Gravitational potential due to a uniform solid sphere

Gravitational potential due to a uniform solid sphere:

1) At a point outside the sphere:

                    Consider a uniform solid sphere having radius  R and  mass  M.  Let  P  be  a point at a distance ‘r' from the centre Of the solid sphere, where the gravitational potential due to the sphere is to be determined.

                      Imagine the solid sphere to be made  up  of  a  large  number  of  thin  concentric spherical shell's of masses m₁, m₂, m₃, ..... etc. Now, the gravitational potential at P due to whole sphere is equal to sum of potentials due to all such shells.

                 .•. V = - [ Gm₁+ Gm+ .....]
                                   r            r

                         = - G ( m₁+ m₂ + .....)
                               r

                 .•. V = - GM
                                r

Where, m₁+m₂ + .... = M = Mass of the sphere.

Intensity  of gravitational field due to solid sphere:

E = - dV   =   - d  ( - GM )
         dr            r        r

.•. E = - GM
               r²

                     Negative sign shows that, the intensity of gravitational field is directed towards the centre O of the sphere.

2) At a point on the surface of the sphere:

At a point on the surface of the sphere, r=R :

V = - GM    and
           R
E = - GM
           R²

3) At a point inside the sphere :

           Consider a uniform point P inside the sphere at a distance‘r' from its centre O , as shown in figure.

Let R = radius of sphere  and
 ϱ = density of the sphere

           Now, with O as a centre and radius OP = r drawn a sphere.

           Then the point P lies on the surface of the solid sphere of the radius ‘r,' and inside the spherical shell of internal radius R and external radius R.

.•. volume of inner solid sphere = 4 π r³
                                                                 3

.•. Mass of inner solid sphere = 4 π r³ϱ
                                                            3

The potential V₁ at P due to inner solid sphere is :

V₁ = - G ( 4πr³ ϱ )  = 4πr² ϱ G     ..... (1)
                     r

               To find the potential V₂ due to the outer spherical shell, draw two concetric spheres of radii ‘x' and ( x + dx ) ,  forming a thin spherical shell of thickness ‘ dx '.

              .•.  Volume of shell = 4πx² dx

              .•. Mass of shell = 4πdx²dx

              .• Potential at P due to this shell

              P  = - G (4πdx²dx )   =   -4πdGx dx
                                  x
              .•. The potential V₂  at P due to thick spherical shell is obtained by integrating above equation between the limits x=r to x=R.

                 V₂ =  -4πϱG ᷊∫  ᷢᷢᷢᷢᷢᷢᷢᷢx dx = -4πϱG [x²/2] ᷊ ᷢᷢᷢᷢᷢᷢᷢᷢ

             .•. V₂ =  -4πϱG [ R² - r²]
                                              2

             .•. V₂ = -2πϱG ( R²-r² )    ......(2)

             .•. Total potential at P is given by :

                  V = V₁+V₂

             .•. V = - 4 π r²ϱ G-2πϱ G(R²-r²)
                          3

             .•. V = - 2πϱ G(2 r² + R² - r²)
                                      3

            .•. V = 2πϱ G(3R²-r²)
                       3

            .•. V =  -G( 4  π R³ϱ ) [3R²-r²/2R³]
                               3

                V = - GM [ 3R³- r²] .......(3) ,
                                      2R³
                Since,  4 πR³ϱ  = M
                             3

           This is an expression for the potential inside a solid sphere, at a distance, from its centre.

            The minimum potential at the centre of the sphere can be obtained by putting r=0.

             V₀= -3GM   ......(4)
                         2R

             The intensity of gravitational field at P inside the solid sphere is:

            E = - dV  =  -  d [-GM ( 3R²-r²)]
                     dr         dr               2R³

            E  = - GM [2r]
                      2R³

       .•. E = - GM   ... .... (5)
                     R³

Gravitational potential due to a spherical shell

Gravitational potential due to a spherical shell:

1) At a point outside the shell:

Consider a uniform spherical shell of radius R.

Let P be a point outside the spherical shell at a distance ‘r ' from the centre O of the shell.

Let ϱ = Mass per unit area of the surface.

As shown in figure, two planes AD and BC cut the shell vertically.

The element between the two planes is a slice ABCD in the form of a ring  of small angular width dϴ with the OP as axis.

Each element of the ring is at a distance , AP = x from the outside point P.

.•. Thickness of the shell, AB = R dϴ 

Radius of the shell, CK = R sinϴ

.•. Surface area of the slice= (2πR sinϴ) (R dϴ )  = 2πR² sinϴ dϴ 

.•. Mass of the slice= Surface area of  slice× Mass per unit area (ϱ)

Figure: Gravitational potential at a point outside the spherical shell.

