Answer:
Weight of a body at a height becomes \[\frac{1}{4}\]of its value on the surface of earth \[\Rightarrow \]g at the height becomes \[\frac{1}{4}\]of its value on surface of earth.
\[g=\frac{GM}{{{R}^{2}}}\] \[\Rightarrow \] \[g\propto \frac{1}{{{R}^{2}}}\] (\[\because \]G, M are constant) \[\Rightarrow \]\[\frac{{{g}_{2}}}{{{g}_{1}}}=\frac{R_{1}^{2}}{R_{2}^{2}}\]\[\Rightarrow \]\[\frac{g}{4g}=\frac{{{(R)}^{2}}}{{{(R+h)}^{2}}}={{\left( \frac{R}{R+h} \right)}^{2}}\] \[\Rightarrow \]\[\frac{1}{2}=\frac{R}{R+h}\]\[\Rightarrow \]\[R+h=2R\]\[\Rightarrow \]\[h=R\] Surface Earth \[{{g}_{1}}=g\] \[{{g}_{2}}=\frac{g}{4}\] \[{{R}_{1}}=R\] \[{{R}_{2}}=R+h\]
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