Its
value at latitude q is given by : gq= g – Rw2 cos2q. What will be the acceleration due to gravity on the surface
of the moon with its radius as 1/4th the radius of the earth
5. Rotation of the earth about its own axis is responsible for
and its mass as 1/80th the mass of earth?
decrease in the value of g with latitude.
Solution :
6. The weight of the body varies along with the value of g (i.e.
W = mg) GM e GM m
g= and g' = ;
7. Inertial mass and gravitational mass- Re 2
R m2
(a) Inertial mass (Mi) defined by Newton's law of motion
2 2
External force applied on the body g' M m æ R e ö 1 æ4ö 1
Mi = F/a = \ = ç ÷÷ = ´ ç ÷ = ; \ g' = g / 5
Acceleration produced in it g M e çè R m ø 80 è 1 ø 5
, Gravitation 213
Example 2.
A mass M splits into two parts m and (M–m), which are R m
then separated by a certain distance. What will be the x
Center of earth
ratio m/M which maximizes the gravitational force
between the two parts?
Solution :
æ 4 px 3 ö
Gm (M - m) Gç d ÷m
F= ; For maxima, ç 3 ÷
x2 \F = è ø
2
dF G m 1 x
= 2 (M - 2m) = 0 or = 3M
dm x M 2 where d = density of earth = ;
Example 3. 4p R 3
Identical point masses each equal to m are placed at GMm
\F = .x [M is total mass of earth]
x = 0, x = 1, x = 2, x = 4, ... The total gravitational force R3
on mass m at x = 0 due to all other masses is Example 6.
(a) infinite (b) (4/3) Gm2 If the radius of the earth were to shrink by two per cent,
(c) (4/3 GM2) (d) zero its mass remaining the same, the acceleration due to
Solution : (c) gravity on the earth’s surface would
o m m m (a) decrease by 2% (b) increase by 2%
(c) increase by 4% (d) decrease by 4%
x=0 x=1 x=2 x=4 Solution : (c)
2 2 2
Gm Gm Gm
F0 = + + + .... GM
12
2 2
42 As g = 2 ; So, if R decreases then g increases.
R
é1 1 1 ù Taking logarithm of both the sides;
F0 = Gm 2 ê + + + ...ú
2 2 2 log g = log G + log M – 2 log R
ë1 2 4 û
It is a G.P. with common ratio 1/4. dg 2 dR
Differentiating it, we get =0+0– ;
é 1 ù
F0 = Gm 2 ê 4Gm 2 g R
\ ú F0 =
ë1 - û 3 dg æ -2 ö 4
Example 4. \ = -2ç ÷=
g è 100 ø 100
Four identical point masses each equal to m are placed
dg 4
at the corners of a square of side a. The force on a point \ % increase in g = ´100 = ´ 100 = 4%
mass m’ placed at the point of intersection of the two g 100
diagonals is Example 7.
(a) (4Gmm¢)/a2 (b) (2Gmm¢)/a2 An apple of mass 0.25 kg falls from a tree. What is the
(c) (Gmm¢)/a2 (d) zero acceleration of the apple towards the earth? Also calculate
Solution : (d) the acceleration of the earth towards the apple.
The forces acting on the mass m¢ placed at the point of Given : Mass of earth = 5.983 × 1024 kg,
intersection of diagonals are balanced by each other as Radius of earth = 6.378 × 106 metre,
shown in figure. Therefore net force is zero. G = 6.67 × 10–11 Nm2kg–2.
Solution :
m m Mass of apple, m = 0.25 kg,
Mass of earth, M = 5.983 × 1024 kg,
F
F m¢ Radius of earth, R = 6.378 × 106 metre
Gravitational constant, G = 6.67 × 10–11 Nm2kg–2
F Let F be the gravitational force of attraction between the
F apple and earth.
m m GmM
Then, F=
Example 5. R2
A tunnel is dug along the diameter of the earth. Determine Let a be the acceleration of apple towards the earth.
the force on a particle of mass m placed in the tunnel at a
distance x from the centre. F GmM GM
a= = = 2
Solution : m mR 2 R
Force on a mass m placed at a distance x from the centre is
equal to the force of gravity due to a mass of spherical 6.67 ´ 10 -11 ´ 5.983 ´ 1024
volume of earth of radius x. Þa= ms-2 = 9.810 ms-2
(6.378 ´ 10 )6 2
NEWTON'S UNIVERSAL LAW OF GRAVITATION
GM e G(pRe3r) 4
Gravitational force is an attractive force between any two point g= = = prGRe ……(iv)
2 2 3
Re Re
masses M1 and M2 separated by any distance r.
where r is the density of earth.
æM M ö This is the relation between universal gravitational constant
It is given by F = G ç 1 2 2 ÷ 'G' and acceleration due to gravity 'g'.
