Question

how much below the surface of earth does the acceleration due to gravity

a) reduce to 36%

b)reduces by 36% of its value on the surface of earth.

Answer 1

$$\begin{gathered} Accelerationduetogravitybelowtheearthsurfaceisgivenby: \\ g\text{'}=g\left(1-\frac{d}{R}\right) \\ wheredisthedepthbelowtheearthsurface \\ a)g\text{'}=36\%ofg \\ g\text{'}=\frac{36}{100}g \\ g\left(1-\frac{d}{R}\right)=\frac{36}{100}g \\ 1-\frac{36}{100}=\frac{d}{R} \\ \frac{64}{100}=\frac{d}{R} \\ d=\frac{64}{100}R \\ thereforebelowthe64\%ofR. \\ b)g\text{'}=g-36\%ofg \\ g\text{'}=g-\frac{36}{100}g \\ g\text{'}=\frac{64}{100}g \\ g\left(1-\frac{d}{R}\right)=\frac{64}{100}g \\ 1-\frac{64}{100}=\frac{d}{R} \\ d=\frac{36}{100}R \\ therefore36\%ofRbelowtheearthsurface \end{gathered}$$