Derive the relationship between #"g"# and #"G"#?

1 Answer
Oct 2, 2017

#GM=gr^2#
r is the distance between object and the center of earth #r=h+R#
h is height of object from the surface and R is the radius of earth.
and
#g=4/3 piGRrho#

Explanation:

Suppose Earth is a sphere of radius #r#. It has mass #M#.
Gravitational Force on the object of mass #m# which is situated at a distance #r# from the center of Earth is
#F=(GMm)/r^2# . . .[A]

If the object is free falling from the height h from the surface of earth (or at a distance #r# from the center of earth) it experience the acceleration #g#

According to Newton's second law
Force on the object due to acceleration #g# is
#F=mg# . . .[B]

Comparing Equation [A] and [B]
#cancelmg=(GMcancelm)/r^2#
#gr^2=GM#
Where #r=h+R#
#R="Radius of Earth"#
#h="height of object from the surface of earth"#

If #h#<<#R#
then we can write #r=R+happroxR#
or
#gR^2=GM# . . .[C]

(When the object is near the surface of earth so we can neglect the height of object comparing with Radius of Earth)

If average density of earth is #rho#
then mass of earth #M="Volume"xx"density"#
#M=4/3piR^3rho#

In equation [c]
#gR^2=G4/3piR^3rho#
#g=(G4/3piR^3rho)/R^2#
#g=4/3piGRrho#