Explain the variation of g with altitude.

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Concept understanding — Variation Of Gravity

Variation of Gravity: Why Your Weight Changes Even When You Don't

Imagine you step on a weighing scale at sea level in Mumbai, then carry that same scale to the top of Mount Everest. The scale would show a smaller number — you'd weigh less. But you haven't lost any mass. What changed?

The force pulling you down — gravity — is not constant everywhere on Earth. It varies. That's what we mean by variation of gravity.

The Core Idea

Gravity is the force with which the Earth pulls objects toward its centre. The strength of this pull depends on two things: the mass of the Earth and your distance from its centre. Since the Earth is not a perfect sphere and it spins, that distance and the effective pull change from place to place.

The acceleration due to gravity, denoted by $g$ , is approximately $9.8 \, \text{m/s}^2$ at sea level. But that's an average. The actual value can be slightly higher or lower depending on where you are.

Why Does Gravity Vary? Three Main Reasons

1. Altitude (Height Above Sea Level)

This is the most intuitive one. As you go higher, you move farther from the Earth's centre. Gravity follows an inverse-square law: double the distance, and the force becomes one-fourth.

The formula for $g$ at a height $h$ above the Earth's surface (where $R$ is Earth's radius, about 6400 km) is:

g_h = \frac{GM}{(R+h)^2}

For small heights compared to $R$ , we can approximate:

g_h \approx g \left(1 - \frac{2h}{R}\right)

Note

This means for every kilometre you go up, $g$ decreases by roughly $0.003 \, \text{m/s}^2$ . That's why at the top of a tall mountain, you weigh about 0.5% less than at sea level.

2. Depth (Going Underground)

What happens if you go down a mine or into the Earth's crust? Intuition might say gravity increases because you're closer to the centre. But the opposite happens.

Inside the Earth, the mass above you pulls upward, partially cancelling the pull from below. For a uniform Earth, only the mass inside the sphere of radius $r$ (your distance from the centre) contributes to gravity at that point.

g_d = \frac{GM'}{r^2}

Where $M'$ is the mass of the sphere of radius $r$ . If Earth had uniform density $\rho$ , then $M' = \frac{4}{3}\pi r^3 \rho$ , giving:

g_d = \frac{4}{3}\pi G \rho r

This means gravity decreases linearly as you go deeper. At the centre of the Earth, $g = 0$ — you'd be weightless, pulled equally in all directions.

Watch out

This linear decrease assumes uniform density. The real Earth has a dense iron core, so the actual variation is more complicated — gravity actually increases slightly as you go down through the crust before eventually decreasing.

3. Rotation of the Earth (Latitude Effect)

The Earth spins once every 24 hours. This rotation creates a centrifugal force that acts outward, away from the axis of rotation. This force effectively reduces the weight you feel.

The effect is strongest at the equator (where the rotational speed is highest, about 1670 km/h) and zero at the poles (where you're on the axis of rotation).

The effective $g$ at latitude $\phi$ is:

g_{\text{eff}} = g - \omega^2 R \cos^2 \phi

Where $\omega$ is Earth's angular speed ( $7.3 \times 10^{-5} \, \text{rad/s}$ ) and $R$ is Earth's radius.

At the equator ( $\phi = 0^\circ$ ), the reduction is about $0.034 \, \text{m/s}^2$ — roughly 0.35% of $g$ .

Tip

| Location | Approximate $g$ (m/s²) | Why? |

|----------|------------------------|------|

| Equator (sea level) | 9.78 | Fastest rotation + bulging equator |

| 45° latitude | 9.81 | Intermediate |

| North Pole | 9.83 | No rotation effect + closer to centre |

4. Shape of the Earth (Oblateness)

The Earth is not a perfect sphere. Because of its rotation, it bulges at the equator and flattens at the poles. The equatorial radius is about 21 km larger than the polar radius.

This means:

At the poles, you're closer to the Earth's centre → stronger gravity
At the equator, you're farther from the centre → weaker gravity …

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Step 1. For an object at height $h$ above the Earth's surface, its distance from Earth's centre is $R_e+h$ , so its acceleration due to gravity is

g'=\frac{GM_e}{(R_e+h)^2}=\frac{GM_e}{R_e^2}\left(1+\frac{h}{R_e}\right)^{-2}=g\left(1+\frac{h}{R_e}\right)^{-2}.

