Physics - Gravitation Question with Solution | TestHub

The acceleration due to gravity at a height $h$ above the earth's surface is given by $g_h=g{\left(1+\frac{h}{R}\right)}^{-2}$. This shows that $g_h<g$ for any $h>0$.

The acceleration due to gravity at a depth $d$ below the earth's surface is given by $g_d=g\left(1-\frac{d}{R}\right)$. This shows that $g_d<g$ for any $d>0$.

Thus, acceleration due to gravity decreases as we go up or down from the earth's surface.

For $h=d$ (small values), $g_h\approx g\left(1-\frac{2h}{R}\right)$ and $g_d=g\left(1-\frac{h}{R}\right)$, so $g_h\ne g_d$.

At $h=640$ km, $g_h=g{\left(1+\frac{640}{6400}\right)}^{-2}=g{\left(\frac{11}{10}\right)}^{-2}=\frac{100}{121}g\approx 0.826g$, which is a decrease of about $17.4\%$, not $10\%$.

At the poles, the latitude $\lambda =90^{\circ }$. The effective gravity is $g^{\prime }=g-R{\omega }^2{\cos }^2\lambda =g-0=g$. There is no reduction in gravity at the poles due to the earth's rotation.

Hence, only the first statement is correct.

Answer: Acceleration due to gravity decreases as we go up or down from earth's surface.