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

1 Answer
Oct 2, 2017

GM=gr^2GM=gr2
r is the distance between object and the center of earth r=h+Rr=h+R
h is height of object from the surface and R is the radius of earth.
and
g=4/3 piGRrhog=43πGRρ

Explanation:

Suppose Earth is a sphere of radius rr. It has mass MM.
Gravitational Force on the object of mass mm which is situated at a distance rr from the center of Earth is
F=(GMm)/r^2F=GMmr2 . . .[A]

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

According to Newton's second law
Force on the object due to acceleration gg is
F=mgF=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