A) 0.9 g
B) 0.99 g
C) 0.8 g
D) 1.01 g
Correct Answer: B
Solution :
\[h=32\,km\], \[R=6400\,km\], so \[h<<R\] \[{g}'=g\left( 1-\frac{2h}{R} \right)\]\[=g\left( 1-\frac{2\times 32}{6400} \right)\]Þ \[{g}'=\frac{99}{100}g=0.99\,g\]
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