Variation In Acceleration Due To Gravity

PhysicsGravitationFor JEE aspirants

The acceleration due to gravity $g$ is not the same everywhere on Earth. Its surface value $g_{0} = GM / R^{2} \approx 9.8 m/s^{2}$ changes with altitude ( $g^{'} \approx g_{0} (1 - 2 h / R)$ for small heights), depth ( $g^{'} = g_{0} (1 - d / R)$ , becoming zero at the centre), and latitude ( $g^{'} = g_{0} - ω^{2} R cos^{2} θ$ from Earth's rotation). Earth's slightly oblate shape adds a small extra difference between pole and equator. This concept covers all four causes of variation, worked examples, and the composite $g$ -vs- $r$ profile for JEE Mains and NEET.

Key Formulas - Quick Reference

At the surface: $g_{0} = \frac{GM}{R ^{2}} \approx 9.8 m/s^{2}$
At altitude $h$ (exact): $g^{'} = g_{0} (1 + \frac{h}{R})^{- 2}$
At altitude $h ≪ R$ (approximation): $g^{'} \approx g_{0} (1 - \frac{2 h}{R})$
At depth $d$ : $g^{'} = g_{0} (1 - \frac{d}{R})$
Inside Earth (distance $r$ from centre): $g = g_{0} \frac{r}{R}$
Outside Earth: $g = g_{0} (\frac{R}{r})^{2}$
At the centre: $g = 0$
Latitude $θ$ (rotation effect): $g^{'} = g_{0} - ω^{2} R cos^{2} θ$
At equator ( $θ = 0$ ): $g_{e q} = g_{0} - ω^{2} R$
At poles ( $θ = 9 0^{\circ}$ ): $g_{p o l e} = g_{0}$ (no rotation effect)
Earth's rotation: $ω = 2 π /86400 rad/s \approx 7.27 \times 1 0^{- 5} rad/s$

1. Relation between G and g

Consider a body of mass $m$ resting on Earth's surface. By Newton's law of gravitation, the Earth (mass $M$ , radius $R$ ) attracts it with a force:

$F = \frac{GM m}{R ^{2}}$

By Newton's second law this force equals $m g$ , so the two expressions must match:

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

Substituting standard values ( $G = 6.67 \times 1 0^{- 11}$ , $M = 5.98 \times 1 0^{24} kg$ , $R = 6.4 \times 1 0^{6} m$ ) gives:

$g \approx 9.8 m/s^{2}$

This is only the reference value at the surface. In reality $g$ varies with position because of altitude, depth, latitude (rotation), and Earth's non-spherical shape.

2. Variation of g with Altitude

At height $h$ above the surface, the distance from Earth's centre is $R + h$ . Treating Earth as spherically symmetric:

$g^{'} = \frac{GM}{( R + h ) ^{2}} = g \cdot \frac{R ^{2}}{( R + h ) ^{2}} = g (1 + \frac{h}{R})^{- 2}$

Figure 1: Variation of

g

with altitude. At height

h

, the distance from Earth's centre is

R + h

and gravity is

g^{'} = g (\frac{R}{R + h})^{2} = g (1 + \frac{h}{R})^{- 2}

. For

h ≪ R

g^{'} \approx g (1 - \frac{2 h}{R})

For small altitudes ( $h ≪ R$ ), apply the binomial approximation $(1 + x)^{- 2} \approx 1 - 2 x$ :

g^{'} \approx g (1 - \frac{2 h}{R})

The fractional decrease is $\frac{g - g ^{'}}{g} = \frac{2 h}{R}$ .

Numerical example: at $h = 100 km$ :

$\frac{2 h}{R} = \frac{2 \times 100}{6400} = 0.03125 \approx 3%$

So $g$ decreases by about 3% at 100 km altitude.

The approximation

g^{'} \approx g (1 - 2 h / R)

is accurate to within 0.1% up to

h \approx 100 km

but breaks down for large altitudes. For a satellite at

h = R

(radius

= 2 R

), the exact formula gives

g^{'} = g /4

, not

g (1 - 2)

which would be nonsense.

