Discussion
The height above the surface of the earth at which the acceleration due to gravity becomes \(\frac{g}{9}\) in terms of radius of earth \(R\) is ( \(g\) is acceleration due to gravity at the surface of the earth)
Answer: D
The height is 2R.
Gravity at height \(h\) is \(g' = g\left(\frac{R}{R + h}\right)^{2}\).
\[ \begin{aligned} \left(\frac{R}{R + h}\right)^{2} &= \frac{1}{9} \\ \frac{R}{R + h} &= \frac{1}{3} \\ R + h &= 3R \Rightarrow h = 2R \end{aligned} \]
The trap is 3R, which is the distance from the earth's centre, not the height above the surface. The small-height formula \(g' = g(1 - \frac{2h}{R})\) cannot be used here, since \(h\) is not small compared with \(R\).
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