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Physics Question on Gravitation

The acceleration due to gravity at a height h above the earths surface is 9ms−2.9m{{s}^{-2}}. If g=10 ms−2g=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 ms−29\text{ }m{{s}^{-2}}

B

8.5 ms−28.5\text{ }m{{s}^{-2}}

C

10ms−210m{{s}^{-2}}

D

9.5 ms−29.5\text{ }m{{s}^{-2}}

Answer

9.5 ms−29.5\text{ }m{{s}^{-2}}

Explanation

Solution

Above the surface of earth at height h, acceleration due to gravity g=g1(1+hR)2g=g\frac{1}{{{\left( 1+\frac{h}{R} \right)}^{2}}} Given that, g=10 ms−2g=10\,m{{s}^{-2}} at surface of earth g=101(1+hR)2g=10\frac{1}{{{\left( 1+\frac{h}{R} \right)}^{2}}} ?(i) Below the surface of earth atdepth h, acceleration due to gravity g=g(1−hR)g=g\left( 1-\frac{h}{R} \right) ∴\therefore g=10(1−hR)g=10\left( 1-\frac{h}{R} \right) ?(ii) From E (i) 9=10(1+hR)−29=10{{\left( 1+\frac{h}{R} \right)}^{-2}} By Binomial theorem 9=10(1−2hR)9=10\left( 1-\frac{2h}{R} \right) ?(iii) ∴\therefore 910−1=−2hR\frac{9}{10}-1=-\frac{2h}{R} or −110=−2hR-\frac{1}{10}=-\frac{2h}{R} R20=h,\frac{R}{20}=h, Now, we put the value of h in E (ii), we have g =10(1−R/20R)g\,=10\left( 1-\frac{R/20}{R} \right) =10(1−120)=10\left( 1-\frac{1}{20} \right) =10(1920)=10\left( \frac{19}{20} \right) =9.5 ms−2=9.5\,m{{s}^{-2}}