A) 0.99 g
B) 0.8 g
C) 1.01 g
D) 0.9 g
E) 9 g
Correct Answer: A
Solution :
\[g=\frac{g}{{{\left( 1+\frac{h}{{{R}_{e}}} \right)}^{2}}}=\frac{g}{{{\left( 1+\frac{32}{6400} \right)}^{2}}}=0.99\,g\]
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