Step 2. For $h\ll R_e$ , a first-order binomial expansion of $(1+x)^{-2}\approx1-2x$ (with $x=h/R_e$ ) gives

g'\approx g\left(1-\frac{2h}{R_e}\right).

Step 3. Since the correction term is subtracted, $g'<g$ always -- $g$ decreases as altitude increases.

Step 4. Numerically, this drop is negligible for everyday heights (a mango falling 15 m sees essentially no change), but becomes significant at satellite altitudes (e.g. at 1600 km, $g'\approx g/1.5$ , roughly a 33% reduction). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. The value of an acceleration of a freely falling body is ______.$
›Reveal solutionSolution

A freely falling body (falling under gravity alone, no air resistance) accelerates at the constant value g ≈ 9.8 m/s^2 downward.

Near the Earth's surface, every freely falling object experiences the same acceleration g, independent of its mass, given approximately by g = GM/R^2 where G is the universal gravitational constant, M is Earth's mass and R is Earth's radius. Its standard value is g ≈ 9.8 m/ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Correct relation between g and density (ρ) of earth is:$
(a) g = (3/4)Rρ
(b) g = (4/3)πRρ
(c) g = (4/3)πGRρ
(d) g = (3/4)πGRρ
›Reveal solutionSolution

$g=GM/R^2$ combined with $M=\tfrac43\pi R^3\rho$ (sphere of density $\rho$ ) gives $g=\tfrac43\pi GR\rho$ .

The acceleration due to gravity at the Earth's surface is:

$g = \frac{GM}{R^2}$

Modeling the Earth as a uniform sphere of radius $R$ and density $\rho$ , its mass is: …

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CBSE 2026Set ANN1 mark
$Q. At the centre of Earth, the value of g is ________ .$
›Reveal solutionSolution

At the Earth's centre the acceleration due to gravity is zero, because no mass is enclosed below that point to produce a net gravitational pull.

Inside a uniform sphere, only the mass contained within the radius r at which the body sits contributes to the gravitational field (the outer shells exert no net force on a point inside them). The value of g at depth d below the surface is

g_d = g (1 - d/R), …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Acceleration due to gravity for the earth is maximum at (A) equator (B) pole (C) centre (D) none of these$
›Reveal solutionSolution
Acceleration due to gravity is maximum at the poles and minimum at the equator.

Two effects reduce $g$ as one moves from the pole to the equator:

Earth's shape: earth is an oblate spheroid, slightly flattened at the poles and bulging at the equator, so the equatorial radius is larger than the polar radius. Since $g \propto 1/R^2$ , a larger radius at the equator means smaller $g$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the center of the earth?$
(a) 100 N
(b) 50 N
(c) 200 N
(d) 400 N
›Reveal solutionSolution

Unlike the inverse-square fall-off above the surface, gravity inside a uniform earth decreases linearly with depth — so at half the radius, both $g$ and the weight are exactly halved.

Assuming the earth is a uniform sphere of radius $R$ , the acceleration due to gravity at depth $d$ below the surface is:

$g_d = g\left(1 - d/R\right)$

'Half way down to the centre' means $d = R/2$ :

…

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CBSE 2025Set sz1 markMCQ
$Q. At what depth below the surface of earth the value of 'g' is same as that of height of 5 km? (A) 10 km (B) 7.5 km (C) 5 km (D) 2.5 km$
›Reveal solutionSolution

Setting the depth expression equal to the height expression gives d = 2h = 10 km.

At height h above the surface: $g_h = g\left(1 - \dfrac{2h}{R}\right)$ .

At depth d below the surface: $g_d = g\left(1 - \dfrac{d}{R}\right)$ . …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. If the earth starts rotating from east to west instead of west to east about its axis, then the value of g on equator (A) will increase (B) will decrease (C) will be same (D) will be zero$
›Reveal solutionSolution

Reversing earth's spin direction doesn't change $g$ at the equator, since the correction term depends on $\omega^2$ .

