Question

The value of acceleration due to gravity at the Earth's equator is less than at the poles because of shape and rotation of the earth

• A. true
• B. false

Answer 1

Answer: (A)

true

Answer 2

The value of acceleration due to gravity ($g$) at the Earth's surface is affected by two primary factors: the Earth's shape and its rotation.

1. Effect of Earth's Shape: The Earth is not a perfect sphere; it is an oblate spheroid, flattened at the poles and bulging at the equator. This means the radius of the Earth ($R$) is greater at the equator ($R_e \approx 6378 \text{ km}$) than at the poles ($R_p \approx 6357 \text{ km}$). The acceleration due to gravity is given by $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant and $M$ is the mass of the Earth. Since $R_e > R_p$, the gravitational acceleration due to the Earth's mass distribution alone would be less at the equator than at the poles.

2. Effect of Earth's Rotation: The Earth rotates about its axis. Due to this rotation, objects on the surface experience a centrifugal force directed away from the axis of rotation. This centrifugal force reduces the effective acceleration due to gravity. The effect is maximum at the equator (where the tangential velocity is highest and the radius of the circular path is maximum) and zero at the poles (which are on the axis of rotation). The effective acceleration due to gravity ($g'$) at a latitude $\lambda$ is approximately given by: $g' = g - \omega^2 R \cos^2 \lambda$ where $g$ is the acceleration due to gravity if the Earth were not rotating, $\omega$ is the angular velocity of the Earth, and $R$ is the radius at that latitude.

   • At the poles ($\lambda = 90^\circ$, $\cos 90^\circ = 0$), $g'_{poles} = g$.
   • At the equator ($\lambda = 0^\circ$, $\cos 0^\circ = 1$), $g'_{equator} = g - \omega^2 R_e$.

   Since $\omega^2 R_e$ is a positive value, the rotation reduces the effective gravity at the equator.

Both the larger radius at the equator (due to shape) and the maximum centrifugal effect at the equator (due to rotation) contribute to the acceleration due to gravity being less at the equator compared to the poles.