Q104 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
Select the correct statement out of the following
- Correct answerAAcceleration due to gravity decreases as we go up or down from earth's surface.
- BAcceleration due to gravity is same at a height 'h' and depth 'd' from earth's surface, if $h = d$.
- CAcceleration due to gravity at 640 km above the earth surface decreases by 10% of its value at earth's surface (earth's radius is 6400 km)
- DThere is reduction in acceleration due to gravity at poles due to rotation of earth as the poles are lying on the axis of rotation and do not revolve.
Solution
The acceleration due to gravity at a height $h$ above the earth's surface is given by $g_h = g\left(1 + \dfrac{h}{R}\right)^{-2}$. 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 $g_d = g\left(1 - \dfrac{d}{R}\right)$. 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), $g_h \approx g\left(1 - \dfrac{2h}{R}\right)$ and $g_d = g\left(1 - \dfrac{h}{R}\right)$, so $g_h \neq g_d$.
At $h = 640$ km, $g_h = g\left(1 + \dfrac{640}{6400}\right)^{-2} = g\left(\dfrac{11}{10}\right)^{-2} = \dfrac{100}{121}g \approx 0.826g$, which is a decrease of about $17.4\%$, not $10\%$.
At the poles, the latitude $\lambda = 90^{\circ}$. The effective gravity is $g' = g - R\omega^2 \cos^2 \lambda = g - 0 = g$. 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.
The acceleration due to gravity at a depth $d$ below the earth's surface is given by $g_d = g\left(1 - \dfrac{d}{R}\right)$. 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), $g_h \approx g\left(1 - \dfrac{2h}{R}\right)$ and $g_d = g\left(1 - \dfrac{h}{R}\right)$, so $g_h \neq g_d$.
At $h = 640$ km, $g_h = g\left(1 + \dfrac{640}{6400}\right)^{-2} = g\left(\dfrac{11}{10}\right)^{-2} = \dfrac{100}{121}g \approx 0.826g$, which is a decrease of about $17.4\%$, not $10\%$.
At the poles, the latitude $\lambda = 90^{\circ}$. The effective gravity is $g' = g - R\omega^2 \cos^2 \lambda = g - 0 = g$. 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.
Hence, the correct answer is option: 'A'.