How does acceleration due to gravity vary with height and depth?

With height h above Earth's surface: g_h = g(1 + h/R)^(-2) ≈ g(1 - 2h/R) for h << R. With depth d below Earth's surface: g_d = g(1 - d/R). Notice that for small distances (h = d << R), the rate of decrease of g with height is twice as fast as the rate of decrease with depth (Δg_h ≈ 2 Δg_d). At Earth's center (d = R), g = 0.

Why is g maximum at the poles and minimum at the equator?

Two reasons: 1. Shape of Earth: The Earth is an oblate spheroid with equatorial radius R_e larger than polar radius R_p by about 21 km. Since g = GM/R^2, smaller radius at poles makes g larger. 2. Earth's rotation: The effective gravity at latitude λ is g' = g - ω^2 R cos^2 λ. At the equator (λ = 0°), centrifugal reduction is maximum (g_eq = g - ω^2 R), while at the poles (λ = 90°), rotation has no effect (g_p = g).

§ 7.2NCERT Class 11 · Physics · Chapter 7

Acceleration Due to Gravity (g) and its Variations

Q: Why is g maximum at the poles and minimum at the equator?

Two reasons: 1. Shape of Earth: The Earth is an oblate spheroid with equatorial radius R_e larger than polar radius R_p by about 21 km. Since g = GM/R^2, smaller radius at poles makes g larger. 2. Earth's rotation: The effective gravity at latitude λ is g' = g - ω^2 R cos^2 λ. At the equator (λ = 0°), centrifugal reduction is maximum (g_eq = g - ω^2 R), while at the poles (λ = 90°), rotation has no effect (g_p = g).

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 8 min read

The acceleration experienced by a freely falling body under the gravitational pull of the Earth is called the acceleration due to gravity, denoted by g. On the surface of the Earth, standard g is approximately 9.8 m/s². However, g is not a universal constant; it varies systematically with altitude, depth, latitude, and Earth's geometric shape.

1. Relation Between g and Universal Constant G

Consider a body of mass m placed on the surface of the Earth of mass M and radius R. By Newton's law of gravitation, the force acting on the body is:

F = G M m / R²

From Newton's second law of motion, F = m g. Equating the two expressions gives the fundamental relation:

g = G M / R²

In terms of average density ρ of the Earth (assuming a uniform sphere of volume V = 4/3 π R³):

g = G [4/3 π R³ ρ] / R² = 4/3 π G ρ R

Anatomy of Variations in Acceleration Due to Gravity

Analytical variations of g across altitude h, depth d, and rotational latitude λ

ALTITUDE (HEIGHT h):                            DEPTH (DEPTH d):
-------------------                            ----------------
gh = g [R / (R + h)]²                          gd = g [1 - (d / R)]
If h << R:                                     Linear decrease with depth:
gh ≈ g [1 - (2h / R)]                          At center of Earth (d = R):
Percentage decrease: (2h / R) * 100%           gd = 0 (weightlessness at center)

ROTATION OF EARTH (LATITUDE λ):                  SHAPE OF EARTH (OBLATE SPHEROID):
-------------------------------                ---------------------------------
g' = g - ω² R cos² λ                          Re > Rp (Equator bulges by ~21 km)
At Equator (λ = 0°):                             g = GM / R²
geq = g - ω² R  (Minimum g)                    gp > geq
At Poles (λ = 90°):                            g is MAXIMUM at poles,
gp = g          (Maximum g)                    MINIMUM at equator

Key JEE Rate-of-Change Rule: For small distances (h = d ≪ R), the rate of decrease of g with height above the surface is twice as large as the rate of decrease of g with depth below the surface (Δg_h = 2 Δg_d).

2. Analytical Derivations of Variations in g

1. Variation with Altitude (Height h Above Surface)

At a height h above the Earth's surface, the distance from the center is r = R + h:

g_h = G M / (R + h)² = [G M / R²] [1 + h/R]⁻² = g (1 + h/R)⁻²

Using binomial expansion for h ≪ R:

g_h ≈ g (1 − 2h/R) (Valid ONLY when h < 5% of R, i.e., h < 320 km)

2. Variation with Depth (Depth d Below Surface)

At a depth d below the surface, the distance from the center is r = R − d. By the shell theorem, the outer spherical shell of thickness d exerts zero net force. Only the inner sphere of radius (R − d) contributes:

M' = ρ [4/3 π (R − d)³] = M (R − d)³ / R³

g_d = G M' / (R − d)² = G M (R − d) / R³ = g (1 − d/R)

Note: Unlike the height formula, the depth formula g_d = g(1 − d/R) is exact for all depths from surface to center (assuming uniform density). At the center of the Earth (d = R), g = 0.

