JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Variation in g With Height

Acceleration due to gravity at the surface of the earth

\[g=\frac{GM}{{{R}^{2}}}\]                                                   ...(i)            

Acceleration due to gravity at height h from the surface of the earth              

\[g'=\frac{GM}{{{(R+h)}^{2}}}\]                                         ...(ii)               

From (i) and (ii) \[g'=g{{\left( \frac{R}{R+h} \right)}^{2}}\]  ...(iii)            

=\[g\frac{{{R}^{2}}}{{{r}^{2}}}\]                                     ...(iv)

[As r = R + h]

(i) As we go above the surface of the earth, the value of g decreases because \[{g}'\propto \frac{1}{{{r}^{2}}}\].

(ii) If \[r=\infty \] then \[{g}'=0\], i.e., at infinite distance from the earth, the value of g becomes zero.

(iii) If \[h<<R\] i.e., height is negligible in comparison to the radius then from equation (iii) we get

\[{g}'=g{{\left( \frac{R}{R+h} \right)}^{2}}\]\[=g{{\left( 1+\frac{h}{R} \right)}^{-2}}\]\[=g\left[ 1-\frac{2h}{R} \right]\]                                   

[As \[h<<R\]]

(iv) If \[h<<R\] then decrease in the value of g with height :

Absolute decrease \[\Delta g=g-{g}'=\frac{2hg}{R}\]

Fractional decrease \[\frac{\Delta g}{g}=\frac{g-{g}'}{g}=\frac{2h}{R}\]

Percentage decrease \[\frac{\Delta g}{g}\times 100%=\frac{2h}{R}\times 100%\]