variation of g

Question

The height at which the acceleration due to gravity becomes g/9 (where g is surface gravity) in terms of Earth's radius R is:

RankGuru · Accepted Answer

1. **Formula for g at height**: $g' = \frac{g}{(1 + h/R)^2}$.
2. **Given condition**: $g' = \frac{g}{9}$.
   $$\frac{g}{9} = \frac{g}{(1 + h/R)^2} \implies (1 + h/R)^2 = 9$$
3. **Square root both sides**: $1 + \frac{h}{R} = 3 \implies \frac{h}{R} = 2$.
4. **Result**: $h = 2R$.

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More variation of g Practice Questions

At small height h ≪ R above Earth, g decreases approximately by factor?