3. Variation of g with Depth

Consider a point at depth $d$ below the surface, so its distance from Earth's centre is $r = R - d$ . By the shell theorem, only the mass inside the sphere of radius $r$ contributes; the outer shell exerts no net force.

Assuming Earth has uniform density $ρ$ , the mass enclosed within radius $r$ is:

$M (r) = \frac{4}{3} π r^{3} ρ$

Gravity at depth $d$ :

$g^{'} = \frac{GM ( r )}{r ^{2}} = \frac{4 π Gρ r}{3} = \frac{4 π Gρ ( R - d )}{3}$

At the surface ( $d = 0$ ), $g = \frac{4 π GρR}{3}$ . Taking the ratio:

g^{'} = g (1 - \frac{d}{R})

Figure 2: Variation of

g

with depth. At depth

d

, only mass inside the inner sphere of radius

R - d

contributes:

g^{'} = g (1 - \frac{d}{R})

. Both

g^{'} = 0

at the centre and

g^{'} = g

at the surface follow immediately.

Numerical example: at depth $d = 10 km$ :

$\frac{d}{R} = \frac{10}{6400} \approx 0.0016 = 0.16%$

So gravity decreases by only about 0.16% at 10 km depth - a much smaller effect than the 3% decrease at 100 km altitude.

Value at the centre

Setting $d = R$ (or equivalently $r = 0$ ):

$g^{'} = g (1 - \frac{R}{R}) = 0$

At Earth's centre, the acceleration due to gravity is zero. This makes physical sense: symmetric pulls from all sides cancel.

General profile: g inside and outside Earth

Combining the results:

Inside ( $r \leq R$ , i.e. $d = R - r \geq 0$ ): $g = g_{0} \frac{r}{R}$ (linear in $r$ )
Outside ( $r \geq R$ ): $g = g_{0} (\frac{R}{r})^{2}$ (inverse-square)

Figure 3:

g

versus

r

from Earth's centre. Inside (

r \leq R

g = g_{0} (r / R)

, linear. Outside (

r \geq R

g = g_{0} (R / r)^{2}

, inverse-square. Maximum at the surface.

The maximum value of

g

is reached exactly at the surface (

r = R

). Both above and below the surface,

g

decreases - but for very different reasons: above, distance grows; below, less mass contributes.

4. Variation of g with Latitude (Rotation Effect)

Earth rotates about its polar axis with angular speed $ω = 2 π /86400 \approx 7.27 \times 1 0^{- 5} rad/s$ . A body of mass $m$ resting at latitude $θ$ on the surface moves in a horizontal circle whose radius (perpendicular distance from the rotation axis) is:

$r = R cos θ$

Figure 4: Rotation reduces apparent gravity. At latitude

θ

, perpendicular distance from axis is

r = R cos θ

. Apparent gravity:

g^{'} = g - ω^{2} R cos^{2} θ

. Maximum reduction at equator (

θ = 0

), zero effect at poles (

θ = 9 0^{\circ}

In the rotating (non-inertial) reference frame of the Earth, the body appears in equilibrium under three horizontal or nearly-horizontal effects:

Real gravitational attraction $m g$ toward Earth's centre
Centrifugal pseudo-force $m ω^{2} r = m ω^{2} R cos θ$ pointing outward from the rotation axis
The reduced apparent weight $m g^{'}$ that we actually measure

Taking the component of the centrifugal force along $m g$ (radially outward) gives:

g^{'} = g - ω^{2} R cos^{2} θ

Special cases

At the poles ( $θ = 9 0^{\circ}$ ): $cos θ = 0$ , so $g^{'} = g$ . Rotation has no effect on gravity at the poles.
At the equator ( $θ = 0$ ): $cos θ = 1$ , so $g^{'} = g - ω^{2} R$ . Rotation reduces gravity by the maximum amount here.

Magnitude of the effect

Compute the fractional reduction at the equator:

$\frac{ω ^{2} R}{g} = \frac{( 7.27 \times 1 0 ^{- 5} ) ^{2} \times 6.4 \times 1 0 ^{6}}{9.8} \approx 0.0034 \approx 0.34%$

So the equatorial reduction due to rotation is small - about 0.034 m/s² - but real and measurable.

