Question

The acceleration due to gravity at a height h above the earths surface is $ 9m{{s}^{-2}}. $ If $ g=10\text{ }m{{s}^{-2}} $ on the earths surface, its value at a point at an equal distance h below the surface of the earth is

• A. $ 9\text{ }m{{s}^{-2}} $
• B. $ 8.5\text{ }m{{s}^{-2}} $
• C. $ 10m{{s}^{-2}} $
• D. $ 9.5\text{ }m{{s}^{-2}} $

Answer 1

The Correct Option is D

Above the surface of earth at height h, acceleration due to gravity $ g=g\frac{1}{{{\left( 1+\frac{h}{R} \right)}^{2}}} $ Given that, $ g=10\,m{{s}^{-2}} $ at surface of earth $ g=10\frac{1}{{{\left( 1+\frac{h}{R} \right)}^{2}}} $ ?(i) Below the surface of earth atdepth h, acceleration due to gravity $ g=g\left( 1-\frac{h}{R} \right) $ $ \therefore $ $ g=10\left( 1-\frac{h}{R} \right) $ ?(ii) From E (i) $ 9=10{{\left( 1+\frac{h}{R} \right)}^{-2}} $ By Binomial theorem $ 9=10\left( 1-\frac{2h}{R} \right) $ ?(iii) $ \therefore $ $ \frac{9}{10}-1=-\frac{2h}{R} $ or $ -\frac{1}{10}=-\frac{2h}{R} $ $ \frac{R}{20}=h, $ Now, we put the value of h in E (ii), we have $ g\,=10\left( 1-\frac{R/20}{R} \right) $ $ =10\left( 1-\frac{1}{20} \right) $ $ =10\left( \frac{19}{20} \right) $ $ =9.5\,m{{s}^{-2}} $