è r ø
Change in the Value of Acceleration due to Gravity (g)
where G is the universal gravitational constant. (i) Due to rotation or latitude of earth : Let us consider a
Its value, G = 6.67 × 10–11 Nm2 kg–2 particle P at rest on the surface of earth at a latitude f.
Dimensions of G are [M–1 L3 T–2]
g ( f ) = g - Re w 2 cos 2 f
(a) The gravitational force is the central force and follows
inverse square law. It acts along the line joining the particles. w
(b) Since the work done by the gravitational force is
independent of the path followed and hence it is a r
conservative force. f
(c) It is the weakest force in nature. It is 1038 times smaller R g¢
than nuclear force and 1036 times smaller than electric force.
Strongest force being nuclear force (for small range)
followed by electric force.
(d) Gravitation is independent of the presence of other bodies
At poles f = 90º, so g(f) = g ...(v)
around it.
At equator f = 0º so g(f) = g –w2Re ...(vi)
ACCELERATION DUE TO GRAVITY (g) (ii) Due to shape of earth : The shape of the earth is not
The force of attraction exerted by earth on a body of mass m is perfectly spherical. It is flattened at poles. The polar radius
the force of gravity. is 21 km less than the equatorial radius. Hence acceleration
due to gravity at poles is greater than at the equator.
So the force of gravity from Newton’s gravitational law is gp = 9.83 ms–2 = value of g at poles
GM e m ge = 9.78 ms–2 = values of g at equator
F= ……(i) i.e. gp > ge
r2 (iii) At a depth 'd' below the earth surface : The acceleration
where Me = mass of earth due to gravity at the surface of earth.
r = distance of the body from the centre of earth. GM e éæ 4 ö ù
g= = êç ÷pR e rú G (from eqn. iv) ...(vii)
The force of gravity can be written as F = mg ……(ii) Re 2
ë è 3 ø û
where g is called the acceleration due to gravity.
From the expression (i) and (ii), we get d
P
GM e R-d
g= ……(iii)
r2 O R
If body is located at the surface of earth i.e., r = Re (radius of
earth) then Earth
Surface of earth
, EBD_7751
212 PHYSICS
If a body is taken at a depth d below the earth surface, then
the body is attracted by inner core of mass M' of earth (Mi )1 a 2
=
radius (Re– d). Then acceleration due to gravity at point P (M i )2 a1
is
(ff applied force on (Mi)1 and (Mi)2 is same.)
GM ¢ G é () p(R e - d)3 r ù (b) Gravitational mass Mg defined by Newton's law of
g¢ = = ë û
gravitation
(R e - d) 2 (R e - d) 2
= G (()p(R e - d)r) ...(viii) Fg W Weight of body
Mg = g = g = Accelerati on due to gravity
From eqns. (vii) and (viii) we have
æ R -d ö æ d ö 8. If two spheres of same material, mass and radius are put in
g¢ = g ç e ÷ = g ç1 - ÷ ...(ix) contact, the gravitational attraction between them is directly
è Re ø è Re ø proportional to the fourth power of the radius.
The value of acceleration due to gravity at the centre of the
earth is zero.
(iv) At a height 'h' above the earth surface : The acceleration R
2 B A
due to gravity at the surface of earth from expression (iv) 1 æ4 ö 4 R
is defined as F= Gç ps ÷ R
4 è3 ø
GM e m ¬ 2R ® m
g= 2
Re
This system may be considered to be made up of two mass
At a height h above the earth surface, the acceleration due each separated by 2R distance.
to gravity is
9. If the earth stops rotating about its axis, the value of g at
GM e GM e the equator will increase by about 0.35%, but that at the
g¢ = = ...(x)
(R e + h ) 2 r2 poles will remain unchanged.
-2
If the earth starts rotating at the angular speed of about 17
æR +hö times the present value, there will be weightlessness on
Þ Þ g ¢ = gç e ÷ [from eqs. (ix) & (x)]
ç R ÷ equator, but g at the poles will remain unchanged. In such a
è e ø
case, the duration of the day will be about 84 min.
If h << Re
10. If the radius of planet decreases by n%, keeping the mass
æ 2h ö unchanged, the acceleration due to gravity on its surface
g ¢ » g ç1 -
è R ÷ø
...(xi)
e
Dg - 2 DR
increases by 2n%. i.e., =
Keep in Memory g R
11. If the mass of the planet increases by m% keeping the
1. The value of the acceleration due to gravity on the moon radius constant, the acceleration due to gravity on its
is about one sixth of that on the earth and on the sun is
Dg DM
about 27 times that on the earth. surface increases by m% i.e., = where R = constant.
g M
æ gE ö 12. If the density of planet decreases by r % keeping the radius
çè gs = 27g E and g M = ÷ø
6 constant, the acceleration due to gravity decreases by r%.