Solved Example 1

Determine the angular speed at which the Earth would have to rotate on its axis so that a person on the equator would weigh 20% of their present weight. Take equatorial radius

R = 6400 km

Solution:

Apparent weight at the equator (latitude $λ = 0$ , $cos λ = 1$ ):

$W^{'} = W - m R ω^{2}$

With $W = m g$ and $W^{'} = 0.20 m g = \frac{1}{5} m g$ :

$\frac{1}{5} m g = m g - m R ω^{2} ⟹ m R ω^{2} = \frac{4}{5} m g$

$ω = \frac{4 g}{5 R} = \frac{4 \times 9.8}{5 \times 6.4 \times 1 0 ^{6}}$

$ω \approx 1.106 \times 1 0^{- 3} rad/s$

Compare with actual $ω \approx 7.27 \times 1 0^{- 5} rad/s$ - Earth would need to spin about 15 times faster to reduce equatorial weight to a fifth.

5. Effect of Earth's Oblate Shape

Earth is not a perfect sphere; centrifugal effects from rotation over geological time have flattened it slightly at the poles and made it bulge at the equator. It is an oblate spheroid:

Equatorial radius: $R_{e} \approx 6378 km$
Polar radius: $R_{p} \approx 6357 km$
Difference: about $21 km$ - the equator is farther from the centre.

Figure 5: Earth is an oblate spheroid. Equatorial radius

R_{e} \approx 6378 km

, polar radius

R_{p} \approx 6357 km

. Because

g = GM / R^{2}

g_{p o l e} > g_{e q u a t or}

from shape alone; rotation further reduces

g_{e q u a t or}

Since $g = GM / R^{2}$ (ignoring rotation), the smaller polar radius gives a larger $g$ at the poles:

$g_{p o l e} > g_{e q u a t or}$

This effect is independent of rotation. Rotation further reduces $g$ at the equator, so the two effects add up. The measured values are:

$g_{p o l e} \approx 9.832 m/s^{2}$
$g_{e q u a t or} \approx 9.780 m/s^{2}$
Difference: about $0.05 m/s^{2}$ (0.5%)

Roughly two-thirds of this difference comes from shape, one-third from rotation.

A body's weight

W = m g

therefore increases slightly as it is moved from the equator to the poles. A commercial airliner or a shipload of goods weighs a fraction of a percent more at the poles than at the equator.

Solved Example 2

At what height above Earth's surface will the value of

g

be reduced to

\frac{1}{4}

of its surface value? Take

R = 6400 km

Solution:

Using the exact expression:

$g^{'} = g (\frac{R}{R + h})^{2} = \frac{g}{4} ⟹ (\frac{R}{R + h})^{2} = \frac{1}{4}$

$\frac{R}{R + h} = \frac{1}{2} ⟹ R + h = 2 R ⟹ h = R$

$h = 6400 km$

At an altitude equal to Earth's radius, gravity is only a quarter of the surface value.

Solved Example 3

At what depth below Earth's surface will the value of

g

be reduced to half of its surface value?

Solution:

$g^{'} = g (1 - \frac{d}{R}) = \frac{g}{2} ⟹ 1 - \frac{d}{R} = \frac{1}{2}$

$d = \frac{R}{2} = 3200 km$

Compare: 3200 km depth halves $g$ , while it takes only about 2650 km of altitude to do the same (using the exact altitude formula). Depth reduction is slower because you also lose enclosed mass, not just distance.

