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The acceleration due to gravity at the Earth's surface depends on

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CDS I 2022 English Previous Year Paper (10-April-2022)
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both its mass and radius.

Understanding Acceleration Due to Gravity on Earth

The acceleration due to gravity, often denoted by 'g', is a fundamental concept in physics. It represents the acceleration experienced by an object falling freely near the surface of a massive body like the Earth, assuming no air resistance. The value of this acceleration is not constant throughout the universe; it depends on the properties of the massive body itself.

Formula for Acceleration Due to Gravity

The acceleration due to gravity (g) on the surface of a planet can be calculated using Newton's Law of Universal Gravitation. The formula is given by:

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

Let's break down what each term in this formula represents:

  • \(G\): This is the Universal Gravitational Constant. It is a constant value approximately equal to \(6.674 \times 10^{-11} \, \text{N(m/kg)}^2\). It does not depend on the specific planet.
  • \(M\): This represents the mass of the planet. In this case, it's the mass of the Earth.
  • \(R\): This represents the radius of the planet. For the Earth's surface, this is the average radius of the Earth.

Dependency of Earth's Surface Gravity

Looking at the formula \(g = \frac{GM}{R^2}\), we can clearly see which physical properties of the Earth influence the value of 'g' at its surface:

  • The value of 'g' is directly proportional to the mass (\(M\)) of the Earth. This means if Earth had more mass (while keeping the radius the same), the acceleration due to gravity on its surface would be greater.
  • The value of 'g' is inversely proportional to the square of the radius (\(R\)) of the Earth. This means if Earth had a larger radius (while keeping the mass the same), the acceleration due to gravity on its surface would be smaller because the surface would be further from the center of mass.
  • The Universal Gravitational Constant (\(G\)) is a constant value and doesn't change.

Therefore, the acceleration due to gravity at the Earth's surface is determined by the combined values of Earth's mass (\(M\)) and Earth's radius (\(R\)).

Factor Influence on 'g' Notes
Earth's Mass (\(M\)) Directly proportional Higher mass means higher 'g'
Earth's Radius (\(R\)) Inversely proportional to \(R^2\) Higher radius means lower 'g'
Gravitational Constant (\(G\)) Constant factor Universal value

Based on the analysis of the formula and the factors involved, the acceleration due to gravity at the Earth's surface depends on both its mass and its radius.

Revision Table: Factors Affecting Surface Gravity

Concept Formula Relation Key Dependency
Acceleration due to gravity (g) \(g \propto M\) Planet's Mass
Acceleration due to gravity (g) \(g \propto \frac{1}{R^2}\) Planet's Radius

Additional Information: Variations in Earth's Gravity

While the formula \(g = \frac{GM}{R^2}\) gives the theoretical value of acceleration due to gravity at the surface, the actual value varies slightly across the Earth's surface. These variations are due to:

  • Earth's shape: The Earth is not a perfect sphere; it is an oblate spheroid, slightly flattened at the poles and bulging at the equator. This means the radius \(R\) is slightly larger at the equator than at the poles, causing 'g' to be slightly lower at the equator.
  • Altitude: As altitude increases, the distance \(R\) from the Earth's center increases, leading to a decrease in 'g'.
  • Local geology: Variations in the density of the Earth's crust below the surface can cause small local variations in 'g'.
  • Earth's rotation: The centrifugal force due to the Earth's rotation reduces the effective gravity, particularly at the equator.

However, the primary factors determining the average acceleration due to gravity at the Earth's surface are still its total mass and average radius, as described by the formula.

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Similar Questions

  1. For which one of the following does the centre of mass lie outside the body?

  2. An object weighs 9 N on the surface of the Earth. What would be its weight, when measured on the surface of a planet where the acceleration due to gravity is 9 times that on the surface of the Earth?


Important Questions from Gravity

  1. Which of the following law states that, "The force between two objects is directly proportional to the product of their masses?"

  2. Which of the following statements about the movement of planets is true?

    A. A planet's orbit is elliptical with the Sun at one of two focal points.

    B. The orbit of a planet is circular with the sun in the center.

    C. The orbit of a planet is elliptical with another planet in one of the two center-points.

    D. The orbit of a planet is circular with another planet in the center.

  3. If the mass of a person is 60 kg on the surface of earth then the same person’s mass on the surface of the moon will be:

  4. The centripetal force required to keep the moon in its orbit is provided by which force?

  5. How is the acceleration due to gravity denoted?

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