A) \[\frac{g}{{{\left( 1\,+\,\frac{h}{R} \right)}^{2}}}\]
B) \[g\left( 1-\frac{2h}{g} \right)\]
C) \[\frac{g}{{{\left( 1-\frac{h}{R} \right)}^{2}}}\]
D) \[g{{\left( 1-\frac{h}{R} \right)}^{{}}}\]
Correct Answer: A
Solution :
\[g=\frac{G{{M}_{e}}}{{{R}^{2}}}\] \[\Rightarrow \] \[g\propto \frac{1}{{{R}^{2}}}\] \[\therefore \] \[\frac{g'}{g}=\frac{{{R}^{2}}}{{{(R+h)}^{2}}}\] Or \[g'=\frac{g}{{{\left( 1+\frac{h}{R} \right)}^{2}}}\]
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