Variation in g With Height
Category : JEE Main & Advanced
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 \[\frac{\Delta g}{g}\times 100%=\frac{2h}{R}\times 100%\]
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