3. Variation with Latitude (Earth's Rotation)

As the Earth rotates with angular speed ω about its polar axis, a body at latitude λ revolves in a circle of radius r = R cos λ. The centrifugal acceleration acts radially outwards from the axis of rotation:

g' = g − ω² R cos² λ

At the Equator (λ = 0°): cos 0° = 1 ⇒ g_eq = g − ω² R (minimum effective gravity).
At the Poles (λ = 90°): cos 90° = 0 ⇒ g_p = g (rotation has zero effect).

3. Comparative Summary of Gravity Variations

Factor	Mathematical Formula	Nature of Graph	Extreme Values
Altitude (h)	g_h = g / (1 + h/R)²	Inverse-square decay (1/r²)	g → 0 as h → ∞
Depth (d)	g_d = g (1 − d/R)	Linear decline (straight line)	g = 0 at center of Earth
Latitude (λ)	g' = g − ω² R cos² λ	Trigonometric (cos² λ)	Max at poles; min at equator
Shape (Oblateness)	R_e − R_p ≈ 21 km	Inverse-square on radius	g_p − g_e ≈ 0.02 m/s²

JEE High-Probability Question: "What should be the angular speed of Earth so that bodies on the equator feel weightless?"
Weightlessness at equator means g_eq = 0 ⇒ g − ω² R = 0 ⇒ ω = √(g / R). Plugging numbers: ω ≈ 1.24 × 10⁻³ rad/s, which is approximately 17 times the present rotation speed (Earth's day would become 1.4 hours)!

4. Solved High-Yield JEE / NEET Practice Questions

Question 1: At what height above the Earth's surface does the acceleration due to gravity become 1/4th of its value on the surface? (Radius of Earth = R)

(A) R / 4

(B) R / 2

(D) 2R

Show answer

Correct Answer: (C) R

Step-by-step Solution:
1. Use the exact altitude formula (since 1/4th is a large change, approximation 1 − 2h/R cannot be used!):
g_h = g / (1 + h/R)².
2. Given g_h = g / 4:
g / 4 = g / (1 + h/R)² ⇒ (1 + h/R)² = 4.
3. Taking the positive square root:
1 + h/R = 2 ⇒ h/R = 1 ⇒ h = R.
At height equal to Earth's radius (6,400 km), gravity drops to 25% of its surface value.

Question 2: If the change in the value of g at a height h above the surface of the Earth is the same as at a depth d below the surface, then (for h << R):

(A) d = h

(B) d = 2h

(D) d = h² / R

Show answer

Correct Answer: (B) d = 2h

Step-by-step Solution:
1. Change in g at height h: Δg_h = g − g_h = g − g(1 − 2h/R) = 2g h / R.
2. Change in g at depth d: Δg_d = g − g_d = g − g(1 − d/R) = g d / R.
3. Equating both changes: 2g h / R = g d / R ⇒ d = 2h.

Question 3: If the Earth suddenly stops rotating about its polar axis, the acceleration due to gravity at the equator will:

(A) Increase by ω² R

(B) Decrease by ω² R

(D) Become zero

Show answer

Correct Answer: (A) Increase by ω² R

Step-by-step Solution:
1. When rotating, g_eq = g − ω² R.
2. If Earth stops rotating (ω = 0), the centrifugal reduction vanishes, so g'_eq = g.
3. Hence, the acceleration due to gravity at the equator will increase by ω² R. At the poles, it remains completely unchanged.

5. Frequently Asked Questions (FAQs)

1. When can we use g_h ≈ g(1 − 2h/R) versus the exact formula?

Use the approximation g_h ≈ g(1 − 2h/R) ONLY when h is very small compared to R (typically h < 320 km or h/R < 5%). For larger heights (e.g., h = R, h = 2R), you MUST use the exact formula g_h = g / (1 + h/R)².

2. What would a graph of g versus distance r from the center of Earth look like?

From r = 0 to r = R (inside Earth), g increases linearly with r (straight line through origin: g ∝ r). At r = R (surface), g reaches its maximum value G M / R². For r > R (outside Earth), g decreases quadratically following an inverse-square curve (g ∝ 1/r²).

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