Acceleration Due to Gravity of the Earth

Table of Contents
1. What is Acceleration due to Gravity?
2. What is Gravity?
3. Derivation of g near Earth's surface
4. Dependence and Important Consequences
5. Variation of g with Height (above the surface)
6. Variation of g with Depth (below the surface)
7. Variation of g due to the Shape of Earth (Latitude)
8. Variation of g due to Earth's Rotation
9. Combined Effects and Typical Numerical Values
10. Important Conclusions
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What is Acceleration due to Gravity?

Acceleration due to gravity is the acceleration gained by a body because of the gravitational attraction of the Earth (or any other gravitating body). It is a vector quantity having magnitude and direction (towards the centre of the Earth). The acceleration due to gravity is denoted by g. Its SI unit is m s^-2. The standard value of g at mean sea level on Earth is approximately 9.80665 m s^-2, commonly quoted as 9.8 m s^-2.

Formula, Unit and Dimensional Formula

What is Gravity?

Gravity is the attractive force between masses. For a mass m near the Earth (mass M and radius R), the gravitational force on m according to Newton's law of universal gravitation is

F = GMm / r²

where r is the distance between the centres of the two masses. If the body is on or near the surface, r ≈ R and the magnitude of acceleration of the test mass is obtained by using Newton's second law (F = ma) and cancelling m. This gives the expression for acceleration due to gravity:

g = GM / R²

Inertial Mass and Gravitational Mass

• Inertial mass is the property that appears in Newton's second law F = ma and measures resistance to acceleration.
• Gravitational mass is the property that appears in the gravitational force F = GMm/r².
• Experimentally, inertial and gravitational masses are equal to a high degree of precision; this equality is the basis of the principle of equivalence used in deriving g = GM/R².

Derivation of g near Earth's surface

Consider a test mass m located at the surface of the Earth of mass M and radius R.

Force due to gravity: F = GMm / R²

Newton's second law: F = m a

Equating the two expressions and cancelling m gives

a = g = GM / R²

Thus g depends on the mass M and radius R of the Earth, not on the test mass.

Dependence and Important Consequences

• All bodies fall with the same acceleration g (neglecting air resistance).
• The value of g at a place depends on the mass and radius distribution of the Earth and on the rotation of the Earth.
• g varies with altitude (height above surface), depth (below surface), latitude (shape of Earth) and due to Earth's rotation.

Variation of g with Height (above the surface)

For a point at height h above the Earth's surface, the distance from the centre of the Earth is r = R + h. The gravitational acceleration there is

g_h = GM / (R + h)² = g · (1 + h/R)^-2

Therefore g decreases with increasing height and tends to zero as h → ∞.

Approximation for small heights (h ≪ R):

(1 + h/R)^-2 ≈ 1 - 2h/R

g_h ≈ g (1 - 2h/R)

This linear approximation is useful for low-altitude changes (for example, a few kilometres above the surface).

Variation of g with Depth (below the surface)

Assuming the Earth to be a sphere of uniform density ρ, the mass enclosed within radius (R - d) (where d is the depth below the surface) is proportional to (R - d)³. The gravitational acceleration at depth d is due only to the mass inside this radius (external shells produce no net force inside a uniform spherical shell).

Let M be the total mass of Earth and g be acceleration on the surface.

M = ρ·(4/3)πR³, M_inside = ρ·(4/3)π(R - d)³

g = GM / R², g_d = G·M_inside / (R - d)²

Substituting M_inside gives

g_d = G·ρ·(4/3)π(R - d)³ / (R - d)² = (4/3)πGρ (R - d)

Since g = (4/3)πGρ R, we obtain

g_d = g · (R - d) / R = g (1 - d/R)

Thus g decreases linearly with depth and becomes zero at the centre (d = R).

Variation of g due to the Shape of Earth (Latitude)

The Earth is not a perfect sphere but an oblate spheroid - its equatorial radius R_e is slightly larger than its polar radius R_p. Because gravitational acceleration is inversely proportional to the square of the distance from the Earth's centre, this geometric difference causes g to vary with latitude. Ignoring rotation, the relation between gravity at pole and equator due to radius alone is

g_p / g_e = R_e² / R_p²

Since R_e > R_p, gravitational acceleration is larger at the poles than at the equator (for the same mass distribution). Therefore a person's weight is slightly greater at the poles than at the equator (other effects aside).

Variation of g due to Earth's Rotation

The Earth rotates about its axis with angular speed ω (ω = 2π / 86400 s ≈ 7.2921×10^-5 s^-1). A body on the Earth's surface experiences a centrifugal acceleration directed outward from the axis of rotation, of magnitude ω²r, where r = R cosθ is the distance from the rotation axis and θ is the latitude measured from the equator. The centrifugal acceleration reduces the apparent gravitational acceleration.

Effective (apparent) gravity g′ at latitude θ is

g′ = g - ω² R cos²θ

At the equator (θ = 0°) the reduction is maximum: g′ = g - ω²R. At the poles (θ = 90°) the centrifugal term is zero and g′ = g. Numerically, ω²R ≈ 0.0337 m s^-2, so the rotation reduces gravity by about 0.034 m s^-2 at the equator.

Combined Effects and Typical Numerical Values

• Standard gravity at sea level and 45° latitude is defined as g_n = 9.80665 m s^-2.
• At the equator, the combined effect of larger radius and centrifugal reduction gives a typical value ≈ 9.78 m s^-2.
• At the poles, typical gravitational acceleration ≈ 9.83 m s^-2.

Important Conclusions

• Acceleration due to gravity near Earth's surface is given by g = GM / R² and is independent of the test mass.
• g decreases with increasing altitude approximately as g_h ≈ g (1 - 2h/R) for h ≪ R.
• g decreases with depth inside the Earth approximately as g_d = g (1 - d/R) for a uniform Earth model.
• g is greater at the poles and smaller at the equator because of Earth's oblate shape and the centrifugal reduction due to rotation; consequently a person's weight is slightly greater at the poles than at the equator.

References for Further Reading

• Textbooks on Newtonian gravitation and classical mechanics covering gravitational field and potential (Class 11 physics and standard undergraduate texts).
• Standard tables of physical constants for precise values of G, mean Earth radius and standard gravity.