.•. Mass of the slice, Mring = 2πR²sinϴ dϴ ϱ

.•. potential at P due to the ring is :

dV = - GMring
                X

      =   - G(2πR²ϱsinϴ dϴ )/x    .....(1)

      In ∆ APO :    x² = r² + R² - 2rR cosϴ

Differentiating above equation, we get :

      2x dx = 2rR sinϴ dϴ      
                           ( since,R and r are constant ) 

      .•. x = rR sinϴ dϴ 
                     dx

Substituting this value of ‘x' in eq. (1) , we get :

dV = - G ( 2πR²ϱsinϴ dϴ ) dx 
                     rR sinϴ dϴ

     

     = - 2πRϱG dx

              r

Integrating above equation for the whole shell between the limits, 

x = ( r- R ) to x = ( r+R ) , we get :

But, 4πR²ϱ = M   ( Mass of the whole shell )

.•.  V = - GM 
                 r     ...... (2)

Thus, for a point outside the shell, the shell behabes as if the whole mass is concentrated at the centre of the shell.

Now, the intensity of gravitational field at a point outside the shell is given by:

E = - dV / dr  

  

   = - d / dr ( - GM / r )

.•. E = - GM / r²      ...... (3)

Negative sign shows that intensity of gravitational field is directed towards the centre O of the shell.

(2) Potential on the surface of the shell :

For a point on the surface of the shell, r= R.

.•. Potential ,  V = - GM / R

Intensity of the field,  E = - GM / R²

Guass’s theorem for gravitation

Guass’s theorem for gravitation:

              Guass's law states that the total  gravitational  flux  over  a closed  surface,  having  a  unique outward  drawn normal to  it  at every point is - 4πG times the  total mass enclosed by that surface.

                The gravitational field E at a distance ‘r' from a point mass M is given by :

                 E= - GM/r²    .....(1)

                 Then the flux (ϕ) of the gravitational field, through the surface of the sphere of radius ‘r' is given by :

                  ϕ = - GM × 4πr² = - 4πGM

                            r²   

                  As shown in  figure, this flux  ϕ  is  due  to  the  normal component of gravitational field 

                 E = - GM cosϴ

                           r²

                  Let , dϕ = small flux through an  element  of  area  dA.

               .•.dϕ = -GM cosϴ dA  

                              r²

               .•. dϕ= E.dA = n. E dA

               .•. Total gravitational flux enclosed by the closed surface is :

           ϕ = ∫ E. dA = ∫ n^. E dA

       .•. ϕ =  ∫ E1.dA +  ∫ E2.dA + ........

       .•. ϕ = -4πG (M1 + M2 + ......)

       .•. ϕ = -4πGM 

       where , M = sum of all masses enclosed by the surface.

This is Gauss's law in gravitation.

                          

Gravitational potential energy

Gravitational potential energy:

               The  amount  of  work  done  in moving  a unit  mass  from  infinity  to  the point  under  consideration ,  in  the gravitational  field  of  the  body  is  called gravitational  potential  at  that  point.

               The  gravitational  potential  is scalar quantity and it's S.I unit is joule/kg.

Gravitational potential due to a point mass :

               Consider a point A at a distance ‘r' from a particle of mass M, as shown in figure.

               Then intensity of Gravitational field at point A is given by:

                E = - GM/r²   (along AM )

               .•.   Difference   of   Gravitational potential between two points A and B at a distance ‘dr' apart is :

                VA-VB= dV = - E dr     .......(1)

                .•. E = - dV/dr                ....... (2)

                This  is  the  relation  between gravitational  field  and  gravitational potential.

                 .•. the gravitational field is defined as  the  negative  gradient  of  gravitational potential.

                  Now, from eq. (1) , we get:

                  dV = - ( - GM/r²) dr = GM/r² dr

                  Hence, potential V at point A ( i.e work  done  in  moving  unit  mass  from infinity to point A) in the gravitational field of M is given by:

                   Thus, gravitational potential is zero at infinity. And  at all other points it is less than zero i.e it has a negative value.

Gravitational potential energy:

                    The  gravitational  potential energy  of  a  body  at  a  point  in  the gravitational field is defined as the amount of work done in bringing the body from infinity to the desired point.

                    Consider a point A at a distance ‘r' from a  body  of  mass M.  A  particle  of mass ‘m' is placed at B , at a distance (r+dr ) in the gravitational field of M.

Gravitational force of attraction acting on mass ‘m' is :

                      F = - GMm/r²  (along AM )

                     Hence, the work done (dW) by the force F in moving the mass‘m' from B to A , through a small distance ‘dr' is given by:

                     dW = - F dr = - (- GM/r²) dr

.•.  total work done in moving the mass ‘m' from ∞ to point A is given by:

                  .•. W = - GMm/r      .....(1)

                  This work done is stored in the body as it's gravitational potential energy (U) and it is given by :

                   U = - GMm/ r       .......(2)

.•. Gravitational P.E (U) =gravitational potential (V) × mass of the body (m)

                   Gravitational potential energy (U) is always negative . It is a scalar quantity and it's SI unit is joule ( J ).

Gravitational field

Gravitational Field

               The gravitational field is the space around a body within which any other mass experience a gravitational force of attraction.

Intensity of gravitational field at a point:

                It is defined as the gravitational force acting on a unit point mass, placed at a given point, in the gravitational field of the body.