At the equator, the effective (measured) value of gravity is reduced from the true gravitational value by the centrifugal effect of the earth's rotation: $g_{eq}=g_{true}-\omega^2R$ , where $\omega$ is the earth's angular speed and $R$ its equatorial radius.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Where will it be profitable to purchase one kilogram of sugar?$
(a) At poles
(b) At equator
(c) At 45 degree latitude
(d) At 40 degree latitude
›Reveal solutionSolution

Because g is minimum at the equator (due to Earth's bulge and rotation), buying sugar 'by weight' there gets you more actual mass for the same weight reading — so it is profitable to buy at the equator.

A spring balance measures WEIGHT (W = mg), not mass directly. Earth's value of g is slightly larger at the poles and slightly smaller at the equator (due to Earth's equatorial bulge and the effect of its rotation reducing effective gravity at the equator).

…

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CBSE 2024Set SET-AP55001 markMCQ
$Q. At height h above the earth's surface, the value of g changes by the same amount as it does at depth x inside the earth (both x and h are much smaller than the earth's radius) when:$
(a) x = h
(b) x = h/2
(c) x = 2h
(d) x = h^2
›Reveal solutionSolution

For small h, R: g at height h is g(1 − 2h/R); g at depth x is g(1 − x/R). Equating the fractional drops gives x = 2h.

At height h above Earth's surface (h << R):

g_h = g / (1 + h/R)^2 ≈ g(1 − 2h/R)

so the fractional decrease is Δg_h/g ≈ 2h/R.

At depth x below Earth's surface (x << R):

g_x = g(1 − x/R)

so the fractional decrease is Δg_x/g = x/R.

For the two changes to be equal:

2h/R = x/R

⟹ x = 2h

…

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CBSE 2024Set SET-NDP60001 markMCQ
$Q. The value of 'g' is maximum at:$
(a) At surface of earth
(b) At height
(c) At depth
(d) At center
›Reveal solutionSolution

g is maximum at the earth's surface and decreases both with height above it and with depth below it.

Above the surface, at height $h$ , $g_h = g\left(1 - \dfrac{2h}{R}\right)$ (for $h \ll R$ ) — this is always less than the surface value $g$ . Below the surface, at depth $d$ , $g_d = g\left(1 - \dfrac{d}{R}\right)$ — this too is always less than $g$ , and $g$ falls all the way to zero at the centre of the earth ( $d = R$ ). So moving away from the surface in either direction (up or down) reduces $g$ ; the surface itsel …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Value of acceleration due to gravity with increasing depth from earth surface :$
(a) increases
(b) decreases
(c) remains unaltered
(d) None
›Reveal solutionSolution

Below the earth's surface, only the mass enclosed within radius (R−d) contributes to gravity, so g falls linearly with depth.

At depth d below the earth's surface, treating the earth as a uniform sphere of radius R and density ρ, only the mass of the sphere of radius (R − d) contributes to the gravitational field at that depth (the shell outside contributes zero net field). This gives:

$g_d = g\left(1 - \frac{d}{R}\right)$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Match the values of Physical quantities given in Column I with the Numerical values given in Column II: Column-I (I) Acceleration due to gravity at the centre of earth (II) Escape speed of earth (III) Universal Gravitational constant (IV) Acceleration due to gravity Column-II (A) 9.8 ms^-2 (B) 6.67 x 10^-11 Nm^2 kg^-2 (C) 0 (D) 11.2 kms^-1$
(a) (I)-D (II)-A (III)-B (IV)-C
(b) (I)-C (II)-A (III)-D (IV)-B
(c) (I)-A (II)-B (III)-D (IV)-C
(d) (I)-C (II)-D (III)-B (IV)-A
›Reveal solutionSolution

(I) g at earth's centre = 0 (C); (II) escape speed = 11.2 km/s (D); (III) G = 6.67x10^-11 N m^2 kg^-2 (B); (IV) g at surface = 9.8 m/s^2 (A). This is option (d).

Matching each physical quantity to its numerical value:

(I) Acceleration due to gravity at the centre of the earth: treating the earth as a uniform sphere, the gravitational field inside a uniform spherical shell is zero everywhere inside it (shell theorem), so at the very centre, contributions from all directions cancel and g = 0 -> matches (C).

(II) Escape speed of earth: v_e = sqrt(2GM/R) ~ 11.2 km/s, the minimum speed needed for an object to escape earth's gravitational field from the surface without further propulsion -> matches (D).

…

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Q.Explain the variation of ggg with altitude.

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Q.Explain the variation of $g$ with altitude.

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