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.