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Variation of g with height
Concepts tested here
- g at a height
All Questions
2012 AIPMT-PRE 1 question
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The height at which the weight of a body becomes $\frac{1}{16}$th, its weight on the surface of earth (radius R), is$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2}$
$\frac{g}{16} = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \Rightarrow \left(1 + \frac{h}{R}\right)^2 = 16$
$1 + \frac{h}{R} = 4 \Rightarrow h = 3R$