                 Consider a body of mass M which sets up the gravitational field in the space surroundings it.→

                  Now, a small test mass ‘m' is placed at a point P at a distance ‘r' from the mass M. The test mass ‘m' experience an attractive gravitational force F.

                 Then the intensity of gravitational field E at point P is given by:  E=F/m

                 The intensity of gravitational field is a vector quantity.

      The direction of E is same as that of F.

                 The magnitude of gravitational field strength at point P is :

                 .•. E= F/m = - GM/r²      .....(1)    

                 since, F= - G Mm/ r²

                 Negative sign shows that the field is directed towards the mass M , In vector form, the intensity of gravitational field can be given as :

                 E= GM/r² r^  ....... (2) 

                 Where, r^= unit vector along the line of force.

                  It's SI unit is N/kg and 

dimensions are [M°L¹T-²].

                        

Acceleration due to gravity (g)

Acceleration due to gravity (g): 

               Consider the earth as a solid sphere of mass M and radius R. Let a small body of mass ’m' is placed at a height ‘h' from the surface of the earth, as shown in figure. According to Newton's law gravitation, the force exerted on a body by the  earth is given by:

                  F = G  Mm            ......(1)
                            (R+h)²

            All . freely   falling    bodies experience  a  constant   acceleration,   due   to Gravitational  pull of the earth. This  is  called  acceleration  due  to gravity  (g).  The  acceleration  that results due to earth's gravity is called acceleration due to gravity (g). 

                It is directed downwards towards the centre of the earth.The value of ‘g’ varies from a minimum at the equator and the maximum at the poles.

Now, according to Newton's second law of motion:

               F=mass×acceleration=m gh  .....(2)

 Where, gh= acceleration due to gravity at a                         height ‘h’.

From eq. (1) and (2) , we get:

            m gh =  G  Mm         
                              (R+h)²

            .•.  gh  =    GM        ......(3)
                            (R+h)²

On the surface of the earth, h=0 and gh=g.

         .•.acceleration due to gravity (g) on the surface of the earth is :

                     g = GM       .....(4).
                             R²  

This is the relation between G and g.

 
Expression for ‘g’ at a height ‘h’ above the earth's surface:

                The Gravitational force exerted by the earth on a body of mass ‘m' lying on its surface is given by:

                 mg = GMm                 .......(1)
                             R²

                 If the body is at a height ‘h' above the earth's surface, then we have:

                 m gh = GMm                .....(2)
                              (R+h)²

                 Where, gh = acceleration due to gravity  at  a height  ‘h'  from  the  earth's surface.

                  Eq. (2) by eq. (1) gives:

      gh =  R²       = R²                = (1+h/R)²
        g    (R+h)²    R²(1+h/R)²

    Expanding and neglecting higher powers:

                  gh/g= (1-2h/R)

                  .•. gh= (1-2h/R)g         .....(3)

                 Hence, as altitude ‘h' increase the value of ‘g' decrease.

Expression for ‘g'  at a depth ‘d' below the earth's surface:

Let  p =  density of the earth.
        M  = Mass of the earth having radius
        R= 4πR³p/3

                .•. The Gravitational attraction on a mass ‘m' lying on earth's surface is:

               mg = GMm/R² = G×4/3πR³p×m/R²
              
                      = 4/3 ×πRGm           ...... (1)

              Where ,  g  =  acceleration  due  to gravity on the earth's surface of the earth.

              Now, the Gravitational attraction on a mass ‘m' at a depth ‘d' is due  to inner spherical part of the earth having radius
(R-d).

              mgd= G 4/3 × π(R-d)³ρ×m  
                                           (R-d)²
          
                       = 4/3π(R-d) Gρm      .....(2)


              Where ,  gd  =  acceleration  due  to gravity at a depth‘d' below the earth's surface.


               Eq. (2) by (1) gives gd/g = (R-d)/ R

               .•. gd = (1-d/R)g               .....(3)

               Hence ,  the  value  of  gd  goes  on decreasing with depth and at the centre‘g' will be zero.

Newton's law of gravitation

Newton's Law of Gravitation:

               Every particle of matter in the universe attracts every other particle with a force which is directly proportion  -al to the square of the distance between the two particle.

               Hence, the gravitational force of attraction between two point masses ‘M’ and ‘m' separated by a distance ‘r' is given by:

    F ∝ Mm    and    F  ∝ 1/r²    .•.  F  ∝  Mm/r²
   
                    F=-G Mm             ..........(1)
                                r²
 

  Where, G= Gravitational Constant ,
                 M and m are two point masses,                         separated by a distance ‛r’.

                 The negative sign shows that the Gravitational force is always attractive.

             Gravitation Constant G : In eq. (1) if, M=m and r=1 then F=G (Numerically).

Definition: The universal G gravitational constant (G) is equal to the force of attraction between two bodies each of unit mass, lying ar a unit distance apaet.

              G=6.67×10-¹¹ N.m²/kg²  in S.I units

              The S.I Unit of G is newton.m²/kg²
              or N.m/kg²

Dimension of G is : [M¹L¹T-²] [L²]
                                                [M²]

                                     =   [M-²L³T-²]