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Dimensions Of Acceleration Due To Gravity

By Shailendra Singh

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Updated on 28 May 2025, 17:00 IST

Dimensions Of Acceleration Due To Gravity: Acceleration due to gravity is indeed the acceleration gained by an object as a result of the gravitational force. The SI unit for it is ms². This really has both a magnitude and a direction. So that, it is a vector quantity.

The symbol g is used to portray gravity’s acceleration. It has a standard value of 9.8 ms2 on the earth’s surface at sea level. The formula for its computation is based on Newton’s Second Law of Motion and Newton’s Law of Universal Gravitation.

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The acceleration due to gravity has been calculated using Newton’s Second Law of Motion and Newton’s Law of Universal Gravitation. Those same two laws result in the most practical form of the acceleration due to the gravity formula: g = G*M/R2, where g is the acceleration due to gravity, G is said to be the universal gravitational constant, M is mass, and R is distance.

Understanding the Dimension of Gravity (g)

The dimension of gravity, commonly denoted as g, refers to the acceleration due to Earth's gravitational force. This fundamental constant, approximately 9.8m/s2 on Earth's surface, plays a critical role in physics and daily life. Its dimensional formula of gravity is expressed as [L1T−2], representing length per time squared. 

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This signifies how gravity influences the rate of change in velocity over time. The value of g varies slightly depending on altitude, latitude, and local geological conditions, but it remains a cornerstone for calculations in mechanics, planetary motion, and engineering applications. By understanding g and its dimension, students and researchers can better explore gravitational phenomena and its impact on real-world systems.

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Formula of Acceleration due to Gravity

f = mg signifies the gravitational force acting on a body.

For which f denotes the force acting on the body, g denotes the acceleration due to gravity, and m denotes the body’s mass.

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f = GmM/(r+h)2 would be the universal law of gravitation.

Here,

  • f = force between two bodies,
  • G = universal gravitational constant (6.67×10-11 Nm2/kg2)
  • m = mass of the object,
  • M = mass of the earth,
  • r = radius of the earth.
  • h = height at which the body is from the surface of the earth.

Because the height (h) is so small in comparison to the radius of the earth, we reframe the equation as follows:

f = GmM/r2

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When equating both the expressions,

mg = GmM/r2

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Now, g = GM/r2

Thus, the dimensional formula of gravity will be g = GM/r2

Regardless of mass, all bodies experience the same gravitational acceleration.

Its own value on Earth is determined by the mass of the earth rather than the mass of the object.

Dimensional Formula of Acceleration Due To Gravity

The dimensional formula of Acceleration due to gravity can be represented as:

[M0 L1 T-2]

  • Here,
  • M = Mass
  • L = Length
  • T = Time

Derivation:

We have, Force = Mass × Acceleration due to gravity

  • Then, Acceleration due to gravity (g) = Force × [Mass]-1 . . . . . (1)
  • Now, dimensional formula of the mass = [M1 L0 T0] . . . . . (2)
  • Similarly, the dimensions of Force = [M1 L1 T-2] . . . . (3)
  • Now, when substitute equation (2) and (3) in equation (1) we get,
  • Acceleration due to gravity = Force × [Mass]-1
  • That is, g = [M1 L1 T-2] × [M1 L0 T0]-1 = [M0 L1 T-2].

Thus, acceleration due to gravity has been dimensionally represented as [M0 L1 T-2].

Dimensions Of Acceleration Due To Gravity FAQs

How is g used in physics?

  • Calculating weight (W=mg)
  • Solving problems in mechanics, such as motion, projectile trajectories, and oscillations
  • Understanding orbital mechanics and satellite motion

Does g depend on the mass of the falling object?

No, g is independent of the mass of the falling object. All objects in free fall accelerate at the same rate in the absence of air resistance.

What is the relationship between g and free fall?

During free fall, the only force acting on an object is gravity. The object accelerates at a constant rate equal to g, neglecting air resistance.

How does Earth's rotation affect g?

Earth's rotation causes a centrifugal force that slightly reduces the effective value of g, especially near the equator. This is why g is slightly less at the equator compared to the poles.

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