Earth attracts every object lying an its surface towards its centre with a force known as gravitational towards its centre with a force known as gravitational pull or gravity.
Whenever force acts on any body it produces acceleration and in case of gravitation this acceleration produced under effect of gravity is known as accelesation due to gravity (g)
Value of accelesation due to gravity is independent of mass of the body and its value near surface of earth is 9.8 ms^{-2}
Expression for acceleration due to gravity
Consider mass of earth to be as M_{E} and its redius be R_{E} Suppose a body of mass M (much smaller then that fo earth) is kept at the earth surface. Force eseerted by earth on the body of mass m is
$F=\frac{GMM_{E}}{R_{E}^{2}}$
The force for the body due to earth produces acceleration due to gravity
(g) in the motion of the body. From Newton's Second law of motion
$f=mg$
from above equations
$g=\frac{GM_{E}}{R_{E}^{2}}$
which is acceleration due to gravity at earth's surface.
Acceleration due to gravity below and above the earth surface
(i)Above earth's surface
An object of mass m is placed at hight h above the earth's active in this object is
$ F=\frac {-GMM_{E}}{(R_{E}+h)^{2}}$
From this it can be concluded that value of g decreases as distance above surface of earth increases now,
$g=\frac{GM_{E}}{R_{E}(1+\frac{h}{R_{E}})^{2}}$
$g=\frac{g_{0}}{(1+\frac{h}{R_{E}})^{2}}$
Where $g_{0}=\frac{GM_{E}}{R_{E}^{2}}$ value of acceleration due to gravity at earth surface
for h<<R
$g=g_0(1+ \frac {h}{R_E})^{-2}$
Expanding by Binomial theriom
$g=g_0(1- \frac {2h}{R_E})$
Above equation tells us that for small hight h above surface of earth. value of g decreases by factor $(1-\frac {2h}{R_E})$
(ii)Below the earth's surface
If one goes inside the earth surface the value of g again decreases
$\rho$ = density of material of earth them
$m=\frac {4}{3}(R_E)^3 \rho$
From this acceleration due to gravity at earth's Surface is
$g_{0}=\frac{G(4/3)R_{E}^{3} \rho}{R_{E}^{2}}$
$g_0=\frac {4}{3}G R_E \rho$ ---(A)
Let g acceleration due to gravity at depth D below earth's surface
Body at depth d will experience force only due to portion of reduce (R_{E}-d) of earth's
outer spherical shell of thickness d will not exeperience any force
M is mass of the portion of earth with radius (R_{E}-d) then
$g=\frac{GM}{r^{2}}$
$M=\frac {4}{3}(R_E-d)^3 \rho$
$g=\frac{G(4/3)(R_{E}-d)^{3}P}{(R_{E}-d)^{2}}$
$g= \frac {4}{3}G (R_E -d) \rho $ ---(B)
Dividing epn (B) by (A)
$ \frac {g}{g_0}= (1- \frac {d}{R_E})$
or $g=g_0 (1- \frac {d}{R_E})$
from above equation is clear that acceeleration due to gravity also decreases with depth.
link to this page by copying the following text Also Read