question_answer

The acceleration due to gravity at a height (take g = 10 on Earth's Surface.) above Earth's surface is\[\text{9 m}/\text{se}{{\text{c}}^{\text{2}}}\]. Its value at a point, at an equal distance below the surface of the Earth is_____\[\text{m}/\text{se}{{\text{c}}^{\text{2}}}\].

Answer:

\[g=10\text{ }m/{{s}^{2}};\]        \[{{g}_{1}}=9\text{ }m/{{s}^{2}};\]           \[{{g}_{2}}=?\text{ }h=d\] \[g{{g}_{1}}=\frac{2h}{R}.g\]                                       ????..(1) \[g{{g}_{2}}=\frac{d}{R}.g\]                                         ???..(2) But \[h=d\] \[\frac{g-{{g}_{1}}}{g-{{g}_{2}}}=\frac{2h\times g}{R\times hg}\times R=2\] \[\Rightarrow \]\[g-{{g}_{1}}=2(g-{{g}_{2}})\] \[\Rightarrow \]\[109=2\times 102{{g}_{2}}\] \[\Rightarrow \]\[1=202{{g}_{2}}\]\[\Rightarrow \]\[\text{ }19=2{{g}_{2}}\] \[\Rightarrow \]\[{{g}_{2}}=\frac{19}{2}=9.5\,m/{{s}^{2}}\]