Common Mistakes to Avoid

Watch out

Using $g^{'} = g (1 - 2 h / R)$ for large altitudes: This is a small- $h$ approximation. For $h$ comparable to $R$ (say, satellites), you must use the exact formula $g^{'} = g R^{2} / (R + h)^{2}$ . The approximation is only valid up to a few hundred km.
Confusing depth and altitude formulas: Altitude uses $g (1 - 2 h / R)$ (with factor 2), depth uses $g (1 - d / R)$ (no factor 2). The physics is different - altitude changes distance, depth changes enclosed mass.
Thinking $g = 0$ at the centre means there is no gravity: Gravity is not switched off at the centre; instead, symmetric pulls from all directions cancel to give zero net force. A body released at the centre stays put; a body dropped down a hypothetical tunnel through Earth would execute simple harmonic motion about the centre.
Forgetting rotation has no effect at the poles: $g^{'} = g - ω^{2} R cos^{2} θ$ ; at the poles $cos θ = 0$ , so $g$ is unchanged. Rotation-related weight loss is a maximum at the equator, zero at the poles.
Adding shape and rotation effects with wrong signs: Both effects make $g$ smaller at the equator relative to the poles. They add - they do not partly cancel.
Using $g$ from surface at a satellite altitude: A satellite at 400 km altitude experiences $g^{'} \approx 8.7 m/s^{2}$ , not $9.8$ . Use the altitude formula whenever the height is more than a few km.
Thinking $g$ increases inside Earth as you go deeper: With uniform density it decreases linearly. In real Earth (denser core), $g$ actually rises slightly for the first few hundred km before decreasing - but for JEE/NEET use the uniform-density result: linear decrease all the way to zero at the centre.

Frequently Asked Questions

How does g vary with altitude above Earth's surface?

At height $h$ above the surface, $g^{'} = g (\frac{R}{R + h})^{2}$ . For small heights ( $h ≪ R$ ), this simplifies to $g^{'} \approx g (1 - \frac{2 h}{R})$ . So $g$ decreases as altitude increases, and the fractional decrease is $2 h / R$ for small $h$ .

How does g vary with depth below Earth's surface?

At depth $d$ , $g^{'} = g (1 - \frac{d}{R})$ , assuming uniform density. Gravity decreases linearly with depth, becoming exactly zero at the centre. Only the mass inside the current radius contributes; the outer shell exerts no net force by the shell theorem.

Why is g zero at the centre of Earth?

At the centre, mass surrounds the body symmetrically in all directions. By the shell theorem, each spherical shell of mass around the centre contributes zero net force at any interior point. Since all shells share the centre as an interior point, the total gravitational force there is zero.

Why is g greater at the poles than at the equator?

Two reasons add up: (1) Earth is an oblate spheroid - the polar radius (~6357 km) is smaller than the equatorial radius (~6378 km), and $g \propto 1/ R^{2}$ , so gravity is larger where the surface is closer to the centre. (2) Earth's rotation reduces apparent gravity at the equator by $ω^{2} R \approx 0.034 m/s^{2}$ ; the effect is zero at the poles. Together they give $g_{p o l e} \approx 9.83$ vs $g_{e q} \approx 9.78 m/s^{2}$ .

What is the effect of Earth's rotation on the value of g at latitude θ?

At latitude $θ$ , apparent gravity is $g^{'} = g - ω^{2} R cos^{2} θ$ , where $ω = 2 π /86400 rad/s$ is Earth's angular speed. The reduction is maximum at the equator ( $θ = 0$ : full $ω^{2} R \approx 0.034 m/s^{2}$ ) and zero at the poles ( $θ = 9 0^{\circ}$ ).

Does the mass of a body change with altitude or latitude?

No. Mass is an intrinsic property that depends only on the amount of matter, and it does not change with location. Only the weight $W = m g$ changes because $g$ changes - a body of 1 kg weighs about $9.83 N$ at the pole but only about $9.78 N$ at the equator, yet its mass is 1 kg everywhere.

At what height is g reduced to half its surface value?

Using $g^{'} = g R^{2} / (R + h)^{2} = g /2$ , we get $(R + h)^{2} = 2 R^{2}$ , so $R + h = R 2$ , giving $h = R (2 - 1) \approx 0.414 R \approx 2650 km$ above the surface.

Does g increase or decrease as we go from the equator toward the pole?

$g$ increases as we move from the equator toward either pole. The rotation-related reduction becomes smaller (proportional to $cos^{2} θ$ ), and Earth's smaller polar radius pushes $g$ up further. Total variation across latitude: about $0.05 m/s^{2}$ (~0.5%).

How does g behave inside Earth for a uniform-density model?

For uniform density, $g (r) = g_{0} (r / R)$ inside Earth. It rises linearly from zero at the centre to $g_{0}$ at the surface. Outside the surface it falls as $g_{0} (R / r)^{2}$ . The maximum value of $g$ therefore occurs exactly at the surface.

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