Derive an expression for the effective acceleration due to gravity $g'$ at a place of latitude $\lambda$, taking into account the Earth's rotation about its own axis. Show that $g'$ is least at the equator and greatest at the poles.

An object of mass $m$ at latitude $\lambda$ on the Earth's surface actually moves in a horizontal circle of radius $R\cos\lambda$ about the Earth's rotation axis (this radius is largest, $R$, at the equator, and shrinks to zero at the poles). Moving in this circle at the Earth's angular speed $\omega = 2\pi/T$ requires a centripetal force of magnitude $m\omega^2(R\cos\lambda)$, directed horizontally toward the rotation axis.\n\nThis centripetal force is supplied out of the true gravitational pull $mg$, so the object's apparent weight $mg'$ (what a spring balance under it would actually read) is the true weight MINUS the component of this required centripetal force that acts along the local vertical (the radially outward direction at that latitude); resolving the horizontal requirement along the local vertical brings in one further factor of $\cos\lambda$: $$mg' = mg - m\omega^2 R\cos^2\lambda \quad \Longrightarrow \quad g' = g - \omega^2 R\cos^2\lambda$$ …

Derive an expression for the effective acceleration due to gravity g' at a place of latitude λ, taking into ac

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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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Q.Derive an expression for the effective acceleration due to gravity g′g'g′ at a place of latitude λ\lambdaλ, taking into account the Earth's rotation about its own axis. Show that g′g'g′ is least at the equator and greatest at the poles.