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Question

The free-fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is

The correct answer is

Earth is approximately an ellipsoid having its equatorial radius greater than its polar radius by 21

Understanding Free-Fall Acceleration Variation

The question asks why the free-fall acceleration, denoted by \(g\), increases as one moves from the equator towards the poles at sea level. This variation in \(g\) is a well-known phenomenon related to the properties of the Earth.

Factors Affecting Free-Fall Acceleration (g)

The acceleration due to gravity \(g\) at a point on the Earth's surface is primarily influenced by two main factors:

  • The distribution of mass within the Earth.
  • The distance of the point from the Earth's center of mass.
  • The rotation of the Earth (centrifugal effect).

The formula for gravitational acceleration at a distance \(r\) from the center of a spherical body of mass \(M\) is given by:

\( g = \frac{GM}{r^2} \)

where \(G\) is the gravitational constant. This formula shows that \(g\) is inversely proportional to the square of the distance from the center (\(g \propto 1/r^2\)). This means a smaller distance (\(r\)) results in a larger acceleration (\(g\)).

Earth's Shape and its Impact on 'g'

Earth is not a perfect sphere. Due to its rotation, it bulges slightly at the equator and is flattened at the poles. This shape is best described as an oblate spheroid, which is approximately an ellipsoid.

In an oblate spheroid, the radius from the center to the surface is greater at the equator than at the poles.

  • Equatorial radius (\(R_e\)) is the distance from the center to a point on the surface at the equator.
  • Polar radius (\(R_p\)) is the distance from the center to a point on the surface at the poles.

The equatorial radius is greater than the polar radius (\(R_e > R_p\)). The difference is significant, approximately 21 kilometers.

Since \(g \propto 1/r^2\), and the distance from the center (\(r\)) is larger at the equator (\(r = R_e\)) than at the poles (\(r = R_p\)), the value of \(g\) is smaller at the equator and larger at the poles. As one moves from the equator towards the poles, the distance from the Earth's center decreases, causing the free-fall acceleration \(g\) to increase.

Analysis of the Options

Let's evaluate the given options based on our understanding:

  • Option 1: Earth is a sphere with same density everywhere. This is incorrect because Earth is not a perfect sphere, and density is not uniformly distributed.
  • Option 2: Earth is a sphere with different density at the polar regions than in the equatorial regions. This is incorrect because Earth is not a sphere. While density variations exist, the primary reason for the large-scale variation in \(g\) from equator to pole at sea level is the Earth's shape.
  • Option 3: Earth is approximately an ellipsoid having its equatorial radius greater than its polar radius by 21. This statement accurately describes the shape of the Earth (approximately an ellipsoid) and the relationship between its equatorial and polar radii (\(R_e > R_p\), with the difference being around 21 km). As explained earlier, a larger radius at the equator leads to a smaller \(g\), and a smaller radius at the poles leads to a larger \(g\), explaining the increase in \(g\) from equator to poles.
  • Option 4: Earth is approximately an ellipsoid having its equatorial radius smaller than its polar radius by 21. This statement incorrectly describes the relationship between the equatorial and polar radii. The equatorial radius is larger, not smaller, than the polar radius.

Therefore, the reason for the increase in free-fall acceleration from the equator toward the poles is Earth's shape as an ellipsoid with a greater equatorial radius than its polar radius.

Location on Earth Distance from Center (\(r\)) Free-Fall Acceleration (\(g\))
Equator Largest (\(R_e\)) Smallest (\(g_{equator}\))
Poles Smallest (\(R_p\)) Largest (\(g_{poles}\))

Revision Table: Factors Affecting 'g'

Factor How it Affects 'g'
Distance from Center (\(r\)) \(g \propto 1/r^2\). Smaller \(r\) means larger \(g\). Earth's shape (ellipsoid) causes \(r\) to be smaller at poles than equator.
Earth's Rotation Causes a centrifugal force outward, reducing effective gravity. This effect is maximum at the equator and zero at the poles, further contributing to smaller effective \(g\) at the equator.
Mass Distribution Local variations in density can affect \(g\), but the overall trend from equator to pole is dominated by shape and rotation.
Altitude As altitude increases, \(r\) increases, and \(g\) decreases. (Not relevant at sea level, but a general factor).

Additional Information on Earth's Gravity

The actual variation in gravity on Earth's surface is complex and influenced by geology, topography, and latitude. However, the major systematic variation with latitude, observed at sea level, is primarily explained by Earth's shape (distance from center) and rotational effects.

The difference between equatorial and polar radii is approximately 21.38 km (6378.137 km at equator vs. 6356.752 km at poles).

The effect of Earth's rotation is also significant. A point on the equator experiences the maximum centrifugal force outwards, which opposes gravity. At the poles, there is no centrifugal force component opposing gravity. This rotational effect further contributes to \(g_{equator} < g_{poles}\).

The free-fall acceleration \(g\) is approximately 9.78 m/s\(^2\) at the equator and 9.83 m/s\(^2\) at the poles.

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Important Questions from Gravitation

  1. Which one of the following statement is true for the relation, \(F= \frac{{G{m_1}{m_2}}}{{{r^2}}}\) ?

    (All symbols have their usual meanings)
  2. A planet has a mass M 1and radius R 1. The value of acceleration due to gravity on its surface is g 1. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?

  3. Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be

  4. Suppose the force of gravitation between two bodies of equal masses is F. If each mass is doubled keeping the distance of separation between them unchanged, the force would become

  5. In a vacuum, a five-rupee coin, a feather of a sparrow bird and a mango are dropped simultaneously from the same height. The time taken by them to reach the bottom is t 1, t 2and t 3respectively. In this situation, we will observe that

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