2. The value of g is minimum on the mercury, among all 13. If the radius of the planet decreases by r% keeping the
planets. density constant, the acceleration due to gravity decreases
3. For h<<R, the rate of decrease of the acceleration due to by r%.
gravity with height is twice as compared to that with depth.
Example 1.
4. The value of g increases with the increase in latitude. Its
value at latitude q is given by : gq= g – Rw2 cos2q. What will be the acceleration due to gravity on the surface
of the moon with its radius as 1/4th the radius of the earth
5. Rotation of the earth about its own axis is responsible for
and its mass as 1/80th the mass of earth?
decrease in the value of g with latitude.
Solution :
6. The weight of the body varies along with the value of g (i.e.
W = mg) GM e GM m
g= and g' = ;
7. Inertial mass and gravitational mass- Re 2
R m2
(a) Inertial mass (Mi) defined by Newton's law of motion
2 2
External force applied on the body g' M m æ R e ö 1 æ4ö 1
Mi = F/a = \ = ç ÷÷ = ´ ç ÷ = ; \ g' = g / 5
Acceleration produced in it g M e çè R m ø 80 è 1 ø 5
, Gravitation 213
Example 2.
A mass M splits into two parts m and (M–m), which are R m
then separated by a certain distance. What will be the x
Center of earth
ratio m/M which maximizes the gravitational force
between the two parts?
Solution :
æ 4 px 3 ö
Gm (M - m) Gç d ÷m
F= ; For maxima, ç 3 ÷
x2 \F = è ø
2
dF G m 1 x
= 2 (M - 2m) = 0 or = 3M
dm x M 2 where d = density of earth = ;
Example 3. 4p R 3
Identical point masses each equal to m are placed at GMm
\F = .x [M is total mass of earth]
x = 0, x = 1, x = 2, x = 4, ... The total gravitational force R3
on mass m at x = 0 due to all other masses is Example 6.
(a) infinite (b) (4/3) Gm2 If the radius of the earth were to shrink by two per cent,
(c) (4/3 GM2) (d) zero its mass remaining the same, the acceleration due to
Solution : (c) gravity on the earth’s surface would
o m m m (a) decrease by 2% (b) increase by 2%
(c) increase by 4% (d) decrease by 4%
x=0 x=1 x=2 x=4 Solution : (c)
2 2 2
Gm Gm Gm
F0 = + + + .... GM
12
2 2
42 As g = 2 ; So, if R decreases then g increases.
R
é1 1 1 ù Taking logarithm of both the sides;
F0 = Gm 2 ê + + + ...ú
2 2 2 log g = log G + log M – 2 log R
ë1 2 4 û
It is a G.P. with common ratio 1/4. dg 2 dR
Differentiating it, we get =0+0– ;
é 1 ù
F0 = Gm 2 ê 4Gm 2 g R
\ ú F0 =
ë1 - û 3 dg æ -2 ö 4
Example 4. \ = -2ç ÷=
g è 100 ø 100
Four identical point masses each equal to m are placed
dg 4
at the corners of a square of side a. The force on a point \ % increase in g = ´100 = ´ 100 = 4%
mass m’ placed at the point of intersection of the two g 100
diagonals is Example 7.
(a) (4Gmm¢)/a2 (b) (2Gmm¢)/a2 An apple of mass 0.25 kg falls from a tree. What is the
(c) (Gmm¢)/a2 (d) zero acceleration of the apple towards the earth? Also calculate
Solution : (d) the acceleration of the earth towards the apple.
The forces acting on the mass m¢ placed at the point of Given : Mass of earth = 5.983 × 1024 kg,
intersection of diagonals are balanced by each other as Radius of earth = 6.378 × 106 metre,
shown in figure. Therefore net force is zero. G = 6.67 × 10–11 Nm2kg–2.
Solution :
m m Mass of apple, m = 0.25 kg,
Mass of earth, M = 5.983 × 1024 kg,
F
F m¢ Radius of earth, R = 6.378 × 106 metre
Gravitational constant, G = 6.67 × 10–11 Nm2kg–2
F Let F be the gravitational force of attraction between the
F apple and earth.
m m GmM
Then, F=
Example 5. R2
A tunnel is dug along the diameter of the earth. Determine Let a be the acceleration of apple towards the earth.
the force on a particle of mass m placed in the tunnel at a
distance x from the centre. F GmM GM
a= = = 2
Solution : m mR 2 R
Force on a mass m placed at a distance x from the centre is
equal to the force of gravity due to a mass of spherical 6.67 ´ 10 -11 ´ 5.983 ´ 1024
volume of earth of radius x. Þa= ms-2 = 9.810 ms-2
(6.378 ´ 10 )6 2
