Mechanics

UNIT 3
GRAVITATION AND CENTRAL FORCE MOTION
Gravitational potential: The work done in moving a unit mass from infinity to any point in the
gravitational field of a body is called the gravitational potential at that point due to the body.
Mathematically, gravitational potential V is expressed as
r r
Gm Gm
V    Edr     2 dr  
 
r r
Gravitational potential energy: The gravitational potential energy of a body of mass m’ at a point
distant r from a body of mass m is
Gmm
U  mV  
r
Hence gravitational potential at a point is equal to the gravitational potential energy of unit mass at that
point.
Inertial mass: The mass of a body as determined by the second law of motion from the cceleration
of the body when it is subjected to a force that is not due to gravity is called inertial mass. Inertial
mass measures an object's resistance to being accelerated by a force (represented by the
relationship F = ma).
Gravitational mass: The mass of a body, measured in terms of the gravitational force of attraction
on it due to another body, usually the earth is called gravitational mass.
Gravitational potential due to a spherical shell: Consider a spherical shell of radius ‘a’ and
surface density (mass per unit area of the surface) ‘σ’.
(i) At a point outside the shell:
Let P be a point outside the spherical shell at a
distance r from its centre O. Draw two planes CD and EF
very close to each other and cutting the shell perpendicular
to OP. Then these planes cut a ring or slice from the shell
of radius CK and thickness CE. Join OC, OE and CP,
such that CP = x. Let angle BOC = θ and angle COE = dθ.
Then,
Thickness of the slice, CE = a dθ
Radius of the slice, CK = a sinθ
 Circumference of the slice = 2π.CK = 2πasinθ
 Surface area of the slice = circumference x thickness = 2πasinθ x a dθ = 2πa2sinθdθ
 Mass of the slice = surface area x surface density = 2πa2sinθdθσ
G 2a 2 sin d
dV  
 Potential at P due to the slice, x (i)
From triangle COP,
CP 2  CK 2  KP 2  CK 2  (OP  OK ) 2  CK 2  OP 2  OK 2  2OP.OK
 OC 2  OP 2  2OP.OC cos 
 x 2  a 2  r 2  2ar cos 
Differentiating both sides, we get
2 xdx  2ar ( sin d )  2ar sin d [ a and r are constant]
ar sin d
 x
dx
Substituting this value of x in equation (i), we get
G 2a 2 sin d 2aG
dV   dx   dx
ar sin d r (ii)
The gravitational potential for the entire shell is therefore
PA r a
2aG 2aG 2aG r a 2aG 4a 2G
V    
r ra
dx   dx   x r a   2 a  
PB
r r r r

, But 4πa2 is the surface area of the whole shell
Therefore mass of the whole shell, m =4πa2σ
Gm
V  (iii)
r
Hence the mass of the whole shell behaves as if it were concentrated at the centre.
(ii) At a point on the surface of the shell:
For a point on the surface of the shell, r = a
Thus from (iii) we get
Gm
V  (iv)
a
(iii) At a point inside the shell: The gravitational potential inside
the sphere can be obtained by integrating equation (ii) within
the limits PB = a – r to PA = a + r, i.e.,
PA ar
2aG 2aG 2aG ar
V   dx    dx   xar   2aG 2r
PB
r r a r r r
4a 2G Gm
 V  4aG   
a a
Hence the gravitational potential at every point inside the spherical shell is the same as on the
surface of the shell itself.
Gravitational field due to a spherical shell:
The gravitational field E is related to the gravitational potential as
dV
E (potential gradient)
dr
Thus,
(i) at a point outside the spherical shell
d  Gm  Gm
E     2
dr  r  r
(ii) at a point on the surface of the spherical shell
Here, r = a
Gm
E 2
 a
(iii) at a point inside the spherical shell
d  Gm  Gm
E    0 
dr  a  ( a is constant)
Hence gravitational intensity inside a spherical shell is zero.
Gravitational potential due to a solid sphere:
Consider a solid sphere of radius ‘a’ and mass M.
(i) At a point outside the solid sphere:
Let P be a point outside the solid sphere at a distance r from
the centre O of the sphere. The sphere can be divided into a number
of concentric thin spherical shells of masses m1, m2, m3, etc.
But the potential at a point outside the shell is
Gm
V  
r
 Potential at point P due to all the shells or in other word, the solid sphere is
 Gm Gm2 Gm3  G GM
V   1    ...   (m1  m2  m3  ...)  
 r r r  r r (i)
Hence in the case of solid sphere also, the whole mass appears to be concentrated at the centre.
(ii) At a point on the surface of the solid sphere:
On the surface of a sphere, r = a